Exercise 9.5

In each of the Exercises 1 to 10, show that the given differential equation is

homogeneous and solve each of them.

1. (x2 + xy).dy = (x2 + y2).dx

Solun:- Given eq. is:- (x2 + xy).dy = (x2 + y2).dx

${"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{x^{2}+y^{2}}{x^{2}+xy}}\t\n\\end{align*}","id":"8-0-0-0","ts":1603963190250,"cs":"fi52R45Ukks6dSW5Xj/qrQ==","size":{"width":104,"height":38}}$ .....(1)

Let F(x,y)= ${"type":"$","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\frac{x^{2}+y^{2}}{x^{2}+xy}$","id":"9-0-0-0","ts":1603963258522,"cs":"abIL/P9T4lEVftzgW7yLVg==","size":{"width":38,"height":24}}$

Replace F(x,y) with F(λx,λy):-

${"code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda^{2}x^{2}+\\lambda^{2}y^{2}}{\\lambda^{2}x^{2}+\\lambda^{2}xy}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{x^{2}+y^{2}}{x^{2}+xy}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","id":"10-0-0-0","font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","ts":1603963352535,"cs":"R23GS9l9Zsv4/sX3sKlfGA==","size":{"width":184,"height":104}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

${"id":"7-0-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={v\\diff{x}{x}+x\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={v+x\\diff{v}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603531746342,"cs":"J8dUoQGZ3uRvtVu21WKu1A==","size":{"width":132,"height":69}}$ .......(2)

From eq. 1 and 2:-

${"font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{x^{2}+v^{2}x^{2}}{x^{2}+xvx}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x^{2}+v^{2}x^{2}}{x^{2}+vx^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x^{2}}{x^{2}}\\left(\\frac{1+v^{2}}{1+v}\\right)}\\\\\n{v+x\\diff{v}{x}}&={\\left(\\frac{1+v^{2}}{1+v}\\right)}\t\n\\end{align*}","id":"11-0-0-0","type":"align*","ts":1604492038682,"cs":"AU4BWEl5x2ivf+JvjfKIPw==","size":{"width":176,"height":164}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{x\\diff{v}{x}}&={\\frac{1+v^{2}}{1+v}-v}\\\\\n{x\\diff{v}{x}}&={\\frac{1+v^{2}-v\\left(1+v\\right)}{1+v}}\\\\\n{x\\diff{v}{x}}&={\\frac{1+v^{2}-v-v^{2}}{1+v}}\\\\\n{x\\diff{v}{x}}&={\\frac{1-v}{1+v}}\t\n\\end{align*}","id":"12-0-0-0","ts":1603964129538,"cs":"TAmYpPsR+7L4ePZ7pfOxkQ==","size":{"width":177,"height":156}}$

Separating the variables:-

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"4-0-1-0-0-0-0-0","code":"\\begin{align*}\n{\\left(\\frac{1+v}{1-v}\\right).dv}&={\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","ts":1603964321364,"cs":"on3oWSrsFTH4sWjosTgkmQ==","size":{"width":149,"height":37}}$

Integrating both sides:-

${"code":"\\begin{align*}\n{\\int_{}^{}\\left(\\frac{1+v}{1-v}\\right).dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"4-0-1-0-1-0","type":"align*","ts":1603964364652,"cs":"93WzAKcNqVlpHEPj/CjZsA==","size":{"width":185,"height":37}}$

${"id":"4-0-1-1-1-0-0-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}\\left[\\frac{-\\left(-1-v\\right)}{1-v}\\right].dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1603964565467,"cs":"88/2U6hZdFMmNAxF+liCXw==","size":{"width":216,"height":37}}$

${"type":"align*","id":"4-0-1-1-1-1-0-0","code":"\\begin{align*}\n{-\\int_{}^{}\\left[\\frac{\\left(-1-v\\right)}{1-v}\\right].dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1603964599017,"cs":"ZbmSOkXCGXGTbRneav9wRw==","size":{"width":218,"height":37}}$

${"type":"align*","id":"4-0-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{-\\int_{}^{}\\left[\\frac{\\left(-1-v+1-1\\right)}{1-v}\\right].dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1603964631231,"cs":"w8D6YeBQrohS7R68b4w1mQ==","size":{"width":273,"height":37}}$

${"code":"\\begin{align*}\n{-\\int_{}^{}\\left[\\frac{\\left(-1+\\left(1-v\\right)-1\\right)}{1-v}\\right].dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"4-0-1-1-1-1-1-1-0","ts":1603964679474,"cs":"8ZT0fVxu5VC/8HjucKPXyg==","size":{"width":285,"height":37}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{-\\int_{}^{}\\left[\\frac{\\left(\\left(1-v\\right)-2\\right)}{1-v}\\right].dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","id":"4-0-1-1-1-1-1-1-1-0","ts":1603964780138,"cs":"leuzzaMMt09+Yrujh6Lc5A==","size":{"width":245,"height":37}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{-\\left[\\int_{}^{}\\frac{\\left(1-v\\right)}{1-v}.dv-\\int_{}^{}\\frac{2}{1-v}.dv\\right]}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","id":"4-0-1-1-1-1-1-1-1-1-0","ts":1603964828923,"cs":"AxU2rDXE2Blv60sswldAig==","size":{"width":309,"height":37}}$

${"id":"4-0-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{-\\int_{}^{}\\frac{\\left(1-v\\right)}{1-v}.dv+\\int_{}^{}\\frac{2}{1-v}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1603964862095,"cs":"fUbcxlnqATWLD4xGmbt7ag==","size":{"width":296,"height":36}}$

${"code":"\\begin{align*}\n{-\\int_{}^{}1.dv+\\int_{}^{}\\frac{2}{1-v}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","id":"4-0-1-1-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1603964926449,"cs":"jQ7WgcBWiLc444tOx9WPyg==","size":{"width":246,"height":36}}$

${"id":"4-0-1-1-1-1-1-1-1-1-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{-\\int_{}^{}1.dv-2\\int_{}^{}\\frac{-1}{1-v}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","ts":1603965000678,"cs":"owagz/wC2dg1bfsUBXuKRQ==","size":{"width":257,"height":36}}$

We know that:-

${"id":"17-0-0-0","code":"$\\int_{}^{}\\frac{f^{\\prime }\\left(x\\right)}{f\\left(x\\right)}.dx=\\log_{}\\left|f\\left(x\\right)\\right|+C$","font":{"size":10,"color":"#222222","family":"Arial"},"type":"$","ts":1603537200909,"cs":"/2tTL13+s6KjbMbGXUqaRw==","size":{"width":184,"height":24}}$

${"type":"align*","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"17-2-0-0","ts":1603537288100,"cs":"meBm3VnK+/FjP8EKFwJoUw==","size":{"width":153,"height":36}}$

${"type":"align*","id":"16-1-0-0-0-0","code":"\\begin{align*}\n{-v-2\\log_{}\\left|1-v\\right|}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603967955979,"cs":"dXbNo2krU3WCw/Ht/p5J/w==","size":{"width":205,"height":16}}$

Put the value of v:-

${"font":{"size":10,"family":"Arial","color":"#222222"},"type":"align*","code":"\\begin{align*}\n{\\frac{-y}{x}-2\\log_{}\\left|1-\\frac{y}{x}\\right|\\,\\,}&={\\log_{}\\left|x\\right|+c}\\\\\n{\\frac{-y}{x}-2\\log_{}\\left|\\frac{x-y}{x}\\right|\\,\\,}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","id":"16-1-1-0-0-0","ts":1603967980094,"cs":"EmW+kKTpL2pgktn9RRPW9w==","size":{"width":228,"height":68}}$

${"code":"\\begin{align*}\n{\\frac{-y}{x}-\\log_{}\\left|\\left(\\frac{x-y}{x}\\right)^{2}\\right|\\,\\,}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","type":"align*","id":"16-1-1-1-0-0-0-0","font":{"family":"Arial","color":"#222222","size":10},"ts":1603967992907,"cs":"lGcWolk2Z83o3REewRj9dw==","size":{"width":248,"height":44}}$

${"id":"16-1-1-1-1-0-0-0","code":"\\begin{align*}\n{\\frac{-y}{x}}&={\\log_{}\\left|x\\right|+\\log_{}\\left|\\left(\\frac{x-y}{x}\\right)^{2}\\right|+c}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#222222","size":10},"ts":1603968004727,"cs":"zHnxSUQxvAOIJUgVN8p4DQ==","size":{"width":241,"height":44}}$

${"code":"\\begin{align*}\n{\\frac{-y}{x}}&={\\log_{}\\left|x\\right|+\\log_{}\\left|\\left(x-y\\right)^{2}\\right|-\\log_{}\\left|x^{2}\\right|+c}\\\\\n{\\frac{-y}{x}}&={\\log_{}\\left|\\frac{x}{x^{2}}\\right|+\\log_{}\\left|\\left(x-y\\right)^{2}\\right|+c}\\\\\n{\\frac{-y}{x}}&={\\log_{}\\left|\\frac{1}{x}\\right|+\\log_{}\\left|\\left(x-y\\right)^{2}\\right|+c}\\\\\n{\\frac{-y}{x}}&={\\log_{}\\left|1\\right|-\\log_{}\\left|x\\right|+\\log_{}\\left|\\left(x-y\\right)^{2}\\right|+c}\\\\\n{\\frac{-y}{x}}&={\\log_{}\\left|\\frac{\\left(x-y\\right)^{2}}{x}\\right|+c}\t\n\\end{align*}","id":"16-1-1-1-1-1-0-0","type":"align*","font":{"family":"Arial","color":"#222222","size":10},"ts":1603968035748,"cs":"s3MT6Y8v/ge/lR09iIkSAg==","size":{"width":292,"height":192}}$

${"type":"align*","font":{"color":"#222222","family":"Arial","size":10},"id":"20-0-0-0","code":"\\begin{align*}\n{\\frac{-y}{x}-c}&={\\log_{}\\left|\\frac{\\left(x-y\\right)^{2}}{x}\\right|}\t\n\\end{align*}","ts":1603968050780,"cs":"rURjV140UFpfaMhtYykpBA==","size":{"width":169,"height":44}}$

${"id":"20-1-0","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{e^{\\frac{-y}{x}-c}}&={\\frac{\\left(x-y\\right)^{2}}{x}}\t\n\\end{align*}","type":"align*","ts":1603968065652,"cs":"Chw0/g4Po8xs4zwKpET1+Q==","size":{"width":120,"height":36}}$

${"id":"22-0-0","font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{e^{-\\left(\\frac{y}{x}+c\\right)}}&={\\frac{\\left(x-y\\right)^{2}}{x}}\t\n\\end{align*}","ts":1603968092283,"cs":"JU81R3TVJN6jjhPwNQp+dA==","size":{"width":132,"height":36}}$

${"type":"align*","id":"22-1-0-0","code":"\\begin{align*}\n{\\frac{1}{e^{\\left(\\frac{y}{x}+c\\right)}}}&={\\frac{\\left(x-y\\right)^{2}}{x}}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1603968108317,"cs":"f7vZkeQBI4OPV0oysOcWLQ==","size":{"width":130,"height":40}}$

${"type":"align*","id":"22-1-1-0-0","code":"\\begin{align*}\n{\\frac{1}{e^{\\frac{y}{x}}.e^{c}}}&={\\frac{\\left(x-y\\right)^{2}}{x}}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603967834849,"cs":"xKqUgQEbyy/OrLef68lj4A==","size":{"width":128,"height":40}}$

${"type":"align*","code":"\\begin{align*}\n{xe^{\\frac{-y}{x}}.e^{-c}}&={\\left(x-y\\right)^{2}}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"id":"22-1-1-1-0-0","ts":1603967882623,"cs":"rDy4LgswlpAO4VVcA8ZgVA==","size":{"width":136,"height":20}}$

Here e-C = c both are constants.

${"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","id":"22-1-1-1-1-0","code":"\\begin{align*}\n{Cxe^{\\frac{-y}{x}}}&={\\left(x-y\\right)^{2}}\t\n\\end{align*}","ts":1603967935675,"cs":"34XS66HBqwiJa3jDywlyFg==","size":{"width":120,"height":20}}$

${"font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\left(x-y\\right)^{2}}&={Cxe^{\\frac{-y}{x}}}\t\n\\end{align*}","type":"align*","id":"22-1-1-1-1-1","ts":1603968187845,"cs":"Up8ZBGBotEjtT3cZjGjlAQ==","size":{"width":120,"height":20}}$

${"type":"align","id":"2-0","code":"\\begin{align}\n{2.\\,y^{\\prime }}&={\\frac{x+y}{x}}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603968267128,"cs":"agCEmHaGfeZlpuFSPjh3pQ==","size":{"width":90,"height":30}}$

Solun:- Given eq. is:-

${"id":"2-1-0","font":{"family":"Arial","color":"#222222","size":10},"type":"align","code":"\\begin{align}\n{y^{\\prime }}&={\\frac{x+y}{x}}\t\n\\end{align*}","ts":1603968296027,"cs":"5a8C0Msmd5qrPEVkkmzHqw==","size":{"width":76,"height":30}}$

${"id":"2-1-1","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{x+y}{x}}\t\n\\end{align*}","ts":1603968413276,"cs":"S7QcN/UL8sMkTNVxPv6a+g==","size":{"width":88,"height":32}}$ .....(1)

Let F(x,y)= ${"id":"9-1","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\frac{x+y}{x}$","type":"$","ts":1603968521046,"cs":"oIFoQ5geBo3wj5EFFznsbg==","size":{"width":28,"height":20}}$

Replace F(x,y) with F(λx,λy):-

${"font":{"size":10,"family":"Arial","color":"#222222"},"id":"10-1","code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda x+\\lambda y}{\\lambda x}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{x+y}{x}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","type":"align*","ts":1603968672349,"cs":"Q7wL2ikwQeMyBoVu5Xlclw==","size":{"width":169,"height":97}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

${"id":"7-1","code":"\\begin{align*}\n{\\diff{y}{x}}&={v\\diff{x}{x}+x\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={v+x\\diff{v}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603531746342,"cs":"X7jUg8dVnbp/K8egrcggsA==","size":{"width":132,"height":69}}$ .......(2)

From eq. 1 and 2:-

${"id":"11-1","code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{x+vx}{x}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x\\left(1+v\\right)}{x}}\\\\\n{v+x\\diff{v}{x}}&={1+v}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","ts":1603968764046,"cs":"AFb9+TO7i9OtMl33SC1sMQ==","size":{"width":144,"height":108}}$

${"code":"\\begin{align*}\n{x\\diff{v}{x}}&={1+v-v}\\\\\n{x\\diff{v}{x}}&={1}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","id":"12-1","ts":1603968801624,"cs":"eA1IEqUW6rf8Pqm1hCz0Ng==","size":{"width":116,"height":69}}$

Separating the variables:-

${"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{dv}&={\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","id":"4-0-1-0-0-1-0","ts":1603968821819,"cs":"c040q25g+BLjXCjaIYpbAQ==","size":{"width":76,"height":32}}$

Integrating both sides:-

${"code":"\\begin{align*}\n{\\int_{}^{}dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"id":"4-0-1-0-0-1-1","type":"align*","ts":1603968848503,"cs":"ofFhn0e+8Jx4VF/H8b3Uog==","size":{"width":116,"height":36}}$

We know that:-

${"type":"align*","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"17-2-1","ts":1603537288100,"cs":"SPfZZt/RBHGE3gtSxHwd1Q==","size":{"width":153,"height":36}}$

${"id":"16-1-0-1","code":"\\begin{align*}\n{v}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","ts":1603968891334,"cs":"UxSYd/esgPdLnYbgdlY23Q==","size":{"width":96,"height":16}}$

Put the value of v:-

${"type":"align*","id":"16-1-1-0-1","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{\\frac{y}{x}}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","ts":1603968922662,"cs":"IxKrU7DtgqdUo2VzPvhqeQ==","size":{"width":104,"height":28}}$

y = xlog|x| + cx

3. (x - y).dy - (x + y).dx = 0

Solun:- Given eq. is:- (x - y).dy - (x + y).dx = 0

(x - y).dy = (x + y).dx

${"id":"8-0-1-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{x+y}{x-y}}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1603969246312,"cs":"AS6ZUrwYq0/Eb1aGKj6UoQ==","size":{"width":88,"height":36}}$ .....(1)

Let F(x,y)= ${"type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"id":"9-0-1-0","code":"$\\frac{x+y}{x-y}$","ts":1603969288062,"cs":"7U1E9YSuo0sP40USwog02w==","size":{"width":28,"height":21}}$

