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Exercise 9.4

For each of the differential equations in Exercises 1 to 10 find the general equation:

{"code":"\\begin{align*}\n{1.\\,\\diff{y}{x}}&={\\frac{1-\\cos x}{1+\\cos x}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"1-0-0","type":"align*","ts":1603624711571,"cs":"4c/jneKjfwtHatm6OSwFDg==","size":{"width":128,"height":33}}

Solun:- Given equation is:-

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{1-\\cos x}{1+\\cos x}.....\\left(1\\right)}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"1-1-0","type":"align*","ts":1603624995644,"cs":"DN8lGsNXpieZa/7C5xgqkg==","size":{"width":168,"height":33}}

Separate variables of eq. 1:-

{"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"2-0-0","code":"\\begin{align*}\n{dy}&={\\frac{1-\\cos x}{1+\\cos x}.dx}\t\n\\end{align*}","ts":1603625096667,"cs":"dadyBI+2m7JAs7aeENRqCw==","size":{"width":128,"height":32}}

Integrating both sides:-

{"id":"2-1-0","code":"\\begin{align*}\n{\\int_{}^{}dy}&={\\int_{}^{}\\left(\\frac{1-\\cos x}{1+\\cos x}\\right).dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1603625141992,"cs":"QNSj8DZ7nxqOGVHJ38Nj7Q==","size":{"width":188,"height":37}}

We know that:-

cos 2x = 2cos2x - 1

cos 2x = 1 - 2sin2x

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}dy}&={\\int_{}^{}\\left(\\frac{2\\sin^{2}\\frac{x}{2}}{2\\cos^{2}\\frac{x}{2}}\\right).dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\tan^{2}\\frac{x}{2}.dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\left(\\sec^{2}\\frac{x}{2}-1\\right).dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\sec^{2}\\frac{x}{2}.dx-\\int_{}^{}1.dx}\t\n\\end{align*}","id":"2-1-1","font":{"family":"Arial","color":"#000000","size":10},"ts":1603625412318,"cs":"Rf1B3xveMT06OIqlFatmRg==","size":{"width":213,"height":168}}

We know that:-

{"id":"3-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\sec^{2}x.dx}&={\\tan x+C}\t\n\\end{align*}","type":"align*","ts":1603625444918,"cs":"0yLhuxlnKqyUhKQd+4gKCA==","size":{"width":168,"height":36}}

{"id":"4-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{y}&={\\frac{\\tan\\frac{x}{2}}{\\frac{1}{2}}-x+C}\\\\\n{y}&={2\\tan\\frac{x}{2}-x+C}\t\n\\end{align*}","type":"align*","ts":1603625570285,"cs":"A+2jl3ayLW7HzM9n/gbfKQ==","size":{"width":140,"height":76}}

{"code":"\\begin{align*}\n{2.\\,\\diff{y}{x}}&={{\\sqrt[]{4-y^{2}}}}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"1-0-1-0-0","type":"align*","ts":1603625662916,"cs":"WlDcITKN1L0fEuWDcE4vKA==","size":{"width":117,"height":32}}, (-2 < y < 2)

Solun:- Given equation is:-

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={{\\sqrt[]{4-y^{2}}}}\t\n\\end{align*}","id":"1-0-1-1","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1603625716064,"cs":"XOwjwU6c4lhrlLsc3B+xJg==","size":{"width":104,"height":32}}........(1)

Separate variables of eq. 1:-

{"id":"2-0-1-0-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\frac{1}{{\\sqrt[]{4-y^{2}}}}.dy}&={dx}\t\n\\end{align*}","ts":1603625785910,"cs":"+Fp1TwXvjHykQ0/SeVWFPw==","size":{"width":125,"height":40}}

Integrating both sides:-

{"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{{\\sqrt[]{4-y^{2}}}}.dy}&={\\int_{}^{}1.dx}\t\n\\end{align*}","type":"align*","id":"2-0-1-1","ts":1603625814949,"cs":"l9G78HZtYzhCiRgqmLkhOA==","size":{"width":176,"height":40}}

We know that:-

{"type":"align*","id":"3-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{{\\sqrt[]{a^{2}-x^{2}}}}.dx}&={\\sin^{-1}\\left(\\frac{x}{a}\\right)+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603625985334,"cs":"S4I86N0flHTLcAFj4rdlig==","size":{"width":238,"height":37}}

{"id":"4-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\sin^{-1}\\left(\\frac{y}{2}\\right)}&={x+C}\t\n\\end{align*}","ts":1604319694393,"cs":"OT734q2Wd9OxUNTLMSEPRA==","size":{"width":132,"height":28}}

⇒ y/2 = sin (x + C)

⇒ y = 2sin (x + C)

{"code":"\\begin{align*}\n{3.\\,\\diff{y}{x}+y}&={1}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"1-0-1-0-1-0-0-0","ts":1603626316740,"cs":"25o0nrzqc8cqz0O0ZicEoA==","size":{"width":96,"height":32}}, (y ≠ 1)

Solun:- Given equation is:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\diff{y}{x}+y}&={1}\t\n\\end{align*}","id":"1-0-1-0-1-1-0-0","type":"align*","ts":1603626528653,"cs":"jRPzTbc/7MbjlFMpo17cgQ==","size":{"width":80,"height":32}}

{"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\diff{y}{x}}&={1-y}\t\n\\end{align*}","type":"align*","id":"1-0-1-0-1-1-1","ts":1603626556974,"cs":"7/6qrVkQp+LuMykaCpj/Dg==","size":{"width":80,"height":32}}........(1)

Separate variables of eq. 1:-

{"id":"2-0-1-0-1-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\frac{1}{1-y}.dy}&={dx}\t\n\\end{align*}","ts":1603626584797,"cs":"6truqx4E2PaAu2QdKLcYSw==","size":{"width":102,"height":34}}

Integrating both sides:-

{"id":"2-0-1-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{1-y}.dy}&={\\int_{}^{}dx}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1603626604960,"cs":"SxtVV3znZakx+oYQ6hyoZA==","size":{"width":141,"height":36}}

We know that:-

{"id":"3-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","type":"align*","ts":1603626665565,"cs":"ah5RIUItzXe3tq4fRY7x8w==","size":{"width":153,"height":36}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"4-1-1-0","code":"\\begin{align*}\n{-\\log_{}\\left|1-y\\right|}&={x+C}\t\n\\end{align*}","ts":1603626998687,"cs":"ECY4Ue0gqyBVhxF1/9Ikpw==","size":{"width":144,"height":16}}

{"id":"4-1-1-1","code":"\\begin{align*}\n{\\log_{}\\left|1-y\\right|}&={-x-C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1603627077981,"cs":"jrW8IpJsnepT3VvmnwcZ0A==","size":{"width":140,"height":16}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{1-y}&={e^{-x-C}}\\\\\n{1-y}&={\\frac{e^{-x}}{e^{c}}}\t\n\\end{align*}","id":"4-1-1-2","ts":1603627158994,"cs":"RDCREZzC+DgVi+6d6vNMQw==","size":{"width":96,"height":56}}

{"id":"4-1-1-3","code":"\\begin{align*}\n{y}&={1-\\frac{e^{-x}}{e^{c}}}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1603627367572,"cs":"8eWq8Cx+eL8Jg9P03OXQyw==","size":{"width":85,"height":33}}

Let -1/ec = A (A is constant)

⇒ y = 1 + Ae-x

4. sec2x.tany.dx + sec2y.tanx.dy = 0

Solun:- Given equation is:- sec2x.tany.dx + sec2y.tanx.dy = 0

sec2x.tany.dx = -sec2y.tanx.dy........(1)

Separate variables of eq. 1:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"5-0-0","code":"\\begin{align*}\n{\\frac{\\sec^{2}x}{\\tan x}.dx}&={-\\frac{\\sec^{2}y}{\\tan y}.dy}\t\n\\end{align*}","ts":1603628704320,"cs":"nglTMCoyd1sgEOWwRsFQBA==","size":{"width":169,"height":37}}

Integrating both sides:-

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"5-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{\\sec^{2}x}{\\tan x}.dx}&={-\\int_{}^{}\\frac{\\sec^{2}y}{\\tan y}.dy}\t\n\\end{align*}","type":"align*","ts":1603628713682,"cs":"kfFnmVnb8h0tohH3IaXJCw==","size":{"width":212,"height":37}}

We know that:-

{"id":"3-1-1-1-0","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*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1603628571744,"cs":"j2Sxbc4AeDW00uL67UTCpg==","size":{"width":200,"height":37}}

⇒ log |tan x| = -log |tan y| + log C

⇒ log |tan x| + log |tan y| = log C

⇒ log |tan x.tan y| = log C

⇒ tan x.tan y = C

5. (ex + e-x).dy - (ex - e-x).dx = 0

Solun:- Given equation is:- (ex + e-x).dy - (ex - e-x).dx = 0

(ex + e-x).dy = (ex - e-x).dx........(1)

Separate variables of eq. 1:-

{"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{dy}&={\\frac{\\left(e^{x}-e^{-x}\\right)}{\\left(e^{x}+e^{-x}\\right)}.dx}\t\n\\end{align*}","type":"align*","id":"5-0-1-0-0","ts":1603629021556,"cs":"E0nGhQGZ//HTbFlGTByFMg==","size":{"width":137,"height":38}}

Integrating both sides:-

{"id":"5-0-1-1","code":"\\begin{align*}\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{\\left(e^{x}-e^{-x}\\right)}{\\left(e^{x}+e^{-x}\\right)}.dx}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603629041780,"cs":"O3DdeUxpkuuhvRzhv3W6mQ==","size":{"width":176,"height":38}}

We know that:-

{"id":"3-1-1-1-1-0","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*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1603628571744,"cs":"08ZuZ0uDCXN+5vitpNef/A==","size":{"width":200,"height":37}}

⇒ y = log |ex + e-x| + C

{"type":"align*","id":"1-0-1-0-1-0-0-1-0","code":"\\begin{align*}\n{6.\\,\\diff{y}{x}}&={\\left(1+x^{2}\\right)\\left(1+y^{2}\\right)}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1603629215108,"cs":"iZh0V3Cg8gcBuFOwEhULpA==","size":{"width":173,"height":32}}

Solun:- Given equation is:-

{"id":"1-0-1-0-1-0-0-1-1","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\left(1+x^{2}\\right)\\left(1+y^{2}\\right)}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603629238568,"cs":"z56VMUsQhbWObNgghPiNBQ==","size":{"width":158,"height":32}}........(1)

Separate variables of eq. 1:-

{"code":"\\begin{align*}\n{\\frac{1}{1+y^{2}}.dy}&={\\left(1+x^{2}\\right).dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"5-0-1-0-1-0-0","type":"align*","ts":1603629310320,"cs":"EA8Z6cI8bQtWLZJZfaZK3Q==","size":{"width":173,"height":36}}

Integrating both sides:-

{"id":"5-0-1-0-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{1+y^{2}}.dy}&={\\int_{}^{}\\left(1+x^{2}\\right).dx}\t\n\\end{align*}","type":"align*","ts":1603629331134,"cs":"aOWAZubuUOESg8I2Q/EVEA==","size":{"width":209,"height":36}}

We know that:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{1+x^{2}}.dx}&={\\tan^{-1}x+C}\t\n\\end{align*}","id":"3-1-1-1-2-0","ts":1603629383364,"cs":"mpOXog2mtvmvdVcezC/kpw==","size":{"width":196,"height":36}}

{"font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{\\tan^{-1}y}&={x+\\frac{x^{3}}{3}+C}\t\n\\end{align*}","type":"align*","id":"6-0","ts":1603629458686,"cs":"vZRpAfRPwjRd/VEb4Hs/Jg==","size":{"width":152,"height":34}}

7. ylog y.dx - x.dy = 0

Solun:- Given equation is:- ylog y.dx - x.dy = 0

⇒ ylog y.dx = x.dy........(1)

Separate variables of eq. 1:-

{"type":"align*","id":"5-0-1-0-1-0-1-0-0","code":"\\begin{align*}\n{\\frac{1}{x}.dx}&={\\frac{1}{y\\log_{}y}.dy}\\\\\n{\\frac{1}{x}.dx}&={\\frac{\\frac{1}{y}}{\\log_{}y}.dy}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603629607608,"cs":"h40m1sOVS9zAfE2O+y3XDw==","size":{"width":132,"height":84}}

Integrating both sides:-

{"id":"5-0-1-0-1-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\int_{}^{}\\frac{\\frac{1}{y}}{\\log_{}y}.dy}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1603630081547,"cs":"qO0vwO5o/zEWMOeyrHYjBA==","size":{"width":161,"height":42}}

We know that:-

{"id":"3-1-1-1-1-1","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*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1603628571744,"cs":"xbWABwSUGEDHUPvc665TDQ==","size":{"width":200,"height":37}}

⇒ log x = log |log y| + log C

⇒ log x - log C = log |log y|

⇒ log x + log c = log |log y|

⇒ log |c.x| = log |log y|

⇒ log y = c.x

⇒ y = ecx

{"id":"7-0-0","code":"\\begin{align*}\n{8.\\,x^{5}\\diff{y}{x}}&={-y^{5}}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603629940313,"cs":"nGcsHebBpX4s//kQ2GuibA==","size":{"width":102,"height":32}}

Solun:- Given equation is:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{x^{5}\\diff{y}{x}}&={-y^{5}}\t\n\\end{align*}","id":"7-1-0","ts":1603629966288,"cs":"HSKnfda4nH/QQUnkcNQf8w==","size":{"width":88,"height":32}}........(1)

Separate variables of eq. 1:-

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"5-0-1-0-1-0-1-0-1-0","code":"\\begin{align*}\n{\\frac{1}{y^{5}}.dy}&={\\frac{-1}{x^{5}}.dx}\t\n\\end{align*}","ts":1603630024104,"cs":"UPpexWuEjKsx2teA9T0yGA==","size":{"width":116,"height":36}}

⇒ y-5.dy = -x-5.dx

Integrating both sides:-

{"code":"\\begin{align*}\n{\\int_{}^{}y^{-5}.dy}&={-\\int_{}^{}x^{-5}.dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"5-0-1-0-1-0-1-1-1-0","ts":1603630145576,"cs":"pW/qLPMpZLGkX/JvXBEofg==","size":{"width":168,"height":36}}

We know that:-

{"code":"\\begin{align*}\n{\\int_{}^{}x^{n}.dx}&={\\frac{x^{n+1}}{n+1}+C}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"3-1-1-1-1-2-0","ts":1603630281755,"cs":"uAX1gKhKdXL9rX4zZpMneg==","size":{"width":156,"height":37}}

{"id":"8-0","type":"align*","code":"\\begin{align*}\n{\\frac{y^{-4}}{4}}&={-\\frac{x^{-4}}{4}+c}\\\\\n{y^{-4}}&={-x^{-4}+4c}\\\\\n{y^{-4}}&={-x^{-4}+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603630573109,"cs":"QeOVLJdtlfRv+havPR2OSw==","size":{"width":121,"height":80}}

⇒ x-4 + y-4 = C

{"id":"7-0-1-0-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{9.\\,\\diff{y}{x}}&={\\sin^{-1}x}\t\n\\end{align*}","ts":1603630844127,"cs":"YNdzCRUgSTvtH9vlfs4w0g==","size":{"width":106,"height":32}}

Solun:- Given equation is:-

{"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\sin^{-1}x}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"7-0-1-1-0","ts":1603630985768,"cs":"sBo/HgwIF7qUSQjmYWbMKA==","size":{"width":92,"height":32}}........(1)

Separate variables of eq. 1:-

⇒ dy = sin-1x.dx

Integrating both sides:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}1.dy}&={\\int_{}^{}\\left(\\sin^{-1}x\\times1\\right).dx}\t\n\\end{align*}","type":"align*","id":"5-0-1-0-1-0-1-1-1-1-0","ts":1603631121953,"cs":"rAhrBDt/Z/txKbC+z/OV2A==","size":{"width":197,"height":36}}

According to ILATE:- u = sin-1x and v = 1

We know that:-

{"id":"3-1-1-1-1-2-1-0","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\left(u.v\\right).dx}&={u\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}\\int_{}^{}v.dx\\right].dx+C}\t\n\\end{align*}","type":"align*","ts":1603631232222,"cs":"Lin4yoO+B8U7GaJV+MfZ4g==","size":{"width":348,"height":37}}

{"type":"align*","code":"\\begin{align*}\n{y}&={\\sin^{-1}x\\int_{}^{}1.dx-\\int_{}^{}\\left[\\diff{\\left(\\sin^{-1}x\\right)}{x}\\int_{}^{}1.dx\\right].dx+C}\\\\\n{y}&={x.\\sin^{-1}x-\\int_{}^{}\\left(\\frac{x}{{\\sqrt[]{1-x^{2}}}}\\right).dx+C}\\\\\n{y}&={x.\\sin^{-1}x-\\frac{1}{2}\\int_{}^{}\\left(\\frac{2x}{{\\sqrt[]{1-x^{2}}}}\\right).dx+C}\t\n\\end{align*}","id":"8-1-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1604321152046,"cs":"tuEerdbxJ72RcnE3lIqdHw==","size":{"width":365,"height":134}}