Replace F(x,y) with F(λx,λy):-

${"id":"10-0-1-0","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda x+\\lambda y}{\\lambda x-\\lambda y}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{x+y}{x-y}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","ts":1603969452148,"cs":"1WBByYvxZuyHEmmyN78zCQ==","size":{"width":169,"height":100}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

${"id":"7-2","code":"\\begin{align*}\n{\\diff{y}{x}}&={v\\diff{x}{x}+x\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={v+x\\diff{v}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603531746342,"cs":"18FbMdJeZxluiuekJBzynA==","size":{"width":132,"height":69}}$ .......(2)

From eq. 1 and 2:-

${"font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","id":"11-0-1-0","code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{x+vx}{x-vx}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x\\left(1+v\\right)}{x\\left(1-v\\right)}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{1+v}{1-v}}\t\n\\end{align*}","ts":1603969578815,"cs":"DOiLl+pTLJu7C370l85fiQ==","size":{"width":144,"height":116}}$

${"code":"\\begin{align*}\n{x\\diff{v}{x}}&={\\frac{1+v}{1-v}-v}\\\\\n{x\\diff{v}{x}}&={\\frac{1+v-v\\left(1-v\\right)}{1-v}}\\\\\n{x\\diff{v}{x}}&={\\frac{1+v-v+v^{2}}{1-v}}\\\\\n{x\\diff{v}{x}}&={\\frac{1+v^{2}}{1-v}}\t\n\\end{align*}","id":"12-0-1-0","type":"align*","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1603969787852,"cs":"vkMM2r4kVFH7RjPe68Soag==","size":{"width":170,"height":156}}$

Separating the variables:-

${"code":"\\begin{align*}\n{\\left(\\frac{1-v}{1+v^{2}}\\right).dv}&={\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","id":"4-0-1-0-0-0-1-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1603969885927,"cs":"S9VQV4w0vm1Hx1+YyOj6Rw==","size":{"width":156,"height":37}}$

Integrating both sides:-

${"font":{"size":10,"family":"Arial","color":"#000000"},"id":"4-0-1-0-0-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\left(\\frac{1-v}{1+v^{2}}\\right).dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","ts":1603969960831,"cs":"1fslWr76yWE7Yg/pQMB8uQ==","size":{"width":192,"height":37}}$

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{1+v^{2}}.dv-\\int_{}^{}\\frac{v}{1+v^{2}}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","id":"4-0-1-0-0-0-1-1-1-0-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1603970077583,"cs":"aWSRwnZPdzAC6zRaPOlygg==","size":{"width":281,"height":36}}$

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{1+v^{2}}.dv-\\frac{1}{2}\\int_{}^{}\\frac{2v}{1+v^{2}}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"4-0-1-0-0-0-1-1-1-1-0","ts":1603970112315,"cs":"iO9P0RrS2U9loq9NOKiN1Q==","size":{"width":298,"height":36}}$

We know that:-

${"id":"17-0-1","code":"$\\int_{}^{}\\frac{f^{\\prime }\\left(x\\right)}{f\\left(x\\right)}.dx=\\log_{}\\left|f\\left(x\\right)\\right|+C$","font":{"size":10,"color":"#222222","family":"Arial"},"type":"$","ts":1603537200909,"cs":"m10sRDBsPXHK4vLTk6x9zQ==","size":{"width":184,"height":24}}$

${"id":"17-2-2-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{1+x^{2}}.dx}&={\\tan^{-1}\\left(x\\right)+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","ts":1603970293931,"cs":"re8QE/WLiN6z9BGnd2xK3Q==","size":{"width":208,"height":36}}$

${"code":"\\begin{align*}\n{\\tan^{-1}\\left(v\\right)-\\frac{1}{2}\\log_{}\\left|1+v^{2}\\right|}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"id":"16-1-0-0-1-0-0","type":"align*","ts":1603970411724,"cs":"RYucEDsb3stVwcrU/IlrPg==","size":{"width":258,"height":32}}$

Put the value of v:-

${"font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)-\\frac{1}{2}\\log_{}\\left|1+\\frac{y^{2}}{x^{2}}\\right|}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","id":"16-1-0-0-1-1-0","type":"align*","ts":1603970516893,"cs":"fcL3EJQmVbKIOLQe/sFi4Q==","size":{"width":282,"height":40}}$

${"id":"3-0-0","font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)-\\log_{}\\left|\\left(1+\\frac{y^{2}}{x^{2}}\\right)^{\\frac{1}{2}}\\right|}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","type":"align*","ts":1603970610608,"cs":"fU2AMrEB1Lca2UGhq0/Ckw==","size":{"width":300,"height":50}}$

${"code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)-\\log_{}\\left|{\\sqrt[]{1+\\frac{y^{2}}{x^{2}}}}\\right|}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","type":"align*","id":"3-1-0","font":{"color":"#222222","family":"Arial","size":10},"ts":1603970660230,"cs":"52xTM9hdG4nyQbt9ij02IA==","size":{"width":284,"height":46}}$

${"id":"3-1-1-0","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)-\\log_{}\\left|{\\sqrt[]{\\frac{x^{2}+y^{2}}{x^{2}}}}\\right|}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","ts":1603970743634,"cs":"/xYdIlm6J6j62f+EiGTSXw==","size":{"width":289,"height":46}}$

${"type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)-\\log_{}\\left|\\frac{{\\sqrt[]{x^{2}+y^{2}}}}{x}\\right|}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","id":"3-1-1-1-0","ts":1603970799895,"cs":"XpCgpilqW86DUIguuzALYg==","size":{"width":289,"height":46}}$

${"font":{"color":"#222222","family":"Arial","size":10},"id":"3-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)-\\left[\\log_{}\\left|{\\sqrt[]{x^{2}+y^{2}}}\\right|-\\log_{}\\left|x\\right|\\right]}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","ts":1603970836436,"cs":"XSG7m+Q5GNInPVv/WAYd5w==","size":{"width":358,"height":28}}$

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"3-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)-\\log_{}\\left|{\\sqrt[]{x^{2}+y^{2}}}\\right|+\\log_{}\\left|x\\right|}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","ts":1603970870519,"cs":"5W6rXqQgd5TAWwI///mfOg==","size":{"width":344,"height":28}}$

${"font":{"color":"#222222","family":"Arial","size":10},"id":"3-1-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)}&={\\log_{}\\left|{\\sqrt[]{x^{2}+y^{2}}}\\right|+c}\t\n\\end{align*}","ts":1603970912504,"cs":"fRR8gzOn5rtk1T5TpJegEA==","size":{"width":221,"height":28}}$

${"id":"3-1-1-1-1-1-1-1","type":"align*","code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)}&={\\frac{1}{2}\\log_{}\\left|x^{2}+y^{2}\\right|+c}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603970969209,"cs":"4RKZCauWVAhqxwv1FrIEIQ==","size":{"width":220,"height":32}}$

4. (x2 - y2).dx + 2xy.dy = 0

Solun:- Given eq. is:- (x2 - y2).dx + 2xy.dy = 0

(x2 - y2).dx = -2xy.dy

${"type":"align*","id":"8-0-1-1-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={-\\left(\\frac{x^{2}-y^{2}}{2xy}\\right)}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1604048286760,"cs":"XdfeFpOdUtc302q53KNKeg==","size":{"width":137,"height":38}}$

${"type":"align*","id":"8-0-1-1-1","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{y^{2}-x^{2}}{2xy}}\t\n\\end{align*}","ts":1604048320896,"cs":"bEwRFs7IkYbNu9vstlohhA==","size":{"width":101,"height":37}}$ .....(1)

Let F(x,y)= ${"type":"$","id":"9-0-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\frac{y^{2}-x^{2}}{2xy}$","ts":1604048356610,"cs":"YyZhc8kZMImo7GZN10MCRg==","size":{"width":36,"height":24}}$

Replace F(x,y) with F(λx,λy):-

${"font":{"family":"Arial","size":10,"color":"#222222"},"id":"10-0-1-1-0","type":"align*","code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda ^{2}y^{2}-\\lambda ^{2}x^{2}}{2\\lambda ^{2}xy}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{y^{2}-x^{2}}{2xy}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","ts":1604048465691,"cs":"ICW8Fwxpnrpq9lUTz0bhyw==","size":{"width":181,"height":104}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

${"id":"7-3","code":"\\begin{align*}\n{\\diff{y}{x}}&={v\\diff{x}{x}+x\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={v+x\\diff{v}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603531746342,"cs":"FEcjXuIpfeBIP1v7NQHEMA==","size":{"width":132,"height":69}}$ .......(2)

From eq. 1 and 2:-

${"id":"11-0-1-1-0","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{v^{2}x^{2}-x^{2}}{2xvx}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{v^{2}x^{2}-x^{2}}{2vx^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x^{2}\\left(v^{2}-1\\right)}{2vx^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{v^{2}-1}{2v}}\t\n\\end{align*}","ts":1604056065613,"cs":"xgfy6Z6cesT9//upqGugzw==","size":{"width":160,"height":156}}$

${"font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{x\\diff{v}{x}}&={\\frac{v^{2}-1}{2v}-v}\\\\\n{x\\diff{v}{x}}&={\\frac{v^{2}-1-v\\left(2v\\right)}{2v}}\\\\\n{x\\diff{v}{x}}&={\\frac{v^{2}-1-2v^{2}}{2v}}\\\\\n{x\\diff{v}{x}}&={\\frac{-1-v^{2}}{2v}}\\\\\n{x\\diff{v}{x}}&={\\frac{-\\left(1+v^{2}\\right)}{2v}}\t\n\\end{align*}","id":"12-0-1-1-0","ts":1604056240178,"cs":"VwBkgfyITKmMAICaWeXBuQ==","size":{"width":157,"height":196}}$

Separating the variables:-

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\left(\\frac{2v}{1+v^{2}}\\right).dv}&={\\frac{-1}{x}.dx}\t\n\\end{align*}","type":"align*","id":"4-0-1-0-0-0-1-0-1-0-0-0","ts":1604056214512,"cs":"YjBXWuagj1pGX0eQLyHOkg==","size":{"width":166,"height":37}}$

Integrating both sides:-

${"id":"4-0-1-0-0-0-1-0-1-0-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\left(\\frac{2v}{1+v^{2}}\\right).dv}&={\\int_{}^{}\\frac{-1}{x}.dx}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1604056294297,"cs":"hThDaNLRtSFBGWDcQJbp9A==","size":{"width":204,"height":37}}$

We know that:-

${"type":"$","id":"17-0-2-0","code":"$\\int_{}^{}\\frac{f^{\\prime }\\left(x\\right)}{f\\left(x\\right)}.dx=\\log_{}\\left|f\\left(x\\right)\\right|+C$","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1604056367268,"cs":"59ARFcaRlshrMIwHcgE+jQ==","size":{"width":184,"height":24}}$

${"code":"\\begin{align*}\n{\\log_{}\\left|1+v^{2}\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-1-0-1-0-0-0","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","ts":1604056938190,"cs":"yn92jcoCjFX74NmO/vhySw==","size":{"width":216,"height":20}}$

Put the value of v:-

${"code":"\\begin{align*}\n{\\log_{}\\left|1+\\frac{y^{2}}{x^{2}}\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","id":"16-1-0-0-1-0-1-0-1-0","ts":1604056959897,"cs":"pBRUcObVCzk1aoa2W7rQ7Q==","size":{"width":224,"height":40}}$

${"font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\log_{}\\left|\\frac{x^{2}+y^{2}}{x^{2}}\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-1-0-1-0-1-1-0-0","type":"align*","ts":1604056983297,"cs":"k/n/6Vy5Nw+RLpDvgxyMpA==","size":{"width":232,"height":40}}$

${"code":"\\begin{align*}\n{\\log_{}\\left|x^{2}+y^{2}\\right|-\\log_{}\\left|x^{2}\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"id":"16-1-0-0-1-0-1-0-1-1-1-0","type":"align*","ts":1604056999877,"cs":"Cx5InCb6q9vX+dKIQL4hxw==","size":{"width":293,"height":20}}$

${"code":"\\begin{align*}\n{\\log_{}\\left|x^{2}+y^{2}\\right|-2\\log_{}\\left|x\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","type":"align*","id":"16-1-0-0-1-0-1-0-1-1-1-1","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1604057014455,"cs":"jtaqbFxf7sDO5t5t4x5f1g==","size":{"width":296,"height":20}}$

⇒ log |x2 + y2| = 2log |x| - log |x| + log |C|

⇒ log |x2 + y2| = log |x| + log |C|

⇒ log |x2 + y2| = log |C.x|

⇒ x2 + y2 = C.x

${"font":{"family":"Arial","size":10,"color":"#222222"},"id":"5","type":"align","code":"\\begin{align}\n{5.\\,x^{2}\\diff{y}{x}}&={x^{2}-2y^{2}+xy}\t\n\\end{align*}","ts":1604057317461,"cs":"uzydyF+J4cyiuwcQeIm9wg==","size":{"width":168,"height":32}}$

Solun:- Given eq. is:-

${"font":{"family":"Arial","size":10,"color":"#222222"},"id":"6-0","type":"align*","code":"\\begin{align*}\n{x^{2}\\diff{y}{x}}&={x^{2}-2y^{2}+xy}\t\n\\end{align*}","ts":1604057420827,"cs":"u4TBbLgYWS7ujb/h0tU59g==","size":{"width":153,"height":32}}$

${"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{x^{2}-2y^{2}+xy}{x^{2}}}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"id":"6-1","ts":1604057471928,"cs":"kEn2Pmn0D6XbExxn47AbHA==","size":{"width":145,"height":34}}$ .....(1)

Let F(x,y)= ${"id":"9-0-1-1-1","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\frac{x^{2}-2y^{2}+xy}{x^{2}}$","ts":1604057627843,"cs":"3qjbxN1GveY2l2yL9wl7Rw==","size":{"width":62,"height":22}}$

Replace F(x,y) with F(λx,λy):-

${"type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda ^{2}x^{2}-2\\lambda ^{2}y^{2}+\\lambda x\\lambda y}{\\lambda ^{2}x^{2}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda ^{2}x^{2}-2\\lambda ^{2}y^{2}+\\lambda ^{2}xy}{\\lambda ^{2}x^{2}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{x^{2}-2y^{2}+xy}{x^{2}}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","id":"10-0-1-1-1","ts":1604057768366,"cs":"5CgoVpKfbY46K3I1lTg31g==","size":{"width":236,"height":141}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

${"id":"7-4","code":"\\begin{align*}\n{\\diff{y}{x}}&={v\\diff{x}{x}+x\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={v+x\\diff{v}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603531746342,"cs":"LUZf4LZHT+HaKU1DPbPNCg==","size":{"width":132,"height":69}}$ .......(2)

From eq. 1 and 2:-

${"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{x^{2}-2v^{2}x^{2}+xvx}{x^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x^{2}-2v^{2}x^{2}+x^{2}v}{x^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x^{2}\\left(1-2v^{2}+v\\right)}{x^{2}}}\t\n\\end{align*}","id":"7-5","type":"align*","ts":1604059491926,"cs":"Mvcv8LEa//sku7ohwQavMw==","size":{"width":205,"height":117}}$

${"type":"align*","code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={1+v-2v^{2}}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"id":"11-0-1-1-1","ts":1604059546875,"cs":"2seHpbHNMvhldgfeI4J/5w==","size":{"width":157,"height":32}}$

${"font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{x\\diff{v}{x}}&={1+v-2v^{2}-v}\\\\\n{x\\diff{v}{x}}&={1-2v^{2}}\t\n\\end{align*}","id":"12-0-1-1-1","type":"align*","ts":1604059573618,"cs":"ty9hXv1xQYyWQshQ5qE6cQ==","size":{"width":157,"height":69}}$

Separating the variables:-

${"type":"align*","id":"4-0-1-0-0-0-1-0-1-0-0-1-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\frac{dv}{1-2v^{2}}}&={\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604059605193,"cs":"Bbn8CMzcBXpsgSXe6y3yLw==","size":{"width":117,"height":33}}$

Integrating both sides:-

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{dv}{1-2v^{2}}}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"4-0-1-0-0-0-1-0-1-0-0-1-0-1","ts":1604059625749,"cs":"3AGbojJYish8rororevfRA==","size":{"width":156,"height":36}}$

${"id":"4-0-1-0-0-0-1-0-1-0-0-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{dv}{2\\left(\\frac{1}{2}-v^{2}\\right)}}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1604062750661,"cs":"gwXE+rzHdGrf3nhJYf/ydw==","size":{"width":176,"height":40}}$