Let 1 - x2 = t2

Differentitate w.r.t. to x:-

⇒ -2x.dx = 2t.dt

⇒ 2x.dx = -2t.dt

{"font":{"color":"#000000","family":"Arial","size":10},"id":"8-1-1","type":"align*","code":"\\begin{align*}\n{y}&={x.\\sin^{-1}x-\\frac{1}{2}\\int_{}^{}\\frac{-2t.dt}{t}+C}\\\\\n{y}&={x.\\sin^{-1}x+\\int_{}^{}1.dt+C}\\\\\n{y}&={x.\\sin^{-1}x+t+C}\\\\\n{y}&={x.\\sin^{-1}x+{\\sqrt[]{1-x^{2}}}+C}\t\n\\end{align*}","ts":1604321245129,"cs":"8zCdCXo5ITSHBHezXcFx6A==","size":{"width":232,"height":124}}

10. extan y.dx + (1 - ex).sec2y.dy = 0

Solun:- Given equation is:- extan y.dx + (1 - ex).sec2y.dy = 0

extan y.dx = -(1 - ex).sec2y.dy........(1)

Separate variables of eq. 1:-

{"type":"align*","code":"\\begin{align*}\n{\\frac{-e^{x}}{1-e^{x}}.dx}&={\\frac{\\sec^{2}y}{\\tan y}.dy}\t\n\\end{align*}","id":"9-0-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1603632577892,"cs":"kTtKBzf8B4kq2x2bGRzzIg==","size":{"width":160,"height":37}}

Integrating both sides:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"9-1-0-0","code":"\\begin{align*}\n{-\\int_{}^{}\\frac{e^{x}}{1-e^{x}}.dx}&={\\int_{}^{}\\frac{\\sec^{2}y}{\\tan y}.dy}\t\n\\end{align*}","ts":1603632632747,"cs":"yQRiBnDIkCtcUoaGZdk5lA==","size":{"width":214,"height":37}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{\\sec^{2}y}{\\tan y}.dy}&={\\int_{}^{}\\frac{-e^{x}}{1-e^{x}}.dx}\t\n\\end{align*}","id":"9-1-1","type":"align*","ts":1603632794732,"cs":"0Emm7ewTedZXnDkW8QJHfw==","size":{"width":200,"height":37}}

We know that:-

{"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"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*}","id":"3-1-1-1-1-2-1-1-0","ts":1603632165907,"cs":"/k62znyUUDU8bijV/ZssfA==","size":{"width":200,"height":37}}

⇒ log |tan y| = log |1 - ex| + log C

⇒ log |tan y| = log |C.(1 - ex)|

⇒ tan y = C.(1 - ex)

For each of the differential equations in Exercises 11 to 14, Find a particular solution satisfying the given condition:

{"type":"align*","code":"\\begin{align*}\n{11.\\,\\left(x^{3}+x^{2}+x+1\\right)\\diff{y}{x}}&={2x^{2}+x}\t\n\\end{align*}","id":"10-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603705079674,"cs":"8flIJgZ0QV7/ff6FPIQkwA==","size":{"width":240,"height":32}}; y = 1 when x = 0

Solun:- Given equation is:-

{"code":"\\begin{align*}\n{\\left(x^{3}+x^{2}+x+1\\right)\\diff{y}{x}}&={2x^{2}+x}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"10-1","ts":1603705174375,"cs":"vOR+0yUuki6beRB0YESNuA==","size":{"width":217,"height":32}}........(1)

Separate variables of eq. 1:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{dy}&={\\frac{2x^{2}+x}{x^{3}+x^{2}+x+1}.dx}\t\n\\end{align*}","id":"9-0-1-0-0","ts":1603705307789,"cs":"dKcW723FbRr3uvYjoNjvlw==","size":{"width":173,"height":36}}

Integrating both sides:-

{"code":"\\begin{align*}\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{2x^{2}+x}{x^{3}+x^{2}+x+1}.dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{x^{2}+x^{2}+x}{x^{2}\\left(x+1\\right)+1\\left(x+1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{x^{2}+x\\left(x+1\\right)}{\\left(x^{2}+1\\right)\\left(x+1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{x^{2}}{\\left(x^{2}+1\\right)\\left(x+1\\right)}.dx+\\int_{}^{}\\frac{x\\left(x+1\\right)}{\\left(x^{2}+1\\right)\\left(x+1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{x^{2}-1+1}{\\left(x^{2}+1\\right)\\left(x+1\\right)}.dx+\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{\\left(x^{2}+1\\right)}{\\left(x^{2}+1\\right)\\left(x+1\\right)}.dx-\\int_{}^{}\\frac{1}{\\left(x^{2}+1\\right)\\left(x+1\\right)}.dx+\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx-\\int_{}^{}\\frac{1}{\\left(x^{2}+1\\right)\\left(x+1\\right)}.dx+\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)}.dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"9-0-1-1-0","type":"align*","ts":1603707091202,"cs":"1134Hj6g77F8BGHAIVU4Xg==","size":{"width":509,"height":302}}

By partial fractions:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\frac{1}{\\left(x+1\\right)\\left(x^{2}+1\\right)}}&={\\frac{A}{x+1}+\\frac{Bx+C}{x^{2}+1}}\t\n\\end{align*}","type":"align*","id":"12-0","ts":1603706529559,"cs":"cXGr1JK9/sNUowM8GpD7Nw==","size":{"width":253,"height":36}}

⇒ 1 = A(x2 + 1) + (Bx+C)(x + 1)......(1)

Put x = -1 in eq. 1:-

⇒ 1 = A(1 + 1)

⇒ A = 1/2

Put x = 0 in eq. 1:-

⇒ 1 = A(0 + 1) + (C)(1)

⇒ 1 = A + C

⇒ C = 1/2

Put x = 1 in eq. 1:-

⇒ 1 = A(1 + 1) + (B + C)(2)

⇒ 1 = 2(1/2) + (B + 1/2)(2)

⇒ 1 = 1 + 2B + 1

⇒ 1 = 2 + 2B

⇒ B = -1/2

{"type":"align*","font":{"color":"#222222","family":"Arial","size":10},"id":"13-0-0","code":"\\begin{align*}\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx-\\left[\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx+\\frac{1}{2}\\int_{}^{}\\frac{1-x}{\\left(x^{2}+1\\right)}.dx\\right]+\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx-\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx-\\left(\\frac{1}{2}\\int_{}^{}\\frac{1-x}{\\left(x^{2}+1\\right)}.dx\\right)+\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx-\\frac{1}{2}\\left[\\int_{}^{}\\frac{1}{\\left(x^{2}+1\\right)}.dx-\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)}.dx\\right]+\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx-\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x^{2}+1\\right)}.dx+\\frac{3}{2}\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx-\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x^{2}+1\\right)}.dx+\\frac{3}{4}\\int_{}^{}\\frac{2x}{\\left(x^{2}+1\\right)}.dx}\t\n\\end{align*}","ts":1604321987028,"cs":"aL0XRqCm7dLXKT6+SzkcQw==","size":{"width":584,"height":208}}

We know that:-

{"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"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*}","id":"3-1-1-1-1-2-1-1-1","ts":1603632165907,"cs":"ukytG/3UwXGKttb3t3FqFg==","size":{"width":200,"height":37}}

{"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{1+x^{2}}.dx}&={\\tan^{-1}x+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"14-1","ts":1603707666740,"cs":"QkDjM7slpWDmLnjr+9RD6w==","size":{"width":196,"height":36}}

{"id":"13-1-0","type":"align*","code":"\\begin{align*}\n{y}&={\\frac{1}{2}\\log_{}\\left|x+1\\right|-\\frac{1}{2}\\tan^{-1}x+\\frac{3}{4}\\log_{}\\left|x^{2}+1\\right|+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603707774846,"cs":"8hIDTYRohNEGDcOQv0HKsg==","size":{"width":338,"height":32}}