${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\frac{1}{2}\\int_{}^{}\\frac{dv}{\\left(\\left(\\frac{1}{{\\sqrt[]{2}}}\\right)^{2}-v^{2}\\right)}}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","id":"4-0-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-1-1","ts":1604062916671,"cs":"asjHywMrNsTdAndFW6O4VA==","size":{"width":228,"height":58}}$

We know that:-

${"id":"17-0-2-1-0","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{a^{2}-x^{2}}.dx}&={\\frac{1}{2a}\\log_{}\\left|\\frac{a+x}{a-x}\\right|+C}\t\n\\end{align*}","ts":1604062982110,"cs":"4P1BFbokLm8ry2jHdXTwdQ==","size":{"width":253,"height":36}}$

${"type":"align*","font":{"color":"#222222","size":10,"family":"Arial"},"id":"17-0-2-1-1","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","ts":1604063033069,"cs":"u252KrgKHZBBEWdY6ZffWQ==","size":{"width":153,"height":36}}$

${"type":"align*","font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\frac{1}{4\\frac{1}{{\\sqrt[]{2}}}}\\log_{}\\left|\\frac{\\frac{1}{{\\sqrt[]{2}}}+v}{\\frac{1}{{\\sqrt[]{2}}}-v}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-1-0-1-0-0-1-0-0-0","ts":1604063147544,"cs":"4OGJhHntsNJu3nr+v8+Wtw==","size":{"width":253,"height":52}}$

${"font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\frac{1}{2{\\sqrt[]{2}}}\\log_{}\\left|\\frac{\\frac{1}{{\\sqrt[]{2}}}+v}{\\frac{1}{{\\sqrt[]{2}}}-v}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","type":"align*","id":"16-1-0-0-1-0-1-0-0-1-0-0-0","ts":1604063194093,"cs":"FvdypH8gl12TfEoK82h/ug==","size":{"width":252,"height":52}}$

Put the value of v:-

${"code":"\\begin{align*}\n{\\frac{1}{2{\\sqrt[]{2}}}\\log_{}\\left|\\frac{\\frac{1}{{\\sqrt[]{2}}}+\\frac{y}{x}}{\\frac{1}{{\\sqrt[]{2}}}-\\frac{y}{x}}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","id":"16-1-0-0-1-0-1-0-0-1-0-2-0","ts":1604063229836,"cs":"UM9xPbHt4sqXxi3kyrPjkg==","size":{"width":258,"height":52}}$

${"code":"\\begin{align*}\n{\\frac{1}{2{\\sqrt[]{2}}}\\log_{}\\left|\\frac{\\frac{x+y{\\sqrt[]{2}}}{x{\\sqrt[]{2}}}}{\\frac{x-y{\\sqrt[]{2}}}{x{\\sqrt[]{2}}}}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","id":"16-1-0-0-1-0-1-0-0-1-0-2-1-0-0","ts":1604063278114,"cs":"Phm3E8ePeXZzq/pAayr3BA==","size":{"width":246,"height":61}}$

${"id":"16-1-0-0-1-0-1-0-0-1-0-2-1-0-1-0","code":"\\begin{align*}\n{\\frac{1}{2{\\sqrt[]{2}}}\\log_{}\\left|\\frac{x+y{\\sqrt[]{2}}}{x-y{\\sqrt[]{2}}}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1604063312586,"cs":"45siGQIOIekt2J8Ke1ymjQ==","size":{"width":261,"height":44}}$

Let -log |C| = c

${"code":"\\begin{align*}\n{\\frac{1}{2{\\sqrt[]{2}}}\\log_{}\\left|\\frac{x+y{\\sqrt[]{2}}}{x-y{\\sqrt[]{2}}}\\right|}&={\\log_{}\\left|x\\right|+c}\t\n\\end{align*}","type":"align*","id":"16-1-0-0-1-0-1-0-0-1-0-2-1-0-1-1","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1604063375018,"cs":"vc2PY4o2xCJ1XGP5msjw/A==","size":{"width":224,"height":44}}$

${"type":"align","font":{"size":10,"family":"Arial","color":"#222222"},"id":"13-0-0","code":"\\begin{align}\n{6.\\,x.dy-y.dx}&={{\\sqrt[]{x^{2}+y^{2}}}.dx}\t\n\\end{align*}","ts":1604063671898,"cs":"/eWNBPlUYzL91o4h3Q79Uw==","size":{"width":208,"height":20}}$

Solun:- Given eq. is:-

${"code":"\\begin{align}\n{x.dy-y.dx}&={{\\sqrt[]{x^{2}+y^{2}}}.dx}\t\n\\end{align}","id":"13-1-0","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1604063704274,"cs":"T19h9wpOD7Uc4IMPT9gvYQ==","size":{"width":193,"height":20}}$

${"type":"align","font":{"size":10,"color":"#222222","family":"Arial"},"id":"13-1-1","code":"\\begin{align}\n{x.dy}&={{\\sqrt[]{x^{2}+y^{2}}}.dx+y.dx}\\\\\n{x.dy}&={\\left(y+{\\sqrt[]{x^{2}+y^{2}}}\\right).dx}\t\n\\end{align*}","ts":1604063764182,"cs":"cBx0G3pvLjiLz6ENIb/IEw==","size":{"width":193,"height":53}}$

${"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{y+{\\sqrt[]{x^{2}+y^{2}}}}{x}}\t\n\\end{align*}","id":"8-0-0-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1604063810270,"cs":"180zy+31pxZxNtYk2VT2zA==","size":{"width":144,"height":37}}$ .....(1)

Let F(x,y)= ${"type":"$","id":"9-0-0-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\frac{y+{\\sqrt[]{x^{2}+y^{2}}}}{x}$","ts":1604063842643,"cs":"qurCR675weoImJYpghRciQ==","size":{"width":61,"height":24}}$

Replace F(x,y) with F(λx,λy):-

${"font":{"family":"Arial","color":"#222222","size":10},"id":"10-0-0-1-0","type":"align*","code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y+{\\sqrt[]{\\lambda^{2}x^{2}+\\lambda^{2}y^{2}}}}{\\lambda x}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y+{\\sqrt[]{\\lambda^{2}\\left(x^{2}+y^{2}\\right)}}}{\\lambda x}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y+\\lambda {\\sqrt[]{x^{2}+y^{2}}}}{\\lambda x}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{y+{\\sqrt[]{x^{2}+y^{2}}}}{x}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","ts":1604063986283,"cs":"57VO+A/7CG4oCq6bfzX8lA==","size":{"width":228,"height":198}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

${"id":"7-0-1","code":"\\begin{align*}\n{\\diff{y}{x}}&={v\\diff{x}{x}+x\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={v+x\\diff{v}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603531746342,"cs":"ScRSgGu7YgO3ambgvvKkzA==","size":{"width":132,"height":69}}$ .......(2)

From eq. 1 and 2:-

${"code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{vx+{\\sqrt[]{x^{2}+v^{2}x^{2}}}}{x}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{vx+{\\sqrt[]{x^{2}\\left(1+v^{2}\\right)}}}{x}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{vx+x{\\sqrt[]{1+v^{2}}}}{x}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x\\left(v+{\\sqrt[]{1+v^{2}}}\\right)}{x}}\\\\\n{v+x\\diff{v}{x}}&={v+{\\sqrt[]{1+v^{2}}}}\t\n\\end{align*}","type":"align*","id":"11-0-0-1-0","font":{"color":"#222222","family":"Arial","size":10},"ts":1604064094418,"cs":"AXM7Xu62p/oPeoOtnExHaw==","size":{"width":209,"height":210}}$

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"12-0-0-1-0","code":"\\begin{align*}\n{x\\diff{v}{x}}&={v+{\\sqrt[]{1+v^{2}}}-v}\\\\\n{x\\diff{v}{x}}&={{\\sqrt[]{1+v^{2}}}}\t\n\\end{align*}","type":"align*","ts":1604064161395,"cs":"vufr+wycDxtOcxk3Qx4CaQ==","size":{"width":166,"height":69}}$

Separating the variables:-

${"id":"4-0-1-0-0-0-0-1-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\frac{dv}{{\\sqrt[]{1+v^{2}}}}}&={\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604064192357,"cs":"HA6sc4R9z+IAkAbCsGqYGQ==","size":{"width":122,"height":37}}$

Integrating both sides:-

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{dv}{{\\sqrt[]{1+v^{2}}}}}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"4-0-1-0-0-0-0-1-1","ts":1604064211735,"cs":"Ix4b1N2mKSqwd6bpkvpGqg==","size":{"width":161,"height":37}}$

We know that:-

${"id":"17-0-0-1","type":"$","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}+a^{2}}}\\right|+C$","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1604064345890,"cs":"QoKay5wbdYLFx/sY2gcs4w==","size":{"width":262,"height":24}}$

${"type":"align*","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"17-2-0-1","ts":1603537288100,"cs":"A1DBoV9o2jjL23JPL6aRsw==","size":{"width":153,"height":36}}$

${"font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","id":"16-1-0-0-0-1-0-0","code":"\\begin{align*}\n{\\log_{}\\left|v+{\\sqrt[]{v^{2}+1}}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604134658362,"cs":"DSxUtURNGoESbDvTezgtsw==","size":{"width":244,"height":26}}$

Put the value of v:-

${"font":{"family":"Arial","color":"#222222","size":10},"id":"16-1-0-0-0-1-1-0-0","type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{y}{x}+{\\sqrt[]{\\frac{y^{2}}{x^{2}}+1}}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604134729534,"cs":"ZI/I7zrFMhIGNhBBmkDuVg==","size":{"width":261,"height":46}}$

${"id":"16-1-0-0-0-1-1-1-0","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{\\log_{}\\left|\\frac{y}{x}+{\\sqrt[]{\\frac{y^{2}+x^{2}}{x^{2}}}}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","type":"align*","ts":1604134768473,"cs":"s9eKLQYB6qOKVUu4g9iQ1Q==","size":{"width":268,"height":46}}$

${"id":"16-1-0-0-0-1-1-1-1-0-0","type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{y}{x}+\\frac{{\\sqrt[]{y^{2}+x^{2}}}}{x}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1604134856522,"cs":"Wfa0qnbBrchwn1VTT5hb+Q==","size":{"width":268,"height":46}}$

${"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{\\log_{}\\left|\\frac{y+{\\sqrt[]{y^{2}+x^{2}}}}{x}\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-1-1-1-1-0-0","type":"align*","ts":1604134890911,"cs":"GcOH3h0Q7wcc33VPay5u9Q==","size":{"width":260,"height":46}}$

${"font":{"color":"#222222","family":"Arial","size":10},"id":"16-1-0-0-0-1-1-1-1-1-1","code":"\\begin{align*}\n{\\log_{}\\left|y+{\\sqrt[]{y^{2}+x^{2}}}\\right|-\\log_{}\\left|x\\right|}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","type":"align*","ts":1604134958382,"cs":"NZr89hXBVYVFOL7FiekmDg==","size":{"width":313,"height":24}}$

${"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|y+{\\sqrt[]{y^{2}+x^{2}}}\\right|}&={2\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\\\\\n{\\log_{}\\left|y+{\\sqrt[]{y^{2}+x^{2}}}\\right|}&={\\log_{}\\left|x^{2}\\right|+\\log_{}\\left|C\\right|}\\\\\n{\\log_{}\\left|y+{\\sqrt[]{y^{2}+x^{2}}}\\right|}&={\\log_{}\\left|Cx^{2}\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-1-1-1-1-2-0","ts":1604135084309,"cs":"Jqwfa8LmwQQUQa9eh9kiMw==","size":{"width":264,"height":85}}$

${"font":{"color":"#222222","family":"Arial","size":10},"id":"20-0-0-1","code":"\\begin{align*}\n{y+{\\sqrt[]{y^{2}+x^{2}}}}&={Cx^{2}}\t\n\\end{align*}","type":"align*","ts":1604135120299,"cs":"kOf0aSHX/Tq4jXgfQ1eusg==","size":{"width":140,"height":20}}$

${"type":"align","code":"\\begin{align}\n{7.\\,\\left\\{x\\cos\\left(\\frac{y}{x}\\right)+y\\sin\\left(\\frac{y}{x}\\right)\\right\\}.ydx}&={\\left\\{y\\sin\\left(\\frac{y}{x}\\right)-x\\cos\\left(\\frac{y}{x}\\right)\\right\\}.xdy}\t\n\\end{align*}","id":"13-0-1-0-0","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1604136851117,"cs":"wrB2fwlcqyWnpnUNBNVvjQ==","size":{"width":456,"height":28}}$

Solun:- Given eq. is:-

${"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align}\n{\\left\\{x\\cos\\left(\\frac{y}{x}\\right)+y\\sin\\left(\\frac{y}{x}\\right)\\right\\}.ydx}&={\\left\\{y\\sin\\left(\\frac{y}{x}\\right)-x\\cos\\left(\\frac{y}{x}\\right)\\right\\}.xdy}\t\n\\end{align}","type":"align*","id":"13-0-1-1-0","ts":1604136877494,"cs":"FUX7KsEdxMunvY3vHmsBLw==","size":{"width":441,"height":28}}$

${"id":"13-0-1-1-1-0","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{y\\left\\{x\\cos\\left(\\frac{y}{x}\\right)+y\\sin\\left(\\frac{y}{x}\\right)\\right\\}}{x\\left\\{y\\sin\\left(\\frac{y}{x}\\right)-x\\cos\\left(\\frac{y}{x}\\right)\\right\\}}}\t\n\\end{align*}","ts":1604136826882,"cs":"K03L8qDm8Bzb/09OfouRtQ==","size":{"width":224,"height":44}}$ .....(1)

Let F(x,y)= ${"id":"9-0-0-1-1-0","code":"$\\frac{y\\left\\{x\\cos\\left(\\frac{y}{x}\\right)+y\\sin\\left(\\frac{y}{x}\\right)\\right\\}}{x\\left\\{y\\sin\\left(\\frac{y}{x}\\right)-x\\cos\\left(\\frac{y}{x}\\right)\\right\\}}$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1604136929064,"cs":"6SXbKjSPp+mdD37kXxLuhA==","size":{"width":118,"height":29}}$

Replace F(x,y) with F(λx,λy):-

${"code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y\\left\\{\\lambda x\\cos\\left(\\frac{\\lambda y}{\\lambda x}\\right)+\\lambda y\\sin\\left(\\frac{\\lambda y}{\\lambda x}\\right)\\right\\}}{\\lambda x\\left\\{\\lambda y\\sin\\left(\\frac{\\lambda y}{\\lambda x}\\right)-\\lambda x\\cos\\left(\\frac{\\lambda y}{\\lambda x}\\right)\\right\\}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y\\left\\{\\lambda \\left(x\\cos\\left(\\frac{y}{x}\\right)+y\\sin\\left(\\frac{y}{x}\\right)\\right)\\right\\}}{\\lambda x\\left\\{\\lambda \\left(y\\sin\\left(\\frac{y}{x}\\right)-x\\cos\\left(\\frac{y}{x}\\right)\\right)\\right\\}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda ^{2}y\\left\\{x\\cos\\left(\\frac{y}{x}\\right)+y\\sin\\left(\\frac{y}{x}\\right)\\right\\}}{\\lambda ^{2}x\\left\\{y\\sin\\left(\\frac{y}{x}\\right)-x\\cos\\left(\\frac{y}{x}\\right)\\right\\}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{y\\left\\{x\\cos\\left(\\frac{y}{x}\\right)+y\\sin\\left(\\frac{y}{x}\\right)\\right\\}}{x\\left\\{y\\sin\\left(\\frac{y}{x}\\right)-x\\cos\\left(\\frac{y}{x}\\right)\\right\\}}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","id":"10-0-0-1-1-0","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1604137399013,"cs":"GX5TEGJI4YRGviofrTSRYw==","size":{"width":316,"height":237}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

${"id":"7-0-2","code":"\\begin{align*}\n{\\diff{y}{x}}&={v\\diff{x}{x}+x\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={v+x\\diff{v}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603531746342,"cs":"5FQOuva0c4wsa37OUEySeg==","size":{"width":132,"height":69}}$ .......(2)