Given y = 1 and x = 0

⇒ 1 = 1/2(log |1|) - 1/2tan-1(0) + 3/4(log |1|) + C

⇒ 1 = (1/2)x0 -(1/2)x0 + (3/4)x0 + C

⇒ C = 1

{"type":"align*","code":"\\begin{align*}\n{y}&={\\frac{1}{2}\\log_{}\\left|x+1\\right|-\\frac{1}{2}\\tan^{-1}x+\\frac{3}{4}\\log_{}\\left|x^{2}+1\\right|+1}\t\n\\end{align*}","id":"15-0","font":{"family":"Arial","color":"#222222","size":10},"ts":1603708199202,"cs":"HHgZ0NY+p2S6e1wMmCwuqQ==","size":{"width":334,"height":32}}

{"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{y}&={\\frac{1}{2}\\log_{}\\left|x+1\\right|+\\frac{3}{4}\\log_{}\\left|x^{2}+1\\right|-\\frac{1}{2}\\tan^{-1}x+1}\\\\\n{y}&={\\frac{1}{4}\\left(2\\log_{}\\left|x+1\\right|+3\\log_{}\\left|x^{2}+1\\right|\\right)-\\frac{1}{2}\\tan^{-1}x+1}\\\\\n{y}&={\\frac{1}{4}\\left(\\log_{}\\left|\\left(x+1\\right)^{2}\\right|+\\log_{}\\left|\\left(x^{2}+1\\right)^{3}\\right|\\right)-\\frac{1}{2}\\tan^{-1}x+1}\\\\\n{y}&={\\frac{1}{4}\\left(\\log_{}\\left|\\left(x+1\\right)^{2}.\\left(x^{2}+1\\right)^{3}\\right|\\right)-\\frac{1}{2}\\tan^{-1}x+1}\t\n\\end{align*}","id":"16-0","ts":1603708219998,"cs":"1TXA9NIAZlwvepBFv0HPcQ==","size":{"width":380,"height":142}}

{"id":"10-2-0-0","code":"\\begin{align*}\n{12.\\,x\\left(x^{2}-1\\right)\\diff{y}{x}}&={1}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1603708348217,"cs":"lqyU24meewigo3NJNSkudw==","size":{"width":142,"height":32}}; y = 0 when x = 2

Solun:- Given equation is:-

{"code":"\\begin{align*}\n{x\\left(x^{2}-1\\right)\\diff{y}{x}}&={1}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"10-2-1-0","type":"align*","ts":1603708403030,"cs":"7DtzJBbqdAKWkrDSPrl15w==","size":{"width":120,"height":32}}........(1)

Separate variables of eq. 1:-

{"code":"\\begin{align*}\n{dy}&={\\frac{1}{x\\left(x^{2}-1\\right)}.dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"10-2-1-1-0","ts":1603708461607,"cs":"ybOAe1+PNK9uBik2aAu3kw==","size":{"width":132,"height":36}}

Integrating both sides:-

{"code":"\\begin{align*}\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{1}{x\\left(x^{2}-1\\right)}.dx}\\\\\n{\\int_{}^{}dy}&={\\int_{}^{}\\frac{1}{x\\left(x-1\\right)\\left(x+1\\right)}.dx}\t\n\\end{align*}","id":"9-0-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1603708543533,"cs":"fHZQ78jP2uOBmU9Ea25EEQ==","size":{"width":213,"height":77}}

By partial fractions:-

{"code":"\\begin{align*}\n{\\frac{1}{x\\left(x-1\\right)\\left(x+1\\right)}}&={\\frac{A}{x}+\\frac{B}{x-1}+\\frac{C}{x+1}}\t\n\\end{align*}","id":"12-1","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603708597109,"cs":"ARRM3FzDOeFoCcExWShMqw==","size":{"width":278,"height":36}}

⇒ 1 = A(x2 - 1) + B(x)(x + 1) + C(x)(x - 1)......(1)

Put x = 0 in eq. 1:-

⇒ 1 = A(-1)

⇒ A = -1

Put x = 1 in eq. 1:-

⇒ 1 = B(1)(2)

⇒ B = 1/2

Put x = -1 in eq. 1:-

⇒ 1 = C(-1)(-2)

⇒ C = 1/2

{"code":"\\begin{align*}\n{\\int_{}^{}dy}&={-\\int_{}^{}\\frac{1}{x}.dx+\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx+\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"id":"13-0-1-0","ts":1603709456810,"cs":"wkmLid5S4nSINhAkQIETHA==","size":{"width":404,"height":36}}

We know that:-

{"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"14-0-1-0","ts":1603707623146,"cs":"a7WDhDH1jLx0yhYijg6SsA==","size":{"width":153,"height":36}}

{"code":"\\begin{align*}\n{y}&={-\\log_{}\\left|x\\right|+\\frac{1}{2}\\log_{}\\left|x-1\\right|+\\frac{1}{2}\\log_{}\\left|x+1\\right|+C}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"id":"13-1-1-0-0","ts":1603709740641,"cs":"hMqNIRl5kS9+XUQ9xWYZGA==","size":{"width":322,"height":32}}

Given y = 0 and x = 2

{"code":"\\begin{align*}\n{0}&={-\\log_{}\\left|2\\right|+\\frac{1}{2}\\log_{}\\left|2-1\\right|+\\frac{1}{2}\\log_{}\\left|2+1\\right|+C}\\\\\n{0}&={-\\log_{}\\left|2\\right|+\\frac{1}{2}\\log_{}\\left|1\\right|+\\frac{1}{2}\\log_{}\\left|3\\right|+C}\\\\\n{0}&={-\\log_{}\\left|2\\right|+\\frac{1}{2}\\times0+\\frac{1}{2}\\log_{}\\left|3\\right|+C}\\\\\n{0}&={-\\log_{}\\left|2\\right|+\\frac{1}{2}\\log_{}\\left|3\\right|+C}\\\\\n{0}&={-2\\log_{}\\left|2\\right|+\\log_{}\\left|3\\right|+2C}\\\\\n{0}&={-\\log_{}\\left|4\\right|+\\log_{}\\left|3\\right|+2C}\\\\\n{0}&={\\log_{}\\left|\\frac{3}{4}\\right|+2C}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","id":"16-1-0","ts":1603710481227,"cs":"mF34HHhlg0R7OjhuRfAh+A==","size":{"width":320,"height":222}}

{"id":"17-0","font":{"size":10,"family":"Arial","color":"#222222"},"type":"align*","code":"\\begin{align*}\n{C}&={\\frac{-1}{2}\\log_{}\\left|\\frac{3}{4}\\right|}\t\n\\end{align*}","ts":1603710509169,"cs":"uu/G4a3phuZps3kjGuQEiA==","size":{"width":108,"height":34}}

{"type":"align*","code":"\\begin{align*}\n{y}&={-\\log_{}\\left|x\\right|+\\frac{1}{2}\\log_{}\\left|x-1\\right|+\\frac{1}{2}\\log_{}\\left|x+1\\right|-\\frac{1}{2}\\log_{}\\left|\\frac{3}{4}\\right|}\\\\\n{y}&={-\\log_{}\\left|x\\right|+\\frac{1}{2}\\left[\\log_{}\\left|x-1\\right|+\\log_{}\\left|x+1\\right|\\right]-\\frac{1}{2}\\log_{}\\left|\\frac{3}{4}\\right|}\\\\\n{y}&={-\\log_{}\\left|x\\right|+\\frac{1}{2}\\log_{}\\left|\\left(x-1\\right)\\left(x+1\\right)\\right|-\\frac{1}{2}\\log_{}\\left|\\frac{3}{4}\\right|}\\\\\n{y}&={-\\log_{}\\left|x\\right|+\\frac{1}{2}\\log_{}\\left|\\left(x^{2}-1\\right)\\right|-\\frac{1}{2}\\log_{}\\left|\\frac{3}{4}\\right|}\t\n\\end{align*}","id":"13-1-1-1","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603710539764,"cs":"HSnCNiS+I/ezvEmE3/P56g==","size":{"width":374,"height":152}}

{"id":"18","type":"align*","code":"\\begin{align*}\n{2y}&={-2\\log_{}\\left|x\\right|+\\log_{}\\left|\\left(x^{2}-1\\right)\\right|-\\log_{}\\left|\\frac{3}{4}\\right|}\\\\\n{2y}&={-\\log_{}\\left|x^{2}\\right|+\\log_{}\\left|\\left(x^{2}-1\\right)\\right|-\\log_{}\\left|\\frac{3}{4}\\right|}\\\\\n{2y}&={\\log_{}\\left|\\frac{x^{2}-1}{x^{2}}\\right|-\\log_{}\\left|\\frac{3}{4}\\right|}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603710566202,"cs":"weVRqUxgReIyC+ir3ARpfQ==","size":{"width":280,"height":118}}

{"type":"align*","code":"\\begin{align*}\n{y}&={\\frac{1}{2}\\log_{}\\left|\\frac{x^{2}-1}{x^{2}}\\right|-\\frac{1}{2}\\log_{}\\left|\\frac{3}{4}\\right|}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"id":"19","ts":1603710613451,"cs":"A0AfQTxYpcCo+UzOojNnMQ==","size":{"width":210,"height":40}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{13.\\,\\cos \\left(\\diff{y}{x}\\right)}&={a\\,\\,\\left(a\\epsilon R\\right)}\t\n\\end{align*}","id":"10-2-0-1-0-0","ts":1603710924580,"cs":"3xzTgcbZkKTB9sRT4mdq/Q==","size":{"width":170,"height":37}}; y = 1 when x = 0

Solun:- Given equation is:-

{"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\cos \\left(\\diff{y}{x}\\right)}&={a}\t\n\\end{align*}","id":"10-2-0-1-1-0-0","type":"align*","ts":1603711023285,"cs":"rjppSfDmt+ZL8MaYOrNxlg==","size":{"width":100,"height":37}}

{"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\cos ^{-1}a}\t\n\\end{align*}","id":"10-2-0-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603711530637,"cs":"sHGhIm5B0hVgwTHdN7dMRg==","size":{"width":92,"height":32}}........(1)

Separate variables of eq. 1:-

{"id":"10-2-1-1-1-0","type":"align*","code":"\\begin{align*}\n{dy}&={\\cos ^{-1}a.dx}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603711588437,"cs":"NFNRtO6ddBEwXZGZ/CYW7Q==","size":{"width":108,"height":17}}

Integrating both sides:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}dy}&={\\int_{}^{}\\cos ^{-1}a.dx}\\\\\n{\\int_{}^{}dy}&={\\cos ^{-1}a\\int_{}^{}1.dx}\t\n\\end{align*}","id":"9-0-1-1-1-1-0","ts":1603711626520,"cs":"X9cx/g/sUiFBU03BWlVgwg==","size":{"width":156,"height":76}}

{"type":"align*","code":"\\begin{align*}\n{y}&={x.\\cos ^{-1}a+C}\t\n\\end{align*}","id":"13-1-1-0-1-0-0","font":{"color":"#222222","family":"Arial","size":10},"ts":1603711662876,"cs":"fn2hD1BDY9ObQRBZtQuGLQ==","size":{"width":124,"height":17}}

Given y = 1 and x = 0

{"type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"id":"16-1-1-0","code":"\\begin{align*}\n{1}&={0\\times \\cos ^{-1}a+C}\t\n\\end{align*}","ts":1603711703874,"cs":"rLAaysUlKlmpVZ0vXD4+Tw==","size":{"width":134,"height":17}}

⇒ C = 1

{"id":"13-1-1-0-1-1-0","font":{"family":"Arial","color":"#222222","size":10},"type":"align*","code":"\\begin{align*}\n{y}&={x.\\cos ^{-1}a+1}\t\n\\end{align*}","ts":1603711734793,"cs":"+YsEOgtpeWJncSbPEptuQA==","size":{"width":120,"height":17}}

{"code":"\\begin{align*}\n{\\frac{y-1}{x}}&={\\cos ^{-1}a}\t\n\\end{align*}","id":"20-0","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603711781060,"cs":"lJ+WUz+QDE54oveAXMUJnw==","size":{"width":109,"height":32}}

{"id":"21","code":"\\begin{align*}\n{\\cos\\left(\\frac{y-1}{x}\\right)}&={a}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603711899706,"cs":"ZzqWMnQplKrny+8jUbl2sA==","size":{"width":118,"height":37}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"10-2-0-1-0-1-0-0","type":"align*","code":"\\begin{align*}\n{14.\\,\\diff{y}{x}}&={y\\tan x}\t\n\\end{align*}","ts":1603711993955,"cs":"2e36ABFacY9ljuxB/C3uww==","size":{"width":113,"height":32}}; y = 1 when x = 0

Solun:- Given equation is:-

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={y\\tan x}\t\n\\end{align*}","id":"10-2-0-1-0-1-1","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1603712069261,"cs":"2bXDo0SOP4YM6JtGbdK/FQ==","size":{"width":90,"height":32}}........(1)

Separate variables of eq. 1:-

{"type":"align*","code":"\\begin{align*}\n{\\frac{dy}{y}}&={\\tan x.dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"10-2-1-1-1-1-0-0","ts":1603712188666,"cs":"0gs7Gfl5uxNSSqBTc8MBJg==","size":{"width":102,"height":36}}

Integrating both sides:-

{"id":"10-2-1-1-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{dy}{y}}&={\\int_{}^{}\\tan x.dx}\t\n\\end{align*}","type":"align*","ts":1603712219375,"cs":"CXyVwYNlwPSCiRwBiOU8qw==","size":{"width":141,"height":36}}

We know that:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"id":"14-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}\\tan x.dx}&={\\log_{}\\left|\\sec x\\right|+C}\t\n\\end{align*}","type":"align*","ts":1603712310810,"cs":"lGSPFvbSB8v5eiMb/akcRA==","size":{"width":194,"height":36}}

{"id":"13-1-1-0-1-0-1-0","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\log_{}y}&={\\log_{}\\left|\\sec x\\right|+C}\t\n\\end{align*}","type":"align*","ts":1603712368301,"cs":"+tB84cTmXNfAK5x0AqtFaw==","size":{"width":148,"height":16}}

Given y = 1 and x = 0

⇒ log |1| = log |sec 0| + C

⇒ C = 0

{"font":{"color":"#222222","family":"Arial","size":10},"id":"13-1-1-0-1-1-1-0","code":"\\begin{align*}\n{\\log_{}y}&={\\log_{}\\left|\\sec x\\right|}\t\n\\end{align*}","type":"align*","ts":1603712462561,"cs":"pWFbHmmdN+p8gzGFLq+IGA==","size":{"width":116,"height":16}}

⇒ y = sec x

15. Find the equation of a curve passing through the point (0,0) and whose differential equation is y’ = exsin x.

Solun:- Given equation is:- y’ = exsin x

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\diff{y}{x}}&={e^{x}.\\sin x}\t\n\\end{align*}","id":"10-2-0-1-0-1-1","ts":1603712704372,"cs":"Hx3RZpSolB0fD05W8oFrIA==","size":{"width":98,"height":32}}........(1)

Separate variables of eq. 1:-

{"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{dy}&={e^{x}\\sin x.dx}\t\n\\end{align*}","id":"10-2-1-1-1-1-0-1-0","ts":1603712748679,"cs":"20hpCFnDhuEY9sQ2eLTGpQ==","size":{"width":109,"height":16}}

Integrating both sides:-

{"code":"\\begin{align*}\n{\\int_{}^{}dy}&={\\int_{}^{}e^{x}\\sin x.dx}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"10-2-1-1-1-1-0-1-1-0","ts":1603712773657,"cs":"4GMZfPkRoJEfqLw+Sf/07w==","size":{"width":148,"height":36}}......(2)

Let {"code":"\\begin{align*}\n{I}&={\\int_{}^{}e^{x}\\sin x.dx}\t\n\\end{align*}","id":"22","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1603713193633,"cs":"of7aqze5o4cXAhSLqQJu4Q==","size":{"width":120,"height":36}}

According to ILATE:- u = sin x and v = ex

We know that:-

{"code":"\\begin{align*}\n{\\int_{}^{}\\left(u.v\\right).dx}&={u\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}\\int_{}^{}v.dx\\right].dx+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"14-0-1-1-1","type":"align*","ts":1603712871007,"cs":"5EMEINpTO4to35KHTx10Qw==","size":{"width":348,"height":37}}

{"id":"13-1-1-0-1-0-1-1-0","font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={\\sin x\\int_{}^{}e^{x}.dx-\\int_{}^{}\\left[\\diff{\\left(\\sin x\\right)}{x}\\int_{}^{}e^{x}.dx\\right].dx+C}\\\\\n{I}&={e^{x}\\sin x-\\int_{}^{}\\left(e^{x}\\cos x\\right).dx+C}\\\\\n{I}&={e^{x}\\sin x-\\left[\\cos x\\int_{}^{}e^{x}.dx-\\int_{}^{}\\left[\\diff{\\left(\\cos x\\right)}{x}\\int_{}^{}e^{x}.dx\\right].dx\\right]+C}\\\\\n{I}&={e^{x}\\sin x-\\left[e^{x}\\cos x-\\int_{}^{}-\\left(e^{x}\\sin x\\right).dx\\right]+C}\\\\\n{I}&={e^{x}\\sin x-e^{x}\\cos x-\\int_{}^{}\\left(e^{x}\\sin x\\right).dx+C}\\\\\n{I}&={e^{x}\\sin x-e^{x}\\cos x-I+C}\t\n\\end{align*}","type":"align*","ts":1603713151062,"cs":"azzZzIWn5uD7qo/ZM+jaZg==","size":{"width":437,"height":224}}