From eq. 1 and 2:-

${"id":"11-0-0-1-1-0","type":"align*","code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{vx\\left\\{x\\cos\\left(\\frac{vx}{x}\\right)+vx\\sin\\left(\\frac{vx}{x}\\right)\\right\\}}{x\\left\\{vx\\sin\\left(\\frac{vx}{x}\\right)-x\\cos\\left(\\frac{vx}{x}\\right)\\right\\}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{v\\left\\{x\\cos\\left(v\\right)+vx\\sin\\left(v\\right)\\right\\}}{\\left\\{vx\\sin\\left(v\\right)-x\\cos\\left(v\\right)\\right\\}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{v\\left\\{x\\left(\\cos\\left(v\\right)+v\\sin\\left(v\\right)\\right)\\right\\}}{\\left\\{x\\left(v\\sin\\left(v\\right)-\\cos\\left(v\\right)\\right)\\right\\}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{v\\left\\{\\cos\\left(v\\right)+v\\sin\\left(v\\right)\\right\\}}{\\left\\{v\\sin\\left(v\\right)-\\cos\\left(v\\right)\\right\\}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{v\\cos\\left(v\\right)+v^{2}\\sin\\left(v\\right)}{v\\sin\\left(v\\right)-\\cos\\left(v\\right)}}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1604137868447,"cs":"BfVRKaSKI98KIiW6OzDjmw==","size":{"width":286,"height":217}}$

${"code":"\\begin{align*}\n{x\\diff{v}{x}}&={\\frac{v\\cos\\left(v\\right)+v^{2}\\sin\\left(v\\right)}{v\\sin\\left(v\\right)-\\cos\\left(v\\right)}-v}\\\\\n{x\\diff{v}{x}}&={\\frac{v\\cos\\left(v\\right)+v^{2}\\sin\\left(v\\right)-v\\left(v\\sin\\left(v\\right)-\\cos\\left(v\\right)\\right)}{v\\sin\\left(v\\right)-\\cos\\left(v\\right)}}\\\\\n{x\\diff{v}{x}}&={\\frac{v\\cos\\left(v\\right)+v^{2}\\sin\\left(v\\right)-v\\left(v\\sin\\left(v\\right)-\\cos\\left(v\\right)\\right)}{v\\sin\\left(v\\right)-\\cos\\left(v\\right)}}\\\\\n{x\\diff{v}{x}}&={\\frac{v\\cos\\left(v\\right)+v^{2}\\sin\\left(v\\right)-v^{2}\\sin\\left(v\\right)+v\\cos\\left(v\\right)}{v\\sin\\left(v\\right)-\\cos\\left(v\\right)}}\\\\\n{x\\diff{v}{x}}&={\\frac{2v\\cos\\left(v\\right)}{v\\sin\\left(v\\right)-\\cos\\left(v\\right)}}\t\n\\end{align*}","id":"12-0-0-1-1-0","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1604138219758,"cs":"G6jMcvHFblpZsxXv91dqsg==","size":{"width":348,"height":216}}$

Separating the variables:-

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"4-0-1-0-0-0-0-1-2-0-0","type":"align*","code":"\\begin{align*}\n{\\frac{\\left[v\\sin v-\\cos v\\right].dv}{2v\\cos v}}&={\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604138417866,"cs":"3wCukUdKc1mN8IB9VU8faQ==","size":{"width":190,"height":33}}$

Integrating both sides:-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{\\left[v\\sin v-\\cos v\\right].dv}{2v\\cos v}}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"id":"4-0-1-0-0-0-0-1-2-1-0","ts":1604138471567,"cs":"kE9JlfZrOd0HuWpGrXMe9w==","size":{"width":229,"height":36}}$

${"code":"\\begin{align*}\n{\\int_{}^{}\\left(\\tan v-\\frac{1}{v}\\right).dv}&={2\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"4-0-1-0-0-0-0-1-2-1-1-0","type":"align*","ts":1604141010388,"cs":"x7OUEFMip5nd4rW0NTBABg==","size":{"width":221,"height":37}}$

We know that:-

${"type":"align*","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"17-2-0-2-0","ts":1603537288100,"cs":"qqpSZLRLvLum/niHThMRBQ==","size":{"width":153,"height":36}}$

${"id":"17-2-0-3-0","font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}\\tan x.dx}&={-\\log_{}\\left|\\cos x\\right|+C}\t\n\\end{align*}","type":"align*","ts":1604140566964,"cs":"KxpGWzX/GKPFM8rtboXtCQ==","size":{"width":210,"height":36}}$

${"type":"align*","id":"16-1-0-0-0-1-0-1-0-0-0-0","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{-\\log_{}\\left|\\cos v\\right|-\\log_{}\\left|v\\right|}&={2\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604141087399,"cs":"CBcOZ3JgsEuS5jhbiMjJ/g==","size":{"width":276,"height":16}}$

${"type":"align*","code":"\\begin{align*}\n{-\\left[\\log_{}\\left|\\cos v\\right|+\\log_{}\\left|v\\right|\\right]}&={\\log_{}\\left|x^{2}\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"id":"16-1-0-0-0-1-0-1-0-0-1-0-0","ts":1604141392182,"cs":"Do8d1JO7TTMJ0ziKWKLNFQ==","size":{"width":278,"height":20}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"id":"16-1-0-0-0-1-0-1-0-0-1-1-0","code":"\\begin{align*}\n{-\\log_{}\\left|v.\\cos v\\right|}&={\\log_{}\\left|x^{2}\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604141443061,"cs":"JPecDfYYJNWIkvlRsks47w==","size":{"width":228,"height":20}}$

Put the value of v:-

${"type":"align*","font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{-\\log_{}\\left|\\frac{y}{x}.\\cos \\left(\\frac{y}{x}\\right)\\right|}&={\\log_{}\\left|x^{2}\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-1-1-1-0-0","ts":1604141583930,"cs":"91gZPyK5R7ci7G6cZ3us/g==","size":{"width":265,"height":29}}$

${"type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"code":"\\begin{align*}\n{-\\log_{}\\left|\\frac{y}{x}.\\cos \\left(\\frac{y}{x}\\right)\\right|-\\log_{}\\left|x^{2}\\right|}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-1-1-1-1-0-0","ts":1604141628625,"cs":"WiteZo7klTX+lmLTmhseEA==","size":{"width":265,"height":29}}$

${"id":"16-1-0-0-0-1-0-1-0-0-1-1-1-1-1-0","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{-\\left[\\log_{}\\left|\\frac{y}{x}.\\cos \\left(\\frac{y}{x}\\right)\\right|+\\log_{}\\left|x^{2}\\right|\\right]}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604141711070,"cs":"zhehvwFCLCQDXKis4pU/CQ==","size":{"width":277,"height":29}}$

${"type":"align*","id":"16-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-0-0","code":"\\begin{align*}\n{-\\left[\\log_{}\\left|\\left(\\frac{y}{x}.\\cos \\left(\\frac{y}{x}\\right)\\right).x^{2}\\right|\\right]}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"ts":1604141824346,"cs":"cBaWhAk80RoVmsVXXN0Vwg==","size":{"width":249,"height":29}}$

${"type":"align*","id":"16-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{-\\left[\\log_{}\\left|xy\\cos \\left(\\frac{y}{x}\\right)\\right|\\right]}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1604141958295,"cs":"LRf6ahfTKjIgmBEP4jYfFg==","size":{"width":204,"height":29}}$

${"code":"\\begin{align*}\n{\\left[\\log_{}\\left|xy\\cos \\left(\\frac{y}{x}\\right)\\right|\\right]}&={-\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","ts":1604141999931,"cs":"7DYuVg8SFYyluHpZix+Ipg==","size":{"width":208,"height":29}}$

${"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|xy\\cos \\left(\\frac{y}{x}\\right)\\right|}&={\\log_{}\\left|C^{-1}\\right|}\\\\\n{\\log_{}\\left|xy\\cos \\left(\\frac{y}{x}\\right)\\right|}&={\\log_{}\\left|\\frac{1}{C}\\right|}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"id":"16-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-1-1-1-0","ts":1604142090763,"cs":"T3Wni30v1f0xawxQxSdcYA==","size":{"width":196,"height":68}}$

${"type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{xy\\cos \\left(\\frac{y}{x}\\right)}&={\\frac{1}{C}}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-1-1-1-1-0","ts":1604142135762,"cs":"a1fN+otW38kucvqhI2spfw==","size":{"width":117,"height":32}}$

Let 1/C = c

${"type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"id":"16-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-1-1-1-1-1","code":"\\begin{align*}\n{xy\\cos \\left(\\frac{y}{x}\\right)}&={c}\t\n\\end{align*}","ts":1604142190958,"cs":"/EACYPVfqZQ8zx3p7Qillg==","size":{"width":105,"height":28}}$

${"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align}\n{8.\\,x\\diff{y}{x}-y+x\\sin\\left(\\frac{y}{x}\\right)}&={0}\t\n\\end{align}","id":"13-0-1-0-1-0-0","type":"align*","ts":1604142362992,"cs":"dQH32ITvZnD3wHQ4ApP1gA==","size":{"width":192,"height":32}}$

Solun:- Given eq. is:-

${"type":"align","font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align}\n{x\\diff{y}{x}-y+x\\sin\\left(\\frac{y}{x}\\right)}&={0}\t\n\\end{align*}","id":"13-0-1-0-1-1-0-0","ts":1604142473778,"cs":"wIsTX8oBcio6+OpO7PMGXw==","size":{"width":177,"height":32}}$

${"font":{"size":10,"family":"Arial","color":"#222222"},"type":"align*","id":"13-0-1-0-1-1-1-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{y-x\\sin\\left(\\frac{y}{x}\\right)}{x}}\t\n\\end{align*}","ts":1604142700125,"cs":"YsfW42KdZB9CgPjGbiX9Iw==","size":{"width":141,"height":36}}$ .....(1)

Let F(x,y)= ${"type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\frac{y-x\\sin\\left(\\frac{y}{x}\\right)}{x}$","id":"9-0-0-1-1-1-0","ts":1604142783833,"cs":"2zYSiPjNlctvcKRAF0iuyA==","size":{"width":62,"height":24}}$

Replace F(x,y) with F(λx,λy):-

${"id":"10-0-0-1-1-1-0","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y-\\lambda x\\sin\\left(\\frac{\\lambda y}{\\lambda x}\\right)}{\\lambda x}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y-\\lambda x\\sin\\left(\\frac{y}{x}\\right)}{\\lambda x}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda \\left(y-x\\sin\\left(\\frac{y}{x}\\right)\\right)}{\\lambda x}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{y-x\\sin\\left(\\frac{y}{x}\\right)}{x}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","ts":1604143450053,"cs":"v3ApmwFQ6mWguxBZaQdtng==","size":{"width":224,"height":205}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

${"id":"7-0-3","code":"\\begin{align*}\n{\\diff{y}{x}}&={v\\diff{x}{x}+x\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={v+x\\diff{v}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603531746342,"cs":"cshMEbMN+CaLpIGgn6vMew==","size":{"width":132,"height":69}}$ .......(2)

From eq. 1 and 2:-

${"font":{"family":"Arial","size":10,"color":"#222222"},"id":"11-0-0-1-1-1-0","code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{vx-x\\sin\\left(\\frac{vx}{x}\\right)}{x}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x\\left[v-\\sin\\left(v\\right)\\right]}{x}}\\\\\n{v+x\\diff{v}{x}}&={v-\\sin v}\t\n\\end{align*}","type":"align*","ts":1604143604758,"cs":"21trRr7cPhxkzgr3CrMMFA==","size":{"width":192,"height":113}}$

${"id":"12-0-0-1-1-1-0","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","code":"\\begin{align*}\n{x\\diff{v}{x}}&={v-\\sin v-v}\\\\\n{x\\diff{v}{x}}&={-\\sin v}\t\n\\end{align*}","ts":1604143753001,"cs":"72ybiQXixOJp+kmRGPrpQg==","size":{"width":138,"height":69}}$

Separating the variables:-

${"id":"4-0-1-0-0-0-0-1-2-0-1-0-0","code":"\\begin{align*}\n{\\frac{dv}{\\sin v}}&={\\frac{-1}{x}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1604143801140,"cs":"7nH6esRcEYToOWo9+rEisw==","size":{"width":108,"height":32}}$

Integrating both sides:-

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{dv}{\\sin v}}&={\\int_{}^{}\\frac{-1}{x}.dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"4-0-1-0-0-0-0-1-2-0-1-1-0","ts":1604143833391,"cs":"tkXjMKkMoK879BC7xoZfpg==","size":{"width":148,"height":36}}$

${"code":"\\begin{align*}\n{\\int_{}^{}\\cos ecv.dv}&={\\int_{}^{}\\frac{-1}{x}.dx}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"4-0-1-0-0-0-0-1-2-0-1-1-1","ts":1604143908704,"cs":"fiJ6nnrNNno/9KRSikx0GA==","size":{"width":180,"height":36}}$

We know that:-

${"type":"align*","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"17-2-0-2-1","ts":1603537288100,"cs":"myIkLkTRGi4PaaxJKy5Fug==","size":{"width":153,"height":36}}$

${"type":"align*","id":"17-2-0-3-1","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\cos ecx.dx}&={\\log_{}\\left|\\cos ecx-\\cot x\\right|+C}\t\n\\end{align*}","ts":1604143961625,"cs":"4BERp3lhZoU9jLjxkBgBzQ==","size":{"width":274,"height":36}}$

${"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\cos ec\\,v-\\cot v\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-0","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1604144008469,"cs":"M3z9s23vqzo7zEChyoZBIA==","size":{"width":272,"height":16}}$

Put the value of v:-

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-1-1-0-0-0","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\cos ec\\,\\left(\\frac{y}{x}\\right)-\\cot \\left(\\frac{y}{x}\\right)\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604146201346,"cs":"EGtVprhg8ZfgWVxDM8gfxA==","size":{"width":328,"height":29}}$

${"font":{"color":"#222222","family":"Arial","size":10},"id":"16-1-0-0-0-1-0-1-0-0-0-1-1-1-0-1","code":"\\begin{align*}\n{\\log_{}\\left|\\cos ec\\,\\left(\\frac{y}{x}\\right)-\\cot \\left(\\frac{y}{x}\\right)\\right|}&={\\log_{}\\left|x^{-1}\\right|+\\log_{}\\left|C\\right|}\\\\\n{\\log_{}\\left|\\cos ec\\,\\left(\\frac{y}{x}\\right)-\\cot \\left(\\frac{y}{x}\\right)\\right|}&={\\log_{}\\left|\\frac{1}{x}\\right|+\\log_{}\\left|C\\right|}\\\\\n{\\log_{}\\left|\\cos ec\\,\\left(\\frac{y}{x}\\right)-\\cot \\left(\\frac{y}{x}\\right)\\right|}&={\\log_{}\\left|\\frac{C}{x}\\right|}\t\n\\end{align*}","type":"align*","ts":1604146340068,"cs":"LwPGriOiD6uBprgegoAr+Q==","size":{"width":330,"height":109}}$

${"type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"id":"15-0","code":"\\begin{align*}\n{\\cos ec\\,\\left(\\frac{y}{x}\\right)-\\cot \\left(\\frac{y}{x}\\right)}&={\\frac{C}{x}}\t\n\\end{align*}","ts":1604146437948,"cs":"4uqfNswtRfsZq8nm4FeWxQ==","size":{"width":193,"height":32}}$

${"font":{"color":"#222222","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\frac{1}{\\sin\\left(\\frac{y}{x}\\right)}-\\frac{\\cos\\left(\\frac{y}{x}\\right)}{\\sin\\left(\\frac{y}{x}\\right)}}&={\\frac{C}{x}}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-1-1-1-0-0","ts":1604146419412,"cs":"JAZE71G0zcnnjvt3E4xKtQ==","size":{"width":174,"height":44}}$

${"font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{1-\\cos\\left(\\frac{y}{x}\\right)}&={\\frac{C}{x}\\sin\\left(\\frac{y}{x}\\right)}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-1-1-1-1-0","type":"align*","ts":1604146490353,"cs":"It7FyRMIv6CmLl4jhrDwvg==","size":{"width":182,"height":32}}$

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"16-1-0-0-0-1-0-1-0-0-0-1-1-1-1-2","type":"align*","code":"\\begin{align*}\n{x\\left[1-\\cos\\left(\\frac{y}{x}\\right)\\right]}&={C\\sin\\left(\\frac{y}{x}\\right)}\t\n\\end{align*}","ts":1604146549370,"cs":"+nHIGNzChEqPAoDyjkhmsA==","size":{"width":202,"height":28}}$

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"13-0-1-0-1-0-1-0","code":"\\begin{align}\n{9.\\,y.dx+x\\log_{}\\left(\\frac{y}{x}\\right).dy-2x.dy}&={0}\t\n\\end{align}","type":"align*","ts":1604220447397,"cs":"pcDRzrTMWSaWL2I6XE2gzA==","size":{"width":248,"height":28}}$

Solun:- Given eq. is:-

${"type":"align","id":"13-0-1-0-1-0-2-0","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align}\n{y.dx+x\\log_{}\\left(\\frac{y}{x}\\right).dy-2x.dy}&={0}\t\n\\end{align*}","ts":1604220473034,"cs":"iphNp6xh8MiigRl3Oh325Q==","size":{"width":232,"height":28}}$

${"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align","code":"\\begin{align}\n{y.dx+\\left[x\\log_{}\\left(\\frac{y}{x}\\right)-2x\\right].dy}&={0}\t\n\\end{align*}","id":"13-0-1-0-1-0-2-1-0","ts":1604220515223,"cs":"TCcGQNc/2tFcdWhGOo3xHw==","size":{"width":224,"height":28}}$

${"id":"13-0-1-0-1-1-1-1-0-0","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{y}{2x-x\\log_{}\\left(\\frac{y}{x}\\right)}}\t\n\\end{align*}","ts":1604220577053,"cs":"YumhzL6407NC9pbZdHZplg==","size":{"width":152,"height":40}}$ .....(1)

Let F(x,y)= ${"code":"$\\frac{y}{2x-x\\log_{}\\left(\\frac{y}{x}\\right)}$","id":"9-0-0-1-1-1-1-0-0","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1604220623176,"cs":"BNwP42EqSPYTNuxXwWv81g==","size":{"width":70,"height":24}}$

Replace F(x,y) with F(λx,λy):-

${"code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y}{2\\lambda x-\\lambda x\\log_{}\\left(\\frac{\\lambda y}{\\lambda x}\\right)}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y}{2\\lambda x-\\lambda x\\log_{}\\left(\\frac{y}{x}\\right)}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y}{\\lambda \\left[2x-x\\log_{}\\left(\\frac{y}{x}\\right)\\right]}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{y}{2x-x\\log_{}\\left(\\frac{y}{x}\\right)}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"type":"align*","id":"10-0-0-1-1-1-1-0-0","ts":1604220707475,"cs":"+OqURp11VTjEVS59JiWYTw==","size":{"width":234,"height":213}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

${"id":"7-0-4","code":"\\begin{align*}\n{\\diff{y}{x}}&={v\\diff{x}{x}+x\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={v+x\\diff{v}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603531746342,"cs":"oVK+FjnoBZEbHnFNq8lT2A==","size":{"width":132,"height":69}}$ .......(2)

From eq. 1 and 2:-

${"id":"11-0-0-1-1-1-1-0-0","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{vx}{2x-x\\log_{}\\left(\\frac{vx}{x}\\right)}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{vx}{2x-x\\log_{}\\left(v\\right)}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{vx}{x\\left[2-\\log_{}\\left(v\\right)\\right]}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{v}{2-\\log_{}\\left(v\\right)}}\t\n\\end{align*}","ts":1604220892161,"cs":"s9/AO02f50MhUOAjUMQ23w==","size":{"width":193,"height":164}}$

${"code":"\\begin{align*}\n{x\\diff{v}{x}}&={\\frac{v}{2-\\log_{}\\left(v\\right)}-v}\\\\\n{x\\diff{v}{x}}&={\\frac{v-v\\left(2-\\log_{}\\left(v\\right)\\right)}{2-\\log_{}\\left(v\\right)}}\\\\\n{x\\diff{v}{x}}&={\\frac{v-2v+v\\log_{}\\left(v\\right)}{2-\\log_{}\\left(v\\right)}}\\\\\n{x\\diff{v}{x}}&={\\frac{v\\log_{}\\left(v\\right)-v}{2-\\log_{}\\left(v\\right)}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"id":"12-0-0-1-1-1-1-0-0","ts":1604221068106,"cs":"qBW3+uFCvzRCjbPYvW+log==","size":{"width":180,"height":165}}$

Separating the variables:-

${"id":"4-0-1-0-0-0-0-1-2-0-1-0-1-0-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\frac{2-\\log_{}\\left(v\\right)}{v\\log_{}\\left(v\\right)-v}.dv}&={\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604221130356,"cs":"eneZqvs+2Jriu98+wyWY5Q==","size":{"width":172,"height":37}}$

Integrating both sides:-

${"id":"4-0-1-0-0-0-0-1-2-0-1-0-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{2-\\log_{}\\left(v\\right)}{v\\log_{}\\left(v\\right)-v}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1604221169311,"cs":"qcQ9UZx2FXlHWqyaddvSbg==","size":{"width":210,"height":37}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"id":"4-0-1-0-0-0-0-1-2-0-1-0-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{-\\left(\\log_{}\\left(v\\right)-2\\right)}{v\\log_{}\\left(v\\right)-v}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","ts":1604221210574,"cs":"YC2NVgjcIqc+8oue3Nx9Og==","size":{"width":225,"height":37}}$

${"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"4-0-1-0-0-0-0-1-2-0-1-0-1-1-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{\\log_{}\\left(v\\right)-2}{v\\left(\\log_{}\\left(v\\right)-1\\right)}.dv}&={-\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604221491120,"cs":"2QwLMifElpAie/leZszDdQ==","size":{"width":236,"height":37}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"4-0-1-0-0-0-0-1-2-0-1-0-1-1-1-1-0-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{\\log_{}\\left(v\\right)-1-1}{v\\left(\\log_{}\\left(v\\right)-1\\right)}.dv}&={-\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604221521384,"cs":"6S4odLstVPjrpZkQqRYFPA==","size":{"width":244,"height":37}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{\\left[\\log_{}\\left(v\\right)-1\\right]-1}{v\\left(\\log_{}\\left(v\\right)-1\\right)}.dv}&={-\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","id":"4-0-1-0-0-0-0-1-2-0-1-0-1-1-1-1-0-1-1-0","ts":1604221558124,"cs":"7HBRDjGOqdctIdK5cdyzBQ==","size":{"width":252,"height":37}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"4-0-1-0-0-0-0-1-2-0-1-0-1-1-1-1-0-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{\\log_{}\\left(v\\right)-1}{v\\left(\\log_{}\\left(v\\right)-1\\right)}.dv-\\int_{}^{}\\frac{1}{v\\left(\\log_{}\\left(v\\right)-1\\right)}.dv}&={-\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604221608852,"cs":"b5eky5RC1x9/ab2WT6qSoA==","size":{"width":396,"height":37}}$

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{v}.dv-\\int_{}^{}\\frac{\\frac{1}{v}}{\\log_{}\\left(v\\right)-1}.dv}&={-\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","id":"4-0-1-0-0-0-0-1-2-0-1-0-1-1-1-1-0-1-1-1-1","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1604221690916,"cs":"643Lq2V00qbV0K1WnoeL1Q==","size":{"width":292,"height":41}}$

We know that:-

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{f^{\\prime }\\left(x\\right)}{f\\left(x\\right)}.dx}&={\\log_{}\\left|f\\left(x\\right)\\right|+C}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"id":"17-2-0-2-2-0-0","type":"align*","ts":1604221733426,"cs":"KkhiiojukNEKPGmZeSdVLA==","size":{"width":200,"height":37}}$

${"code":"\\begin{align*}\n{\\log_{}\\left|v\\right|-\\log_{}\\left|\\log_{}\\left|v\\right|-1\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-0","ts":1604221784835,"cs":"MdBHoTQjRI4yvE7aB7JGow==","size":{"width":300,"height":16}}$

Put the value of v:-

${"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{y}{x}\\right|-\\log_{}\\left|\\log_{}\\left|\\frac{y}{x}\\right|-1\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-1-0","ts":1604221826891,"cs":"CNjz7wXWBXU6ZcIkgO0HOw==","size":{"width":322,"height":29}}$

${"font":{"color":"#222222","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{y}{x}\\right|-\\log_{}\\left|\\log_{}\\left|\\frac{y}{x}\\right|-1\\right|+\\log_{}\\left|x\\right|}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-1-1-0","ts":1604221949313,"cs":"av2nkpoIf22TE4vZO0XEFg==","size":{"width":308,"height":29}}$

${"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{\\frac{y}{x}\\times x}{\\log_{}\\left|\\frac{y}{x}\\right|-1}\\right|}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-1-1-1-0-0","ts":1604222009526,"cs":"DsNOiqjTKTJ3Q8pBZLWchw==","size":{"width":181,"height":44}}$

${"type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\log_{}\\left|\\frac{y}{\\log_{}\\left|\\frac{y}{x}\\right|-1}\\right|}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-1-1-1-1-0","ts":1604222033257,"cs":"/rTkcgw/sVEifX2ipstuew==","size":{"width":181,"height":44}}$

${"font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\frac{y}{\\log_{}\\left|\\frac{y}{x}\\right|-1}}&={C}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-1-1-1-2-0-0-0","type":"align*","ts":1604222072939,"cs":"wEGNP/g+3tjSlZa1Gzjtkg==","size":{"width":114,"height":36}}$

${"type":"align*","code":"\\begin{align*}\n{\\frac{y}{C}}&={\\log_{}\\left|\\frac{y}{x}\\right|-1}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-1-1-1-2-1","font":{"family":"Arial","color":"#222222","size":10},"ts":1604222111343,"cs":"K/8vNsV/1XTEam1MwCZjmg==","size":{"width":117,"height":29}}$

Let 1/C = c

${"type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-1-1-1-2-1","code":"\\begin{align*}\n{y.c}&={\\log_{}\\left|\\frac{y}{x}\\right|-1}\t\n\\end{align*}","ts":1604222141719,"cs":"JbHutCHK1C+XZ7O5t8u2kw==","size":{"width":120,"height":29}}$

${"font":{"color":"#222222","family":"Arial","size":10},"type":"align","code":"\\begin{align}\n{10.\\,\\left(1+e^{\\frac{x}{y}}\\right).dx+e^{\\frac{x}{y}}\\left(1-\\frac{x}{y}\\right).dy}&={0}\t\n\\end{align*}","id":"13-0-1-0-1-0-1-1-0-0","ts":1604222282785,"cs":"J+MBehfsrOSjCOEoOgGLTQ==","size":{"width":266,"height":37}}$

Solun:- Given eq. is:-

${"code":"\\begin{align}\n{\\left(1+e^{\\frac{x}{y}}\\right).dx+e^{\\frac{x}{y}}\\left(1-\\frac{x}{y}\\right).dy}&={0}\t\n\\end{align}","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"id":"13-0-1-0-1-0-1-1-1-0","ts":1604222318025,"cs":"9wekMaZs0Exreceqj2NkqA==","size":{"width":244,"height":37}}$

${"type":"align","id":"13-0-1-0-1-0-1-1-1-1","font":{"size":10,"color":"#222222","family":"Arial"},"code":"\\begin{align}\n{\\left(1+e^{\\frac{x}{y}}\\right).dx}&={-e^{\\frac{x}{y}}\\left(1-\\frac{x}{y}\\right).dy}\\\\\n{\\left(1+e^{\\frac{x}{y}}\\right).dx}&={e^{\\frac{x}{y}}\\left(\\frac{x}{y}-1\\right).dy}\t\n\\end{align*}","ts":1604222399250,"cs":"Vc4q3/nm3ohkOXjHu0WTNg==","size":{"width":228,"height":80}}$

${"id":"13-0-1-0-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{\\diff{x}{y}}&={\\frac{e^{\\frac{x}{y}}\\left(\\frac{x}{y}-1\\right)}{1+e^{\\frac{x}{y}}}}\t\n\\end{align*}","type":"align*","ts":1604222590562,"cs":"f+3a7qnf+fxwjgoKvz59MQ==","size":{"width":129,"height":52}}$ .....(1)

Let F(y,x)= ${"font":{"size":10,"family":"Arial","color":"#000000"},"id":"9-0-0-1-1-1-1-1","type":"$","code":"$\\frac{e^{\\frac{x}{y}}\\left(\\frac{x}{y}-1\\right)}{1+e^{\\frac{x}{y}}}$","ts":1604222640431,"cs":"532EvvLlh4Kxy0XsDSu7Qw==","size":{"width":57,"height":36}}$

Replace F(y,x) with F(λy,λx):-

${"code":"\\begin{align*}\n{F\\left(\\lambda y,\\lambda x\\right)}&={\\frac{e^{\\frac{\\lambda x}{\\lambda y}}\\left(\\frac{\\lambda x}{\\lambda y}-1\\right)}{1+e^{\\frac{\\lambda x}{\\lambda y}}}}\\\\\n{F\\left(\\lambda y,\\lambda x\\right)}&={\\lambda^{0}\\left(\\frac{e^{\\frac{x}{y}}\\left(\\frac{x}{y}-1\\right)}{1+e^{\\frac{x}{y}}}\\right)}\\\\\n{F\\left(\\lambda y,\\lambda x\\right)}&={\\lambda^{0}.F\\left(y,x\\right)}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","id":"10-0-0-1-1-1-1-1","ts":1604222868666,"cs":"r0JZNMvISkU6UHTkd+0fqg==","size":{"width":214,"height":144}}$

Then F(y,x) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute x = vy

Differentiate with respect to y:-

${"id":"7-0-5","code":"\\begin{align*}\n{\\diff{x}{y}}&={v\\diff{y}{y}+y\\diff{v}{y}}\\\\\n{\\diff{x}{y}}&={v+y\\diff{v}{y}}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","ts":1604222953594,"cs":"qhd0Y7+okQ23f/WBEOen9Q==","size":{"width":128,"height":76}}$ .......(2)

From eq. 1 and 2:-

${"id":"11-0-0-1-1-1-1-1","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{v+y\\diff{v}{y}}&={\\frac{e^{\\frac{vy}{y}}\\left(\\frac{vy}{y}-1\\right)}{1+e^{\\frac{vy}{y}}}}\\\\\n{v+y\\diff{v}{y}}&={\\frac{e^{v}\\left(v-1\\right)}{1+e^{v}}}\t\n\\end{align*}","type":"align*","ts":1604223081243,"cs":"JhdhX0eydJheyXHbA4eCng==","size":{"width":170,"height":94}}$

${"font":{"color":"#222222","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{y\\diff{v}{y}}&={\\frac{e^{v}\\left(v-1\\right)}{1+e^{v}}-v}\\\\\n{y\\diff{v}{y}}&={\\frac{e^{v}\\left(v-1\\right)-v\\left(1+e^{v}\\right)}{1+e^{v}}}\\\\\n{y\\diff{v}{y}}&={\\frac{v.e^{v}-e^{v}-v-v.e^{v}}{1+e^{v}}}\\\\\n{y\\diff{v}{y}}&={\\frac{-e^{v}-v}{1+e^{v}}}\\\\\n{y\\diff{v}{y}}&={\\frac{-\\left(e^{v}+v\\right)}{1+e^{v}}}\t\n\\end{align*}","id":"12-0-0-1-1-1-1-1","ts":1604223330915,"cs":"Nf6T52pLL+ZzJWNf48va3g==","size":{"width":200,"height":201}}$

Separating the variables:-

${"code":"\\begin{align*}\n{\\frac{1+e^{v}}{\\left(e^{v}+v\\right)}.dv}&={\\frac{-1}{y}.dy}\t\n\\end{align*}","id":"4-0-1-0-0-0-0-1-2-0-1-0-1-0-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1604223489146,"cs":"V7N/3jKWbSmyzTlBKv7MVg==","size":{"width":153,"height":36}}$

Integrating both sides:-

${"id":"4-0-1-0-0-0-0-1-2-0-1-0-1-0-1-1","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1+e^{v}}{\\left(e^{v}+v\\right)}.dv}&={\\int_{}^{}\\frac{-1}{y}.dy}\t\n\\end{align*}","type":"align*","ts":1604223517704,"cs":"bYrAR/uAKvbka03mmh6x3Q==","size":{"width":192,"height":36}}$

We know that:-

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{f^{\\prime }\\left(x\\right)}{f\\left(x\\right)}.dx}&={\\log_{}\\left|f\\left(x\\right)\\right|+C}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"id":"17-2-0-2-2-1","type":"align*","ts":1604221733426,"cs":"whnaY1feQOeSFHW+cNnzkQ==","size":{"width":200,"height":37}}$

${"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"18","ts":1604223664208,"cs":"A4wfC55ETMW7u+3PGQtckQ==","size":{"width":153,"height":36}}$