⇒ 2I = ex(sin x - cos x) + C

{"font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{e^{x}\\left(\\sin x-\\cos x\\right)}{2}+C}\t\n\\end{align*}","id":"13-1-1-0-1-0-1-1-1","ts":1603713353478,"cs":"qH3cLMmAR3fsbPEexsJaqw==","size":{"width":177,"height":33}}

Put the value of I in eq. 2:-

{"code":"\\begin{align*}\n{y}&={\\frac{e^{x}\\left(\\sin x-\\cos x\\right)}{2}+C}\t\n\\end{align*}","type":"align*","id":"10-2-1-1-1-1-0-1-1-1","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603713423210,"cs":"sZe3q+7J01RqHw6PX4ab3g==","size":{"width":177,"height":33}}

Given y = 0 and x = 0 because curve passing through (0,0)

{"font":{"family":"Arial","color":"#222222","size":10},"type":"align*","code":"\\begin{align*}\n{0}&={\\frac{e^{0}\\left(\\sin0-\\cos0\\right)}{2}+C}\\\\\n{0}&={\\frac{\\left(0-1\\right)}{2}+C}\t\n\\end{align*}","id":"13-1-1-0-1-1-1-1-0","ts":1603713645845,"cs":"6Mi2g7aMm2tRHOc6F4YNSQ==","size":{"width":174,"height":73}}

⇒ C = 1/2

{"id":"23-0","code":"\\begin{align*}\n{y}&={\\frac{e^{x}\\left(\\sin x-\\cos x\\right)}{2}+\\frac{1}{2}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603716476673,"cs":"kYxE1rZB2S8dgvrnhdxf8w==","size":{"width":180,"height":33}}

⇒ 2y - 1 = ex(sin x - cos x)

16. For the differential equation {"type":"align*","id":"23-1","code":"\\begin{align*}\n{xy\\diff{y}{x}}&={\\left(x+2\\right)\\left(y+2\\right)}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1603789818066,"cs":"Kno4lf4Te9f2PQfU+b+Zcw==","size":{"width":158,"height":32}}, find the solution curve passing through the point (1,-1).

Solun:- Given equation is:-

{"type":"align*","id":"23-2-0","code":"\\begin{align*}\n{xy\\diff{y}{x}}&={\\left(x+2\\right)\\left(y+2\\right)}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1603789818066,"cs":"MM2CR6ZG/HReo+5eGZJOUQ==","size":{"width":158,"height":32}}........(1)

Separate variables of eq. 1:-

{"type":"align*","code":"\\begin{align*}\n{\\frac{y}{\\left(y+2\\right)}.dy}&={\\frac{\\left(x+2\\right)}{x}.dx}\t\n\\end{align*}","id":"23-3-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1603789932684,"cs":"lKBUQv4Ba9rv9e2tRTGdwg==","size":{"width":177,"height":37}}

Integrating both sides:-

{"id":"24-0-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{y}{y+2}.dy}&={\\int_{}^{}\\frac{x+2}{x}.dx}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603790055764,"cs":"919+PUXfT18xVDOw2L8WHw==","size":{"width":192,"height":36}}

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}\\frac{y+2-2}{y+2}.dy}&={\\int_{}^{}\\frac{x}{x}.dx+\\int_{}^{}\\frac{2}{x}.dx}\t\n\\end{align*}","id":"24-1-0","ts":1603790149381,"cs":"r6We4YqYCGFfTRqjXPfjew==","size":{"width":272,"height":36}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"24-2-0","code":"\\begin{align*}\n{\\int_{}^{}\\frac{y+2}{y+2}.dy-\\int_{}^{}\\frac{2}{y+2}.dy}&={\\int_{}^{}\\frac{x}{x}.dx+\\int_{}^{}\\frac{2}{x}.dx}\t\n\\end{align*}","ts":1603790179339,"cs":"oWZKQMeZmtihKavaaMs2Xg==","size":{"width":348,"height":36}}

{"type":"align*","id":"24-2-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}1.dy-\\int_{}^{}\\frac{2}{y+2}.dy}&={\\int_{}^{}1.dx+\\int_{}^{}\\frac{2}{x}.dx}\t\n\\end{align*}","ts":1603790206345,"cs":"Vu2BXx2tnYWmEKUq75a+5g==","size":{"width":300,"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*}","type":"align*","id":"25-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1603790273494,"cs":"rc5dhSmS2gHxg2H/lrXyag==","size":{"width":200,"height":37}}

{"id":"24-2-1-1-0-0","code":"\\begin{align*}\n{y-2\\log_{}\\left|y+2\\right|}&={x+2\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1603790372220,"cs":"6CN2dylhYPnoO3qL6n9hTw==","size":{"width":237,"height":16}}

Given y = -1 and x = 1 because curve passing through (1,-1)

⇒ -1 - 2log |-1+2| = 1 + 2log |1| + C

⇒ -1 - 2log |1| = 1 + C

⇒ -1 = 1 + C

⇒ C = -2

⇒ y - 2log |y + 2| = x + 2log |x| - 2

⇒ y - log |(y + 2)2| = x + log |x2| - 2

⇒ y - x = log |(y + 2)2| + log |x2| - 2

⇒ y - x + 2 = log |x2(y + 2)2|

17. Find the equation of a curve passing through the point (0,-2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Solun:- Given:- y(slope of its tangent) = x

{"id":"23-2-1-0-0","type":"align*","code":"\\begin{align*}\n{y\\diff{y}{x}}&={x}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603792101109,"cs":"wUQ5myneGVOkTUzVxxioQQ==","size":{"width":62,"height":32}}........(1)

Separate variables of eq. 1:-

{"type":"align*","id":"23-2-1-1-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{y.dy}&={x.dx}\t\n\\end{align*}","ts":1603792157886,"cs":"mpy9JM+TygYMYe+Cz1F6ww==","size":{"width":84,"height":16}}

Integrating both sides:-

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}y.dy}&={\\int_{}^{}x.dx}\t\n\\end{align*}","id":"23-2-1-1-1","ts":1603792199712,"cs":"avuLFcvjPeQaudJ6VhlVCA==","size":{"width":124,"height":36}}

We know that:-

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"25-1-0","code":"\\begin{align*}\n{\\int_{}^{}x^{n}.dx}&={\\frac{x^{n+1}}{n+1}+C}\t\n\\end{align*}","type":"align*","ts":1603792247872,"cs":"blWqqMLThqjorApCglBv0w==","size":{"width":156,"height":37}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"24-2-1-1-0-1-0-0","code":"\\begin{align*}\n{\\frac{y^{2}}{2}}&={\\frac{x^{2}}{2}+C}\t\n\\end{align*}","type":"align*","ts":1603792405963,"cs":"3DF9iWbDbhIEy9KKG6rWGA==","size":{"width":96,"height":34}}

Given x = 0 and y = -2 because curve passing through (0,-2)

{"code":"\\begin{align*}\n{\\frac{4}{2}}&={\\frac{0}{2}+C}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"id":"24-2-1-1-0-1-1","ts":1603792472451,"cs":"znVXQb9zO0XKSe7QetCfnA==","size":{"width":81,"height":32}}

⇒ 2 = 0 + C

⇒ C = 2

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\frac{y^{2}}{2}}&={\\frac{x^{2}}{2}+2}\t\n\\end{align*}","id":"24-2-1-1-0-1-0-1","ts":1603792520716,"cs":"qZVZN7HWf2H5gJcskq4iKg==","size":{"width":92,"height":34}}

⇒ y2 - x2 = 4

18. At any point (x,y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3). Find the equation of the curve given that it passes through (-2,1).