${"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|v+e^{v}\\right|}&={-\\log_{}\\left|y\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-1-0","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1604223746566,"cs":"ee6G4GroTVNOVvMwiPC1HA==","size":{"width":212,"height":16}}$

Put the value of v:-

${"type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{\\log_{}\\left|\\frac{x}{y}+e^{\\frac{x}{y}}\\right|}&={-\\log_{}\\left|y\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-1-1","ts":1604224878638,"cs":"9v2UlRns7rAciGo3lm+0JA==","size":{"width":226,"height":36}}$

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-1-2-0","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{x}{y}+e^{\\frac{x}{y}}\\right|+\\log_{}\\left|y\\right|}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"ts":1604224918248,"cs":"StfdoBOzvA/tO5c9kMFjJA==","size":{"width":212,"height":36}}$

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-1-2-1-0","code":"\\begin{align*}\n{\\log_{}\\left|\\left(\\frac{x}{y}+e^{\\frac{x}{y}}\\right).y\\right|}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","ts":1604224949343,"cs":"3cDzsVR7r46A8CeBXtkUzg==","size":{"width":190,"height":37}}$

${"font":{"family":"Arial","color":"#222222","size":10},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|x+y.e^{\\frac{x}{y}}\\right|}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-1-2-1-1-0","ts":1604224980642,"cs":"l1bjoyT2DCR6d6ARoWdVDg==","size":{"width":160,"height":21}}$

${"type":"align*","code":"\\begin{align*}\n{x+y.e^{\\frac{x}{y}}}&={C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-1-2-1-1-1","ts":1604225017702,"cs":"0q/xxPZ5uQTkUQd5cmHo/A==","size":{"width":93,"height":18}}$

For each of the Differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:

${"code":"\\begin{align}\n{11.\\,\\left(x+y\\right).dy+\\left(x-y\\right).dx}&={0}\t\n\\end{align}","id":"13-0-1-0-1-0-1-1-0-1-0-0-0","type":"align*","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1604225267073,"cs":"RoDD42HciqgPHMrpA6+4jw==","size":{"width":216,"height":16}}$ ; y = 1 when x = 1

Solun:- Given eq. is:-

${"code":"\\begin{align}\n{\\left(x+y\\right).dy+\\left(x-y\\right).dx}&={0}\t\n\\end{align}","font":{"size":10,"color":"#222222","family":"Arial"},"id":"13-0-1-0-1-0-1-1-0-1-0-1","type":"align*","ts":1604225413193,"cs":"7acpcW2DDA5jI4CYifTMfg==","size":{"width":192,"height":16}}$

${"type":"align","code":"\\begin{align}\n{\\left(x+y\\right).dy}&={-\\left(x-y\\right).dx}\\\\\n{\\left(x+y\\right).dy}&={\\left(y-x\\right).dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"id":"13-0-1-0-1-0-1-1-0-1-0-2","ts":1604225480528,"cs":"RpzW5UZduBkahbfNLWGXLg==","size":{"width":177,"height":36}}$

${"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{y-x}{x+y}}\t\n\\end{align*}","id":"13-0-1-0-1-1-1-1-0-1-0","font":{"family":"Arial","color":"#222222","size":10},"ts":1604225530153,"cs":"Gp9IIF8jGq/ZSq/GCkcDxw==","size":{"width":88,"height":36}}$ .....(1)

Let F(x,y)= ${"type":"$","code":"$\\frac{y-x}{x+y}$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"9-0-0-1-1-1-1-0-1-0","ts":1604225562631,"cs":"xhhIRl0SIJQ9hc/Dy7eiDQ==","size":{"width":28,"height":21}}$

Replace F(x,y) with F(λx,λy):-

${"type":"align*","code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y-\\lambda x}{\\lambda x+\\lambda y}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{y-x}{x+y}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","id":"10-0-0-1-1-1-1-0-1-0","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1604225698188,"cs":"33ruTjxRXUDjHq4X29+xHg==","size":{"width":169,"height":100}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

From eq. 1 and 2:-

${"id":"11-0-0-1-1-1-1-0-1-0","type":"align*","code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{vx-x}{x+vx}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x\\left(v-1\\right)}{x\\left(1+v\\right)}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{v-1}{1+v}}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1604225800458,"cs":"t+MTZQpy4urxC24Gvj/lBw==","size":{"width":144,"height":116}}$

${"code":"\\begin{align*}\n{x\\diff{v}{x}}&={\\frac{v-1}{1+v}-v}\\\\\n{x\\diff{v}{x}}&={\\frac{v-1-v\\left(1+v\\right)}{1+v}}\\\\\n{x\\diff{v}{x}}&={\\frac{v-1-v-v^{2}}{1+v}}\\\\\n{x\\diff{v}{x}}&={\\frac{-1-v^{2}}{1+v}}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","id":"12-0-0-1-1-1-1-0-1-0","ts":1604225998178,"cs":"EWj3hRaF20YfQ4Rc/hyUqQ==","size":{"width":170,"height":156}}$

${"type":"align*","code":"\\begin{align*}\n{x\\diff{v}{x}}&={\\frac{-\\left(1+v^{2}\\right)}{1+v}}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"id":"19-0-0","ts":1604226018932,"cs":"+trfoLKbShvH8RRHZfUTfw==","size":{"width":129,"height":37}}$

Separating the variables:-

${"font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{\\frac{1+v}{\\left(1+v^{2}\\right)}.dv}&={\\frac{-1}{x}.dx}\t\n\\end{align*}","type":"align*","id":"19-1-0","ts":1604226202062,"cs":"0X/gDYrV3yf9wca6FSTYVg==","size":{"width":156,"height":36}}$

Integrating both sides:-

${"id":"19-2-0","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1+v}{\\left(1+v^{2}\\right)}.dv}&={\\int_{}^{}\\frac{-1}{x}.dx}\t\n\\end{align*}","type":"align*","ts":1604226229107,"cs":"zp0O6GJwD8HELoGMDIK21w==","size":{"width":194,"height":36}}$

${"type":"align*","id":"19-2-1-0","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{\\left(1+v^{2}\\right)}.dv+\\int_{}^{}\\frac{v}{\\left(1+v^{2}\\right)}.dv}&={\\int_{}^{}\\frac{-1}{x}.dx}\t\n\\end{align*}","ts":1604226293439,"cs":"tqBtKcx0S76LTdc1dxST8w==","size":{"width":317,"height":36}}$

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{\\left(1+v^{2}\\right)}.dv+\\frac{1}{2}\\int_{}^{}\\frac{2v}{\\left(1+v^{2}\\right)}.dv}&={\\int_{}^{}\\frac{-1}{x}.dx}\t\n\\end{align*}","id":"19-2-1-1","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1604226323985,"cs":"Zp2R6vGJxbztEsgJDEOQuw==","size":{"width":334,"height":36}}$

We know that:-

${"code":"\\begin{align*}\n{\\int_{}^{}\\frac{f^{\\prime }\\left(x\\right)}{f\\left(x\\right)}.dx}&={\\log_{}\\left|f\\left(x\\right)\\right|+C}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"id":"17-2-0-2-2-0-1-0","type":"align*","ts":1604221733426,"cs":"Px2DSLJh2iq+KRznNJhhjQ==","size":{"width":200,"height":37}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{1+x^{2}}.dx}&={\\tan^{-1}x+C}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"id":"17-2-0-2-2-0-2-0","ts":1604226414871,"cs":"1/t4yg+znvs+uLmJn+v2jA==","size":{"width":196,"height":36}}$

${"type":"align*","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-0","code":"\\begin{align*}\n{\\tan^{-1}v+\\frac{1}{2}\\log_{}\\left|1+v^{2}\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"ts":1604226536755,"cs":"WSf6bYcm1vlacdpFKbGIfQ==","size":{"width":300,"height":32}}$

Put the value of v:-

${"type":"align*","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-1-0","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)+\\frac{1}{2}\\log_{}\\left|1+\\frac{y^{2}}{x^{2}}\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604226596516,"cs":"RJsOLdBHZBEQIzoguyrEYg==","size":{"width":334,"height":40}}$ ......(3)

Given y = 1 and x = 1

${"type":"align*","code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{1}{1}\\right)+\\frac{1}{2}\\log_{}\\left|1+\\frac{1}{1}\\right|}&={-\\log_{}\\left|1\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-2","ts":1604226913934,"cs":"sw33rNYeGCA0WHBke/BBSw==","size":{"width":329,"height":37}}$

⇒ tan-1(1) +1/2log |2| = log |C|

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-1-1-1-0-1-0","code":"\\begin{align*}\n{\\frac{\\Pi}{4}+\\frac{1}{2}\\log_{}\\left|2\\right|}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","type":"align*","ts":1604227379455,"cs":"FRD8oaLRCF1zACyxg9COaw==","size":{"width":158,"height":32}}$

Put the value of log |C| in eq. 3:-

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-1-1-0-0","code":"\\begin{align*}\n{\\tan^{-1}\\left(\\frac{y}{x}\\right)+\\frac{1}{2}\\log_{}\\left|1+\\frac{y^{2}}{x^{2}}\\right|}&={-\\log_{}\\left|x\\right|+\\frac{\\Pi}{4}+\\frac{1}{2}\\log_{}\\left|2\\right|}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","ts":1604227544498,"cs":"6OQuYGN/GWk+aQsUFHDOQQ==","size":{"width":384,"height":40}}$

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-1-1-1-0-0","code":"\\begin{align*}\n{2\\tan^{-1}\\left(\\frac{y}{x}\\right)+\\log_{}\\left|1+\\frac{y^{2}}{x^{2}}\\right|}&={-2\\log_{}\\left|x\\right|+\\frac{2\\Pi}{4}+\\frac{2}{2}\\log_{}\\left|2\\right|}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","ts":1604227609712,"cs":"9FCZJmMfnctAaSjmp6Gjbw==","size":{"width":396,"height":40}}$

${"code":"\\begin{align*}\n{2\\tan^{-1}\\left(\\frac{y}{x}\\right)+\\log_{}\\left|\\frac{x^{2}+y^{2}}{x^{2}}\\right|}&={-2\\log_{}\\left|x\\right|+\\frac{\\Pi}{2}+\\log_{}\\left|2\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-1-1-1-1-0","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1604227663750,"cs":"frxr1ByHhlLptFKhzIhQFw==","size":{"width":378,"height":40}}$

${"font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{2\\tan^{-1}\\left(\\frac{y}{x}\\right)+\\log_{}\\left|x^{2}+y^{2}\\right|-\\log_{}\\left|x^{2}\\right|}&={-2\\log_{}\\left|x\\right|+\\frac{\\Pi}{2}+\\log_{}\\left|2\\right|}\t\n\\end{align*}","type":"align*","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-0-0","ts":1604227699606,"cs":"Ba0HrDVftvLTRtbXCBchtQ==","size":{"width":440,"height":32}}$

${"type":"align*","font":{"color":"#222222","family":"Arial","size":10},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-1-0-0","code":"\\begin{align*}\n{2\\tan^{-1}\\left(\\frac{y}{x}\\right)+\\log_{}\\left|x^{2}+y^{2}\\right|-2\\log_{}\\left|x\\right|}&={-2\\log_{}\\left|x\\right|+\\frac{\\Pi}{2}+\\log_{}\\left|2\\right|}\t\n\\end{align*}","ts":1604227724098,"cs":"UJCX+8+T4JikGPqSthbngQ==","size":{"width":444,"height":32}}$

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-1-1-1-1-1-1-1","font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{2\\tan^{-1}\\left(\\frac{y}{x}\\right)+\\log_{}\\left|x^{2}+y^{2}\\right|}&={\\frac{\\Pi}{2}+\\log_{}\\left|2\\right|}\t\n\\end{align*}","ts":1604227780615,"cs":"BtpUnGIwWIqACSrZdaQsoA==","size":{"width":288,"height":32}}$

${"id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-0","type":"align","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align}\n{12.\\,x^{2}.dy+\\left(xy+y^{2}\\right).dx}&={0}\t\n\\end{align*}","ts":1604227947700,"cs":"dfbrwewLf+pFBsu4OT3vAQ==","size":{"width":198,"height":20}}$ ; y = 1 when x = 1

Solun:- Given eq. is:-

${"type":"align","id":"13-0-1-0-1-0-1-1-0-1-0-0-1-1","code":"\\begin{align}\n{x^{2}.dy+\\left(xy+y^{2}\\right).dx}&={0}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"ts":1604227930225,"cs":"XStc1zDuKKJEgd7BJ7DMfQ==","size":{"width":176,"height":20}}$

${"id":"13-0-1-0-1-0-1-1-0-1-0-0-1-1","code":"\\begin{align}\n{x^{2}.dy}&={-\\left(xy+y^{2}\\right).dx}\t\n\\end{align}","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","ts":1604227998479,"cs":"qXdjZ698Nd7K/Kn6GLrCrA==","size":{"width":160,"height":20}}$

${"id":"13-0-1-0-1-1-1-1-0-1-1-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{-\\left(xy+y^{2}\\right)}{x^{2}}}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","ts":1604230614624,"cs":"b8tlLtZ5WVmM7YrL8qsXZQ==","size":{"width":129,"height":37}}$ .....(1)

Let F(x,y)= ${"type":"$","id":"9-0-0-1-1-1-1-0-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\frac{-\\left(xy+y^{2}\\right)}{x^{2}}$","ts":1604230634901,"cs":"XvnDvYX7Y/cZFSPqQKsA1w==","size":{"width":56,"height":24}}$

Replace F(x,y) with F(λx,λy):-

${"id":"10-0-0-1-1-1-1-0-1-1-0","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{-\\left(\\lambda x\\lambda y+\\lambda ^{2}y^{2}\\right)}{\\lambda ^{2}x^{2}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{-\\left(\\lambda ^{2}xy+\\lambda ^{2}y^{2}\\right)}{\\lambda ^{2}x^{2}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{-\\lambda ^{2}\\left(xy+y^{2}\\right)}{\\lambda ^{2}x^{2}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{-\\left(xy+y^{2}\\right)}{x^{2}}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","ts":1604230729701,"cs":"acVKIzYdIXp6TK4NM7M9MQ==","size":{"width":212,"height":197}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

From eq. 1 and 2:-

${"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{-\\left(xvx+v^{2}x^{2}\\right)}{x^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{-\\left(vx^{2}+v^{2}x^{2}\\right)}{x^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{-x^{2}\\left(v+v^{2}\\right)}{x^{2}}}\\\\\n{v+x\\diff{v}{x}}&={-v-v^{2}}\t\n\\end{align*}","type":"align*","id":"11-0-0-1-1-1-1-0-1-1-0","ts":1604230807295,"cs":"lWOem6z9gXWU+VpqIGacIA==","size":{"width":190,"height":160}}$

${"id":"12-0-0-1-1-1-1-0-1-1-0","code":"\\begin{align*}\n{x\\diff{v}{x}}&={-v-v^{2}-v}\\\\\n{x\\diff{v}{x}}&={-2v-v^{2}}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"type":"align*","ts":1604230903729,"cs":"zfjrF+PnW03WyLzU1nx1cQ==","size":{"width":134,"height":69}}$

Separating the variables:-

${"code":"\\begin{align*}\n{\\frac{1}{-2v-v^{2}}.dv}&={\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"21-0-0-0-0","type":"align*","ts":1604231035981,"cs":"iMbz29cPsw7+Ffu3IWH8WA==","size":{"width":152,"height":33}}$

Integrating both sides:-

${"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{-2v-v^{2}}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","id":"21-0-0-1-0","type":"align*","ts":1604231087013,"cs":"BheeigK8AKz0fFplHcLhQw==","size":{"width":192,"height":36}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"21-0-0-1-1-0-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{2v+v^{2}}.dv}&={-\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604231111659,"cs":"t95LICPO27Crkzmq4h7bvQ==","size":{"width":193,"height":36}}$

${"id":"21-0-0-1-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{v^{2}+2v+\\left(1\\right)^{2}-\\left(1\\right)^{2}}.dv}&={-\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604231172761,"cs":"hXgGjB/Tm/tYqDzgRFrChA==","size":{"width":285,"height":40}}$

${"id":"21-0-0-1-1-2","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{\\left(v+1\\right)^{2}-\\left(1\\right)^{2}}.dv}&={-\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1604231209024,"cs":"s6mXpZC46XH2W4SPStgg5Q==","size":{"width":244,"height":40}}$