Solun:- Slope of the line segment joining the point of contact of the curve and the point (-4,-3) is:-

We know that:-

Slope = {"code":"$\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"27","ts":1603793679230,"cs":"+RdcQoiHe9EVJJClHXkVQA==","size":{"width":37,"height":20}}

Slope = {"font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\frac{y+3}{x+4}$","id":"28","type":"$","ts":1603793637126,"cs":"Uny6ryViYYT4oP35cFhyyw==","size":{"width":28,"height":21}}

Given:- slope of its tangent = 2(Slope of joining line)

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"23-2-1-0-1-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={2\\left(\\frac{y+3}{x+4}\\right)}\t\n\\end{align*}","ts":1603793756144,"cs":"bL6YwCS8Q1ssRbhjum7sXA==","size":{"width":120,"height":37}}........(1)

Separate variables of eq. 1:-

{"id":"23-2-1-1-0-1-0-0-0","code":"\\begin{align*}\n{\\frac{1}{\\left(y+3\\right)}.dy}&={\\frac{2}{\\left(x+4\\right)}.dx}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1603794654183,"cs":"rXKtbFZxM9I/AfLpRoLJcQ==","size":{"width":177,"height":36}}

Integrating both sides:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{\\left(y+3\\right)}.dy}&={\\int_{}^{}\\frac{2}{\\left(x+4\\right)}.dx}\t\n\\end{align*}","id":"23-2-1-1-0-1-0-1-0","ts":1603794673071,"cs":"o4nOf+PSGLCFbvuYOp6jjA==","size":{"width":216,"height":36}}

We know that:-

{"id":"25-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","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*}","ts":1603794467184,"cs":"sjzvi1nP8/JjalmbwoNcxQ==","size":{"width":200,"height":37}}

{"id":"24-2-1-1-0-1-0-2-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|y+3\\right|}&={2\\log_{}\\left|x+4\\right|+C}\t\n\\end{align*}","ts":1603794719428,"cs":"tx6CQrL3ifunGcrzahKn9Q==","size":{"width":198,"height":16}}

Given x = -2 and y = 1 because curve passing through (-2,1)

⇒ log |4| = 2log |2| + C

⇒ log |4| = log |22| + C

⇒ log |4| = log |4| + C

⇒ C = 0

⇒ log |y+3| = log |(x+4)2| + 0

⇒ log |y+3| - log |(x+4)2| = 0

{"code":"\\begin{align*}\n{\\log_{}\\left|\\frac{y+3}{\\left(x+4\\right)^{2}}\\right|}&={0}\t\n\\end{align*}","id":"29","font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","ts":1603795051228,"cs":"vD8Ai9VXuWAV3utR1mEQow==","size":{"width":124,"height":44}}

{"id":"30","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{\\frac{y+3}{\\left(x+4\\right)^{2}}}&={e^{0}}\\\\\n{\\frac{y+3}{\\left(x+4\\right)^{2}}}&={1}\t\n\\end{align*}","ts":1603795099419,"cs":"GA8NrggBkNmP5+kpcdLkfA==","size":{"width":96,"height":84}}

⇒ y + 3 = (x + 4)2

⇒ y = (x + 4)2 - 3

19. The volume of spherical balloon being inflated changes at a constant rate. If initially, its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Solun:- Given that volume of the spherical balloon being inflated changes at a constant rate.

{"code":"\\begin{align*}\n{\\diff{v}{t}}&={k}\t\n\\end{align*}","id":"23-2-1-0-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1603796427740,"cs":"ecWr9TY50OvPzVNKsETIxQ==","size":{"width":52,"height":32}}

We know that volume of a sphere is (4/3)Ï€r3.

{"code":"\\begin{align*}\n{\\diff{\\left(\\frac{4}{3}\\Pi r^{3}\\right)}{t}}&={k}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"23-2-1-0-1-2","type":"align*","ts":1603796566302,"cs":"Z+4NxMzXK36PMM9aeQJd5g==","size":{"width":97,"height":37}}

{"code":"\\begin{align*}\n{\\frac{4}{3}\\Pi \\diff{\\left(r^{3}\\right)}{t}}&={k}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"31-0","ts":1603796621044,"cs":"sRAOMA+YgzAb/msGUFCJMQ==","size":{"width":100,"height":36}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"id":"31-1-0","type":"align*","code":"\\begin{align*}\n{\\frac{4}{3}\\Pi \\times3r^{2}\\diff{r}{t}}&={k}\t\n\\end{align*}","ts":1603796714533,"cs":"B1iWf/ueSONODNAOQSjsYA==","size":{"width":120,"height":32}}

{"id":"31-1-1-0-0","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{4\\Pi r^{2}\\diff{r}{t}}&={k}\t\n\\end{align*}","type":"align*","ts":1603796752831,"cs":"PbDmPpbKhIKgp7UqsDcRdw==","size":{"width":85,"height":32}}........(1)

Separate variables of eq. 1:-

{"id":"31-1-1-1-0","type":"align*","code":"\\begin{align*}\n{4\\Pi r^{2}.dr}&={k.dt}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603796811257,"cs":"iH1N2NpFE4J9bPM+QYf9Uw==","size":{"width":105,"height":17}}

Integrating both sides:-

{"code":"\\begin{align*}\n{\\int_{}^{}4\\Pi r^{2}.dr}&={\\int_{}^{}k.dt}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"31-1-1-1-1","ts":1603796840915,"cs":"VoEuWl8PUTW1PDXEqDmcfg==","size":{"width":144,"height":36}}

We know that:-

{"id":"25-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}x^{n}.dx}&={\\frac{x^{n+1}}{n+1}+C}\t\n\\end{align*}","ts":1603796898508,"cs":"gafda6jacf9DGTzdqhI1iA==","size":{"width":156,"height":37}}

{"type":"align*","code":"\\begin{align*}\n{\\frac{4\\Pi r^{3}}{3}}&={kt+C}\t\n\\end{align*}","font":{"color":"#222222","size":11,"family":"Arial"},"id":"32-0-0","ts":1603797563220,"cs":"AJrlcZI4FrIJJpI5xlfTKg==","size":{"width":121,"height":40}}……(2)

Given t = 0 and r = 3 units

⇒ (4Ï€)x(9) = k(0) + C

⇒ C = 36Ï€……….(3)

Given t = 3 and r = 6 units

⇒ 288Ï€ = k(3) + C

⇒ 288Ï€ = 3k + C

From eq. 3:-

⇒ 288Ï€ = 3k + 36Ï€

⇒ 252Ï€ = 3k

⇒ k = 84Ï€

From eq. 2:-

Put the value of k and C:-

{"type":"align*","code":"\\begin{align*}\n{\\frac{4\\Pi r^{3}}{3}}&={84\\Pi t+36\\Pi}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"id":"32-1-0-0","ts":1603797800199,"cs":"j7ze4cPKLA3fEqt64PRggg==","size":{"width":141,"height":34}}

{"code":"\\begin{align*}\n{\\frac{r^{3}}{3}}&={21t+9}\t\n\\end{align*}","id":"32-1-1","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603797862287,"cs":"YFoYYDjuFCV/bS1qhsg0WQ==","size":{"width":89,"height":34}}

⇒ r3 = 3(21t + 9)

⇒ r3 = 9(7t + 3)

⇒ r = [9(7t + 3)]1/3

20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (loge2 = 0.6931).

Solun:- Let p,t, and r represent principal, time, and rate.

{"id":"23-2-1-0-1-1-1-0","code":"\\begin{align*}\n{\\diff{p}{t}}&={\\left(\\frac{r}{100}\\right)p}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1603798729501,"cs":"WyH7xNUeWneAVUIlSHxJzQ==","size":{"width":101,"height":32}}........(1)

Separate variables of eq. 1:-

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"33-0","code":"\\begin{align*}\n{\\frac{dp}{p}}&={\\frac{r}{100}.dt}\t\n\\end{align*}","ts":1603798822140,"cs":"8hrZzweAYRf7h4Yg7JxBRw==","size":{"width":96,"height":36}}

Integrating both sides:-

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{dp}{p}}&={\\int_{}^{}\\frac{r}{100}.dt}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"33-1","ts":1603798845453,"cs":"L6XLD/3fqYbQX9XT6PQCRw==","size":{"width":134,"height":36}}

We know that:-

{"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"25-1-1-1-1-0","ts":1603798874608,"cs":"6VCxF+nfHUhTV+jhmbJ1AA==","size":{"width":153,"height":36}}

⇒ log |p| = rt/100 + C……(2)

Given t = 0 and p = 100 Rs

⇒ log |100| = r(0)/100 + C

⇒ C = log |100|……….(3)

Given t = 10 and p = 200 Rs

⇒ log |200| = r(10)/100 + C

⇒ log |200| = r/10 + C

From eq. 3:-

⇒ log |200| = r/10 + log |100|

⇒ log |200| - log |100| = r/10

{"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{200}{100}\\right|}&={\\frac{r}{10}}\t\n\\end{align*}","id":"34-0","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603799556454,"cs":"VmMD6SpE2qEAcf2Cowpo8w==","size":{"width":108,"height":34}}

⇒ log |2| = r/10

⇒ r = 10log |2|

Given log |2| = 0.6931

⇒ r = 10x0.6931

⇒ r = 6.93%

21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).