We know that:-

${"type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x^{2}-a^{2}}.dx}&={\\frac{1}{2a}\\log_{}\\left|\\frac{x-a}{x+a}\\right|+C}\t\n\\end{align*}","id":"17-2-0-2-2-0-1-1-0","ts":1604232015470,"cs":"zMoXIqo4j9IP3JKRXjkpGw==","size":{"width":253,"height":36}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"id":"17-2-0-2-2-0-2-1-0","ts":1604229039148,"cs":"BqLSY6+wwCE4GkUrfnZ9NA==","size":{"width":153,"height":36}}$

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-0-0","code":"\\begin{align*}\n{\\frac{1}{2}\\log_{}\\left|\\frac{\\left(v+1\\right)-1}{\\left(v+1\\right)+1}\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","ts":1604232148310,"cs":"F87IGoYJb+6UW0Oyv2+mEQ==","size":{"width":272,"height":37}}$

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-0","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\frac{1}{2}\\log_{}\\left|\\frac{v}{v+2}\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604234294389,"cs":"luJ4NSYGqasjDJ2gixHQYw==","size":{"width":232,"height":32}}$

Put the value of v:-

${"font":{"color":"#222222","size":10,"family":"Arial"},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-0-0","type":"align*","code":"\\begin{align*}\n{\\frac{1}{2}\\log_{}\\left|\\frac{\\frac{y}{x}}{\\frac{y}{x}+2}\\right|}&={-\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604234316632,"cs":"yms1dqOjIl2LELGYHcXNIw==","size":{"width":237,"height":42}}$

${"font":{"size":10,"family":"Arial","color":"#222222"},"type":"align*","code":"\\begin{align*}\n{\\frac{1}{2}\\log_{}\\left|\\frac{\\frac{y}{x}}{\\frac{y}{x}+2}\\right|}&={\\log_{}\\left|\\frac{C}{x}\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-1-0","ts":1604235285485,"cs":"MDgw1IsXLE52suqG67Pqew==","size":{"width":170,"height":42}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-1-1-0","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{\\frac{y}{x}}{\\frac{y}{x}+2}\\right|}&={2\\log_{}\\left|\\frac{C}{x}\\right|}\t\n\\end{align*}","ts":1604235317910,"cs":"W7Yrh/iel9TxZF3sKKQcrQ==","size":{"width":166,"height":42}}$

${"type":"align*","font":{"family":"Arial","color":"#222222","size":10},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-1-1-1-0","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{\\frac{y}{x}}{\\frac{y}{x}+2}\\right|}&={\\log_{}\\left|\\left(\\frac{C}{x}\\right)^{2}\\right|}\t\n\\end{align*}","ts":1604235349203,"cs":"qKD6TWuxK87fHGuOJ17Mog==","size":{"width":185,"height":44}}$

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","code":"\\begin{align*}\n{\\frac{\\frac{y}{x}}{\\frac{y}{x}+2}}&={\\left(\\frac{C}{x}\\right)^{2}}\t\n\\end{align*}","ts":1604235380005,"cs":"Sfxak/BWzGVGRFRrRjYhqQ==","size":{"width":116,"height":44}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{\\frac{y}{y+2x}}&={\\frac{C^{2}}{x^{2}}}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-1-1-1-1-1-0","ts":1604235733466,"cs":"uehl6i234mDj9+EhESRqPQ==","size":{"width":97,"height":37}}$ ......(3)

Given y = 1 and x = 1

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-1-1-1-2","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{\\frac{\\frac{1}{1}}{\\frac{1}{1}+2}}&={\\left(\\frac{C}{1}\\right)^{2}}\t\n\\end{align*}","type":"align*","ts":1604235515241,"cs":"PGL1p2OskEGjPlW0G1ZWbg==","size":{"width":116,"height":44}}$

${"font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\frac{1}{3}}&={C^{2}}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-1-1","type":"align*","ts":1604235550364,"cs":"/kSd/93TpHDsybqT4pKtQg==","size":{"width":54,"height":32}}$

${"type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{C^{2}}&={\\frac{1}{3}}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-1-1-1-0-1-1","ts":1604235652141,"cs":"v7H1sbG1yX8tq8O/ztz6qQ==","size":{"width":54,"height":32}}$

Put the value of log |C| in eq. 3:-

${"code":"\\begin{align*}\n{\\frac{y}{y+2x}}&={\\frac{\\frac{1}{3}}{x^{2}}}\t\n\\end{align*}","type":"align*","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-1-1-1-1-1-1-0-0","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1604235810596,"cs":"w5SyHge0SkxjVaxxtfw8MQ==","size":{"width":94,"height":40}}$

${"font":{"family":"Arial","color":"#222222","size":10},"type":"align*","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-1-1-1-1-1-1-1","code":"\\begin{align*}\n{\\frac{y}{y+2x}}&={\\frac{1}{3x^{2}}}\t\n\\end{align*}","ts":1604235837754,"cs":"zZyQyyj/gxVHA8V4sGKxlw==","size":{"width":102,"height":34}}$

${"code":"\\begin{align*}\n{\\frac{x^{2}y}{y+2x}}&={\\frac{1}{3}}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-1-1-0-0-1-1-1-1-1-1-2","type":"align*","ts":1604235862844,"cs":"+zRcHy9lboqG4PvywuJChw==","size":{"width":86,"height":37}}$

⇒ 3x2y = y + 2x

${"type":"align","code":"\\begin{align}\n{13.\\,\\left[x\\sin^{2}\\left(\\frac{y}{x}\\right)-y\\right].dx+x.dy}&={0}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-0","ts":1604236098764,"cs":"iGVWihk5FL7Us8Cqpn1jyw==","size":{"width":244,"height":28}}$ ; y = π/4 when x = 1

Solun:- Given eq. is:-

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-1","type":"align","code":"\\begin{align}\n{\\left[x\\sin^{2}\\left(\\frac{y}{x}\\right)-y\\right].dx+x.dy}&={0}\t\n\\end{align*}","ts":1604236159289,"cs":"2pkSbZUfYGoV3WOaFUtzcg==","size":{"width":221,"height":28}}$

${"code":"\\begin{align}\n{\\left[x\\sin^{2}\\left(\\frac{y}{x}\\right)-y\\right].dx}&={-x.dy}\t\n\\end{align}","font":{"color":"#222222","family":"Arial","size":10},"id":"25-0","type":"align*","ts":1604236334461,"cs":"PeJdPQ+M8JdlCvTTWCXZQQ==","size":{"width":206,"height":28}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"id":"13-0-1-0-1-1-1-1-0-1-1-1-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{-\\left[x\\sin^{2}\\left(\\frac{y}{x}\\right)-y\\right]}{x}}\\\\\n{\\diff{y}{x}}&={\\frac{y-x\\sin^{2}\\left(\\frac{y}{x}\\right)}{x}}\t\n\\end{align*}","ts":1604236492091,"cs":"N0KG01hFZ/hziPunyucUgg==","size":{"width":173,"height":80}}$ .....(1)

Let F(x,y)= ${"id":"9-0-0-1-1-1-1-0-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\frac{y-x\\sin^{2}\\left(\\frac{y}{x}\\right)}{x}$","ts":1604236542807,"cs":"85Uuvzvdypq3IHRAftN0mQ==","size":{"width":66,"height":24}}$

Replace F(x,y) with F(λx,λy):-

${"code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y-\\lambda x\\sin^{2}\\left(\\frac{\\lambda y}{\\lambda x}\\right)}{\\lambda x}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda \\left[y-x\\sin^{2}\\left(\\frac{y}{x}\\right)\\right]}{\\lambda x}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{y-x\\sin^{2}\\left(\\frac{y}{x}\\right)}{x}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","id":"10-0-0-1-1-1-1-0-1-1-1-0","type":"align*","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1604236635019,"cs":"aYkmSi0eovGQXLstmyVLfg==","size":{"width":229,"height":164}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

From eq. 1 and 2:-

${"code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{vx-x\\sin^{2}\\left(\\frac{vx}{x}\\right)}{x}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{vx-x\\sin^{2}\\left(v\\right)}{x}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x\\left[v-\\sin^{2}\\left(v\\right)\\right]}{x}}\\\\\n{v+x\\diff{v}{x}}&={v-\\sin^{2}\\left(v\\right)}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","id":"11-0-0-1-1-1-1-0-1-1-1-0","ts":1604236714682,"cs":"vdb4JY58AMrp/tnsbCFM/Q==","size":{"width":198,"height":157}}$

${"code":"\\begin{align*}\n{x\\diff{v}{x}}&={v-\\sin^{2}\\left(v\\right)-v}\\\\\n{x\\diff{v}{x}}&={-\\sin^{2}\\left(v\\right)}\t\n\\end{align*}","id":"12-0-0-1-1-1-1-0-1-1-1-0","font":{"size":10,"family":"Arial","color":"#222222"},"type":"align*","ts":1604236746107,"cs":"f4urhl0QZ0tMmzJxnl6ytg==","size":{"width":157,"height":69}}$

Separating the variables:-

${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\frac{-1}{\\sin^{2}\\left(v\\right)}.dv}&={\\frac{1}{x}.dx}\t\n\\end{align*}","id":"21-0-0-0-1-0","ts":1604236809178,"cs":"ae4RbGI+yaCAICJLZ9JrPw==","size":{"width":140,"height":37}}$

Integrating both sides:-

${"type":"align*","id":"21-0-0-0-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{-\\int_{}^{}\\frac{1}{\\sin^{2}\\left(v\\right)}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604236843655,"cs":"CwPfbTIMQ5qOsbLMks50Gw==","size":{"width":193,"height":38}}$

${"id":"21-0-0-1-1-0-1-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{-\\int_{}^{}\\cos ec^{2}v.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604236887019,"cs":"SHP67Vz2woTJhCqtfLB/cw==","size":{"width":190,"height":36}}$

We know that:-

${"type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"id":"17-2-0-2-2-0-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\cos ec^{2}x.dx}&={-\\cot x+C}\t\n\\end{align*}","ts":1604236998037,"cs":"NxljTImnuKVPrZsmG0rGhA==","size":{"width":197,"height":36}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"id":"17-2-0-2-2-0-2-1-1-0","ts":1604229039148,"cs":"n8h2o6cE2mZdinHJ7d7Vpw==","size":{"width":153,"height":36}}$

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-0-1-0-0","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\cot v}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604237028911,"cs":"mRue4748quvDp2fQzqgNXA==","size":{"width":157,"height":16}}$

Put the value of v:-

${"type":"align*","code":"\\begin{align*}\n{\\cot \\left(\\frac{y}{x}\\right)}&={\\log_{}\\left|x\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-0-1-1-1-0-0","font":{"color":"#222222","family":"Arial","size":10},"ts":1604237070618,"cs":"kXFwWPvuLh58P8r0usRNfg==","size":{"width":184,"height":28}}$ ......(3)

Given y = π/4 and x = 1

${"id":"26-0","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","code":"\\begin{align*}\n{\\cot \\left(\\frac{\\Pi}{4}\\right)}&={\\log_{}\\left|1\\right|+\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604237197286,"cs":"J2IvxmVal7SrhAIguVHUBg==","size":{"width":190,"height":37}}$

${"type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"id":"26-1-0","code":"\\begin{align*}\n{1}&={\\log_{}\\left|C\\right|}\t\n\\end{align*}","ts":1604237229368,"cs":"NSvqvAcE4zOQyMEzKzXtQQ==","size":{"width":73,"height":16}}$

⇒ C = e

Put the value of C in eq. 3:-

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-0-1-1-1-1","type":"align*","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\cot \\left(\\frac{y}{x}\\right)}&={\\log_{}\\left|x\\right|+\\log_{}\\left|e\\right|}\\\\\n{\\cot \\left(\\frac{y}{x}\\right)}&={\\log_{}\\left|x.e\\right|}\t\n\\end{align*}","ts":1604237463884,"cs":"RfEws+g2M/WRUMZcC2jthw==","size":{"width":180,"height":62}}$

${"code":"\\begin{align}\n{14.\\,\\diff{y}{x}-\\frac{y}{x}+\\cos ec\\left(\\frac{y}{x}\\right)}&={0}\t\n\\end{align}","font":{"color":"#222222","family":"Arial","size":10},"id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-1-0","type":"align*","ts":1604304994835,"cs":"ETWuECLntOdxTC4iOUH+BA==","size":{"width":204,"height":32}}$ ; y = 0 when x = 1

Solun:- Given eq. is:-

${"type":"align","font":{"family":"Arial","color":"#222222","size":10},"id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-2","code":"\\begin{align}\n{\\diff{y}{x}-\\frac{y}{x}+\\cos ec\\left(\\frac{y}{x}\\right)}&={0}\t\n\\end{align*}","ts":1604305096781,"cs":"u2NBQFf7dLuZhSbazHYFOw==","size":{"width":181,"height":32}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-3","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{y}{x}-\\cos ec\\left(\\frac{y}{x}\\right)}\t\n\\end{align*}","ts":1604305164224,"cs":"vT0Vomkuma/2fzVH0+FB4w==","size":{"width":153,"height":32}}$ .....(1)

Let F(x,y)= ${"type":"$","code":"$\\frac{y}{x}-\\cos ec\\left(\\frac{y}{x}\\right)$","id":"9-0-0-1-1-1-1-0-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1604305245406,"cs":"aoI3441857RqeOSloBK9GQ==","size":{"width":98,"height":18}}$

Replace F(x,y) with F(λx,λy):-

${"font":{"color":"#222222","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda y}{\\lambda x}-\\cos ec\\left(\\frac{\\lambda y}{\\lambda x}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{y}{x}-\\cos ec\\left(\\frac{y}{x}\\right)\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","id":"10-0-0-1-1-1-1-0-1-1-1-1-0","ts":1604305327635,"cs":"aVwOHxmkb6KCmYzQu3xv4g==","size":{"width":230,"height":94}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

From eq. 1 and 2:-

${"code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{vx}{x}-\\cos ec\\left(\\frac{vx}{x}\\right)}\\\\\n{v+x\\diff{v}{x}}&={v-\\cos ec\\left(v\\right)}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"id":"11-0-0-1-1-1-1-0-1-1-1-1-0","ts":1604305390514,"cs":"fK4u6jy2Yretjb9rmAXG4A==","size":{"width":205,"height":69}}$

${"code":"\\begin{align*}\n{x\\diff{v}{x}}&={v-\\cos ec\\left(v\\right)-v}\\\\\n{x\\diff{v}{x}}&={-\\cos ec\\left(v\\right)}\t\n\\end{align*}","id":"12-0-0-1-1-1-1-0-1-1-1-1","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","ts":1604305486573,"cs":"X0OSUoK4VrZk+MyY7Z/t+Q==","size":{"width":166,"height":69}}$

Separating the variables:-

${"font":{"color":"#000000","size":10,"family":"Arial"},"id":"21-0-0-0-1-2-0-0","type":"align*","code":"\\begin{align*}\n{\\frac{1}{\\cos ec\\left(v\\right)}.dv}&={\\frac{-1}{x}.dx}\t\n\\end{align*}","ts":1604305520929,"cs":"xqOxER0+aqO/zzd/Yg+a8w==","size":{"width":160,"height":36}}$

Integrating both sides:-

${"id":"21-0-0-0-1-2-1-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{\\cos ec\\left(v\\right)}.dv}&={\\int_{}^{}\\frac{-1}{x}.dx}\t\n\\end{align*}","ts":1604305563239,"cs":"v5qBrlrF3tmYIqYPy9331Q==","size":{"width":200,"height":36}}$

${"id":"21-0-0-1-1-0-1-1","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{-\\int_{}^{}\\sin v.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","type":"align*","ts":1604305719315,"cs":"1XhY/MNin4onLRRFz7AfCg==","size":{"width":168,"height":36}}$

We know that:-

${"id":"17-2-0-2-2-0-1-1-1-1","code":"\\begin{align*}\n{\\int_{}^{}\\sin x.dx}&={-\\cos x+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","ts":1604305680885,"cs":"B1Ty4VZHj5u+r6g+yHkW7w==","size":{"width":174,"height":36}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"id":"17-2-0-2-2-0-2-1-1-1","ts":1604229039148,"cs":"qQDfMkMdLh9LBKN8YbSvfg==","size":{"width":153,"height":36}}$

${"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-0-1-0-1","type":"align*","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{-\\left(-\\cos v\\right)}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","ts":1604305772086,"cs":"UPqUWcaQh1cGE5Z3l9u/JA==","size":{"width":164,"height":16}}$

⇒ cos v = log |x| + C

Put the value of v:-

${"font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{\\cos\\left(\\frac{y}{x}\\right)}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-0-1-1-1-0-1-0","type":"align*","ts":1604305843889,"cs":"TNw9P+aFugN5V6h+RUqIKw==","size":{"width":152,"height":28}}$ ......(3)

Given y = 0 and x = 1

⇒ cos (0) = log |1| + C

⇒ 1 = C

⇒ C = 1

Put the value of C in eq. 3:-

${"code":"\\begin{align*}\n{\\cos\\left(\\frac{y}{x}\\right)}&={\\log_{}\\left|x\\right|+1}\\\\\n{\\cos\\left(\\frac{y}{x}\\right)}&={\\log_{}\\left|x\\right|+\\log_{}\\left|e\\right|}\\\\\n{\\cos\\left(\\frac{y}{x}\\right)}&={\\log_{}\\left|x.e\\right|}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"id":"16-1-0-0-0-1-0-1-0-0-0-1-0-1-0-0-1-0-1-0-0-1-1-1-0-1-0","type":"align*","ts":1604306004094,"cs":"MhYZFy+GFJkF7zmCNX+nUA==","size":{"width":180,"height":96}}$

${"type":"align","id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-1-1-0","code":"\\begin{align}\n{15.\\,2xy+y^{2}-2x^{2}\\diff{y}{x}}&={0}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1604306168929,"cs":"pDp8KXPfyh1vTVk/Mcgomg==","size":{"width":177,"height":32}}$ ; y = 2 when x = 1

Solun:- Given eq. is:-

${"type":"align","id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-1-1-1","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align}\n{2xy+y^{2}-2x^{2}\\diff{y}{x}}&={0}\t\n\\end{align*}","ts":1604306198265,"cs":"r+D/lzO8UhwkL9sMVJxNkw==","size":{"width":154,"height":32}}$

${"type":"align","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align}\n{-2x^{2}\\diff{y}{x}}&={-\\left(2xy+y^{2}\\right)}\t\n\\end{align*}","id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-1-1-2-0","ts":1604306231090,"cs":"og8Glnv2Q8LV2S4cHTu6aw==","size":{"width":166,"height":32}}$

${"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{2xy+y^{2}}{2x^{2}}}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-1-1-2-1","ts":1604306355339,"cs":"JK75AXDAPVRKPhdhCYIuAQ==","size":{"width":110,"height":34}}$ .....(1)

Let F(x,y)= ${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"9-0-0-1-1-1-1-0-1-1-1-1-1","code":"$\\frac{2xy+y^{2}}{2x^{2}}$","type":"$","ts":1604306408569,"cs":"Ho9GAjJlpY65AMbI+hvhGA==","size":{"width":42,"height":22}}$

Replace F(x,y) with F(λx,λy):-

${"font":{"size":10,"color":"#222222","family":"Arial"},"code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{2\\lambda x\\lambda y+\\lambda ^{2}y^{2}}{2\\lambda ^{2}x^{2}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{2\\lambda ^{2}xy+\\lambda ^{2}y^{2}}{2\\lambda ^{2}x^{2}}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda ^{2}\\left(2xy+y^{2}\\right)}{2\\lambda ^{2}x^{2}}}\t\n\\end{align*}","id":"10-0-0-1-1-1-1-0-1-1-1-1-1","type":"align*","ts":1604306567234,"cs":"rdfayo7oWqQcn02+KLMWGQ==","size":{"width":186,"height":117}}$

${"type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"id":"27","code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{2xy+y^{2}}{2x^{2}}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","ts":1604306591237,"cs":"OQ6q1btzfh67AdumcOF43g==","size":{"width":192,"height":61}}$

Then F(x,y) is a homogeneous function of zero degrees so this equation is called a

homogeneous equation.

Substitute y = vx

Differentiate with respect to x:-

From eq. 1 and 2:-

${"id":"11-0-0-1-1-1-1-0-1-1-1-1-1","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{2xvx+v^{2}x^{2}}{2x^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{2vx^{2}+v^{2}x^{2}}{2x^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x^{2}\\left(2v+v^{2}\\right)}{2x^{2}}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{2v+v^{2}}{2}}\t\n\\end{align*}","ts":1604309517129,"cs":"aftdJ86/q2ExJGLEsSweYw==","size":{"width":172,"height":157}}$

${"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{x\\diff{v}{x}}&={\\frac{2v+v^{2}}{2}-v}\\\\\n{x\\diff{v}{x}}&={\\frac{2v+v^{2}-2v}{2}}\\\\\n{x\\diff{v}{x}}&={\\frac{v^{2}}{2}}\t\n\\end{align*}","id":"12-0-0-1-1-1-1-0-1-1-1-2","type":"align*","ts":1604309575704,"cs":"GsxhhAnkQglFS+MBhD2Mtg==","size":{"width":145,"height":113}}$

Separating the variables:-

${"code":"\\begin{align*}\n{\\frac{2}{v^{2}}.dv}&={\\frac{1}{x}.dx}\t\n\\end{align*}","id":"21-0-0-0-1-2-0-1-0-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1604309608693,"cs":"xLbM3ArJnHdRzC5iZDgQ4w==","size":{"width":104,"height":32}}$

Integrating both sides:-

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"21-0-0-0-1-2-0-1-0-1","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{2}{v^{2}}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1604309639084,"cs":"MhriPRkwvwKG5CLt/P419g==","size":{"width":144,"height":36}}$

We know that:-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"id":"17-2-0-2-2-0-2-1-1-2","ts":1604229039148,"cs":"ju0gpOYMfLDhVA+eeFC/WA==","size":{"width":153,"height":36}}$

${"font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}x^{n}.dx}&={\\frac{x^{n+1}}{n+1}+C}\t\n\\end{align*}","id":"30","type":"align*","ts":1604309708013,"cs":"x0ElO/y+ts9EsQSNkDKXKA==","size":{"width":156,"height":37}}$

${"id":"31-0-0","type":"align*","code":"\\begin{align*}\n{\\frac{-2}{v}}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1604309757207,"cs":"p80UowpuVWFQwocdlALYJg==","size":{"width":120,"height":32}}$

Put the value of v:-

${"code":"\\begin{align*}\n{\\frac{-2x}{y}}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"31-0-1-0","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","ts":1604309787427,"cs":"Lh7S8jwXL9rSw44eYw91XQ==","size":{"width":129,"height":34}}$ ......(3)

Given y = 2 and x = 1

${"code":"\\begin{align*}\n{\\frac{-2}{2}}&={\\log_{}\\left|1\\right|+C}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"id":"33","type":"align*","ts":1604309842579,"cs":"LLNZHfPcQp/0p69GqXeXBQ==","size":{"width":120,"height":32}}$

⇒ C = -1

Put the value of log |C| in eq. 3:-

${"type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"id":"31-0-1-1","code":"\\begin{align*}\n{\\frac{-2x}{y}}&={\\log_{}\\left|x\\right|-1}\t\n\\end{align*}","ts":1604309887091,"cs":"yYOr+x2AfQFGoN6+8RA2AQ==","size":{"width":125,"height":34}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{\\frac{2x}{y}}&={1-\\log_{}\\left|x\\right|}\t\n\\end{align*}","id":"31-0-1-2","ts":1604309962755,"cs":"kAORmTH82uhQ87GuYje7mw==","size":{"width":113,"height":34}}$

${"code":"\\begin{align*}\n{y}&={\\frac{2x}{1-\\log_{}\\left|x\\right|}}\t\n\\end{align*}","id":"34","font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","ts":1604309998480,"cs":"PGuxgyz8ZAf0BmQRE0cZZA==","size":{"width":104,"height":36}}$

16. A homogeneous differential equation of the form ${"id":"35","type":"align","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align}\n{\\diff{x}{y}}&={h\\left(\\frac{x}{y}\\right)}\t\n\\end{align*}","ts":1604310447259,"cs":"H9pB3wx7PniiWxNdWjGz7Q==","size":{"width":93,"height":37}}$ can be solved by

making the substitution.

(A) y = vx (B) v = yx (C) x = vy (D) x = v

Solun:- Given eq. is:-

${"id":"35","type":"align","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align}\n{\\diff{x}{y}}&={h\\left(\\frac{x}{y}\\right)}\t\n\\end{align*}","ts":1604310447259,"cs":"H9pB3wx7PniiWxNdWjGz7Q==","size":{"width":93,"height":37}}$

This type of homogeneous equation is solved by the substitution of x = vy.

The correct answer is C.

17. Which of the following is a homogeneous differential equation?

(A) (4x+6y+5).dy - (3y+2x+4).dx = 0

(B) (xy).dx - (x3+ y3).dy = 0

(C) (x3+2y2).dx + 2xy.dy = 0

(D) y2.dx + (x2-xy-y2).dy = 0

Solun:- The correct answer is D because power of all the terms all are same.

Download PDF of Exercise 9.5

NCERT Maths solutions for 12th

Important Note

Exercise 9.5

In each of the Exercises 1 to 10, show that the given differential equation is

homogeneous and solve each of them.

1. (x2 + xy).dy = (x2 + y2).dx

${"type":"align","id":"2-0","code":"\\begin{align}\n{2.\\,y^{\\prime }}&={\\frac{x+y}{x}}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603968267128,"cs":"agCEmHaGfeZlpuFSPjh3pQ==","size":{"width":90,"height":30}}$

${"id":"2-1-0","font":{"family":"Arial","color":"#222222","size":10},"type":"align","code":"\\begin{align}\n{y^{\\prime }}&={\\frac{x+y}{x}}\t\n\\end{align*}","ts":1603968296027,"cs":"5a8C0Msmd5qrPEVkkmzHqw==","size":{"width":76,"height":30}}$

3. (x - y).dy - (x + y).dx = 0

4. (x2 - y2).dx + 2xy.dy = 0

${"font":{"family":"Arial","size":10,"color":"#222222"},"id":"5","type":"align","code":"\\begin{align}\n{5.\\,x^{2}\\diff{y}{x}}&={x^{2}-2y^{2}+xy}\t\n\\end{align*}","ts":1604057317461,"cs":"uzydyF+J4cyiuwcQeIm9wg==","size":{"width":168,"height":32}}$

${"type":"align","font":{"size":10,"family":"Arial","color":"#222222"},"id":"13-0-0","code":"\\begin{align}\n{6.\\,x.dy-y.dx}&={{\\sqrt[]{x^{2}+y^{2}}}.dx}\t\n\\end{align*}","ts":1604063671898,"cs":"/eWNBPlUYzL91o4h3Q79Uw==","size":{"width":208,"height":20}}$

${"code":"\\begin{align}\n{x.dy-y.dx}&={{\\sqrt[]{x^{2}+y^{2}}}.dx}\t\n\\end{align}","id":"13-1-0","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1604063704274,"cs":"T19h9wpOD7Uc4IMPT9gvYQ==","size":{"width":193,"height":20}}$

${"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align}\n{8.\\,x\\diff{y}{x}-y+x\\sin\\left(\\frac{y}{x}\\right)}&={0}\t\n\\end{align}","id":"13-0-1-0-1-0-0","type":"align*","ts":1604142362992,"cs":"dQH32ITvZnD3wHQ4ApP1gA==","size":{"width":192,"height":32}}$

${"type":"align","font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align}\n{x\\diff{y}{x}-y+x\\sin\\left(\\frac{y}{x}\\right)}&={0}\t\n\\end{align*}","id":"13-0-1-0-1-1-0-0","ts":1604142473778,"cs":"wIsTX8oBcio6+OpO7PMGXw==","size":{"width":177,"height":32}}$

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"13-0-1-0-1-0-1-0","code":"\\begin{align}\n{9.\\,y.dx+x\\log_{}\\left(\\frac{y}{x}\\right).dy-2x.dy}&={0}\t\n\\end{align}","type":"align*","ts":1604220447397,"cs":"pcDRzrTMWSaWL2I6XE2gzA==","size":{"width":248,"height":28}}$

${"type":"align","id":"13-0-1-0-1-0-2-0","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align}\n{y.dx+x\\log_{}\\left(\\frac{y}{x}\\right).dy-2x.dy}&={0}\t\n\\end{align*}","ts":1604220473034,"cs":"iphNp6xh8MiigRl3Oh325Q==","size":{"width":232,"height":28}}$

${"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align","code":"\\begin{align}\n{y.dx+\\left[x\\log_{}\\left(\\frac{y}{x}\\right)-2x\\right].dy}&={0}\t\n\\end{align*}","id":"13-0-1-0-1-0-2-1-0","ts":1604220515223,"cs":"TCcGQNc/2tFcdWhGOo3xHw==","size":{"width":224,"height":28}}$

For each of the Differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:

${"code":"\\begin{align}\n{\\left(x+y\\right).dy+\\left(x-y\\right).dx}&={0}\t\n\\end{align}","font":{"size":10,"color":"#222222","family":"Arial"},"id":"13-0-1-0-1-0-1-1-0-1-0-1","type":"align*","ts":1604225413193,"cs":"7acpcW2DDA5jI4CYifTMfg==","size":{"width":192,"height":16}}$

${"type":"align","id":"13-0-1-0-1-0-1-1-0-1-0-0-1-1","code":"\\begin{align}\n{x^{2}.dy+\\left(xy+y^{2}\\right).dx}&={0}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"ts":1604227930225,"cs":"XStc1zDuKKJEgd7BJ7DMfQ==","size":{"width":176,"height":20}}$

${"id":"13-0-1-0-1-0-1-1-0-1-0-0-1-1","code":"\\begin{align}\n{x^{2}.dy}&={-\\left(xy+y^{2}\\right).dx}\t\n\\end{align}","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","ts":1604227998479,"cs":"qXdjZ698Nd7K/Kn6GLrCrA==","size":{"width":160,"height":20}}$

${"code":"\\begin{align}\n{\\left[x\\sin^{2}\\left(\\frac{y}{x}\\right)-y\\right].dx}&={-x.dy}\t\n\\end{align}","font":{"color":"#222222","family":"Arial","size":10},"id":"25-0","type":"align*","ts":1604236334461,"cs":"PeJdPQ+M8JdlCvTTWCXZQQ==","size":{"width":206,"height":28}}$

${"type":"align","id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-1-1-1","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align}\n{2xy+y^{2}-2x^{2}\\diff{y}{x}}&={0}\t\n\\end{align*}","ts":1604306198265,"cs":"r+D/lzO8UhwkL9sMVJxNkw==","size":{"width":154,"height":32}}$

${"type":"align","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align}\n{-2x^{2}\\diff{y}{x}}&={-\\left(2xy+y^{2}\\right)}\t\n\\end{align*}","id":"13-0-1-0-1-0-1-1-0-1-0-0-1-0-1-0-1-1-2-0","ts":1604306231090,"cs":"og8Glnv2Q8LV2S4cHTu6aw==","size":{"width":166,"height":32}}$

making the substitution.

(A) y = vx (B) v = yx (C) x = vy (D) x = v

${"id":"35","type":"align","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align}\n{\\diff{x}{y}}&={h\\left(\\frac{x}{y}\\right)}\t\n\\end{align*}","ts":1604310447259,"cs":"H9pB3wx7PniiWxNdWjGz7Q==","size":{"width":93,"height":37}}$

17. Which of the following is a homogeneous differential equation?

(A) (4x+6y+5).dy - (3y+2x+4).dx = 0

(B) (xy).dx - (x3+ y3).dy = 0

(C) (x3+2y2).dx + 2xy.dy = 0

(D) y2.dx + (x2-xy-y2).dy = 0

Post a Comment

Post a Comment

About Us

Contact Form

Important Note

Differential Equations(Ex. 9.5)

Exercise 9.5

In each of the Exercises 1 to 10, show that the given differential equation is

homogeneous and solve each of them.

1. (x2 + xy).dy = (x2 + y2).dx

3. (x - y).dy - (x + y).dx = 0

4. (x2 - y2).dx + 2xy.dy = 0

For each of the Differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:

; y = 1 when x = 1

; y = 1 when x = 1

; y = π/4 when x = 1

; y = 0 when x = 1

; y = 2 when x = 1

16. A homogeneous differential equation of the form can be solved by

making the substitution.

(A) y = vx (B) v = yx (C) x = vy (D) x = v

17. Which of the following is a homogeneous differential equation?

(A) (4x+6y+5).dy - (3y+2x+4).dx = 0

(B) (xy).dx - (x3+ y3).dy = 0

(C) (x3+2y2).dx + 2xy.dy = 0

(D) y2.dx + (x2-xy-y2).dy = 0

You might like

Post a Comment

Post a Comment

Contact Form