Solun:- Let p, and t represent principal and time.

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\diff{p}{t}}&={\\left(\\frac{5}{100}\\right)p}\t\n\\end{align*}","type":"align*","id":"23-2-1-0-1-1-1-1","ts":1603874620809,"cs":"9jrbJCV4/sQF8D4nIziOjw==","size":{"width":106,"height":37}}........(1)

Separate variables of eq. 1:-

{"id":"33-2-0","code":"\\begin{align*}\n{\\frac{dp}{p}}&={\\frac{5}{100}.dt}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603874656989,"cs":"jlrCfezWPkAfK9zLS+u1hg==","size":{"width":96,"height":36}}

Integrating both sides:-

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"33-3","code":"\\begin{align*}\n{\\int_{}^{}\\frac{dp}{p}}&={\\int_{}^{}\\frac{5}{100}.dt}\t\n\\end{align*}","ts":1603874681159,"cs":"dA+CxxjTzVJ1P2sxmJ72BQ==","size":{"width":134,"height":36}}

We know that:-

{"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"25-1-1-1-1-1","ts":1603798874608,"cs":"VTngKXE7GUpBya2ROAjvgQ==","size":{"width":153,"height":36}}

⇒ log |p| = 5t/100 + C……(2)

Given t = 0 and p = 1000 Rs

⇒ log |1000| = 5(0)/100 + C

⇒ C = log |1000|……….(3)

Given t = 10, after 10 years

⇒ log |p| = 5(10)/100 + C

⇒ log |p| = 1/2 + C

From eq. 3:-

⇒ log |p| = 1/2 + log |1000|

⇒ log |p| - log |1000| = 0.5

{"id":"34-1-0","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"code":"\\begin{align*}\n{\\log_{}\\left|\\frac{p}{1000}\\right|}&={0.5}\t\n\\end{align*}","ts":1603875133396,"cs":"kBU/E6o81oKfkL2kLelMyA==","size":{"width":113,"height":29}}

⇒ p/1000 = e0.5

⇒ p = 1000e0.5

Given e0.5 = 1.648

⇒ p = 1000x1.648

⇒ p = 1648.

22. In culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Solun:- Let n, and t represent the count of bacteria and time.

{"code":"\\begin{align*}\n{\\diff{n}{t}}&\\propto{n}\\\\\n{\\diff{n}{t}}&={kn}\t\n\\end{align*}","id":"36","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1603877518396,"cs":"7Ahvt6aPTS/zqKUiyakJlg==","size":{"width":64,"height":69}}........(1)

Separate variables of eq. 1:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\frac{dn}{n}}&={k.dt}\t\n\\end{align*}","id":"33-2-1-0-0","ts":1603877671489,"cs":"B1vURrpUiTYnNdv3uHZXow==","size":{"width":74,"height":32}}

Integrating both sides:-

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{dn}{n}}&={\\int_{}^{}k.dt}\\\\\n{\\int_{}^{}\\frac{dn}{n}}&={k\\int_{}^{}1.dt}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"33-2-1-1-0","ts":1603877876457,"cs":"cFdDxK3SGJdlL3lvkElDIA==","size":{"width":121,"height":76}}

We know that:-

{"code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x}.dx}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"25-1-1-1-1-2","ts":1603798874608,"cs":"81p27S11ocWOr21yHDK0Jg==","size":{"width":153,"height":36}}

⇒ log |n| = kt + C……(2)

Given t = 0 and let n = n0 bacteria

⇒ log |n0| =k(0) + C

⇒ C = log |n0|……….(3)

Given t = 2 hour, after 2 hours

{"code":"\\begin{align*}\n{n}&={n_{0}+\\frac{10}{100}n_{0}}\\\\\n{n}&={\\frac{110}{100}n_{0}}\\\\\n{n}&={\\frac{11}{10}n_{0}}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"37","type":"align*","ts":1603879645668,"cs":"EhwzEfPl4jTWCbst2/saIQ==","size":{"width":112,"height":106}}

{"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{11}{10}n_{0}\\right|}&={2k+C}\t\n\\end{align*}","id":"38-0","ts":1603880125544,"cs":"qFlGbibsPwJwPnIcPKcwvw==","size":{"width":141,"height":34}}

From eq. 3:-

{"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{11}{10}n_{0}\\right|}&={2k+\\log_{}\\left|n_{0}\\right|}\t\n\\end{align*}","id":"38-1-0","ts":1603880155586,"cs":"oAjUTtWyVBnOdG3c6jc7DQ==","size":{"width":177,"height":34}}

{"font":{"color":"#222222","size":10,"family":"Arial"},"id":"38-1-1-0","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{11}{10}n_{0}\\right|-\\log_{}\\left|n_{0}\\right|}&={2k}\t\n\\end{align*}","type":"align*","ts":1603880172411,"cs":"wLgQ55kP6rOytA32DFtJGg==","size":{"width":177,"height":34}}

{"code":"\\begin{align*}\n{\\log_{}\\left|\\frac{\\frac{11}{10}n_{0}}{n_{0}}\\right|}&={2k}\t\n\\end{align*}","id":"38-1-1-1","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603880200307,"cs":"HIbxSnPL25M9JBUfH3kwSw==","size":{"width":112,"height":44}}

{"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{11}{10}\\right|}&={2k}\t\n\\end{align*}","id":"39","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603880226039,"cs":"wZhA1A20ONB8/dx4hbITVA==","size":{"width":93,"height":34}}

{"id":"39","code":"\\begin{align*}\n{k}&={\\frac{1}{2}\\log_{}\\left|\\frac{11}{10}\\right|}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","ts":1603880269658,"cs":"52YD9jrWVUtXYXwRdt/Eyw==","size":{"width":100,"height":34}}

From eq. 2 and eq. 3:-

{"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|n\\right|}&={\\frac{1}{2}\\log_{}\\left|\\frac{11}{10}\\right|t+\\log_{}\\left|n_{0}\\right|}\t\n\\end{align*}","id":"40-0","ts":1603880422645,"cs":"7qUeiu0PKnBtwqdEbbHG2g==","size":{"width":208,"height":34}}

n0 = 1,00,000 at t = 0

{"font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|n\\right|}&={\\frac{1}{2}\\log_{}\\left|\\frac{11}{10}\\right|t+\\log_{}\\left|1,00,000\\right|}\t\n\\end{align*}","id":"40-1-0","ts":1603880486147,"cs":"aIFtu9qJP1fBQMSlzEDawg==","size":{"width":253,"height":34}}

At t = t and n = 2,00,000

{"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|2,00,000\\right|}&={\\frac{1}{2}\\log_{}\\left|\\frac{11}{10}\\right|t+\\log_{}\\left|1,00,000\\right|}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"id":"40-1-1","ts":1603880542826,"cs":"4Bw1LECf1kxjiadR4KiZ+A==","size":{"width":305,"height":34}}

{"code":"\\begin{align*}\n{\\log_{}\\left|2,00,000\\right|-\\log_{}\\left|1,00,000\\right|}&={\\frac{1}{2}\\log_{}\\left|\\frac{11}{10}\\right|t}\t\n\\end{align*}","type":"align*","id":"41-0","font":{"family":"Arial","color":"#222222","size":10},"ts":1603880593158,"cs":"XHYHhYvpggvmoly79+cmtA==","size":{"width":305,"height":34}}

{"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|\\frac{2,00,000}{1,00,000}\\right|}&={\\frac{1}{2}\\log_{}\\left|\\frac{11}{10}\\right|t}\t\n\\end{align*}","id":"41-1","ts":1603880634524,"cs":"Dk9TGq+J8db38bDMbhahhw==","size":{"width":201,"height":36}}

{"id":"41-2-0","code":"\\begin{align*}\n{\\log_{}\\left|2\\right|}&={\\frac{1}{2}\\log_{}\\left|\\frac{11}{10}\\right|t}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603880662054,"cs":"0LyPsz5ablGlbpCiiBiTjQ==","size":{"width":138,"height":34}}

{"code":"\\begin{align*}\n{2\\log_{}\\left|2\\right|}&={\\log_{}\\left|\\frac{11}{10}\\right|t}\t\n\\end{align*}","id":"41-2-1-0","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603880689765,"cs":"s47QnV2r7DHr3oVA2khesQ==","size":{"width":134,"height":34}}

{"id":"41-2-1-1-0","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"code":"\\begin{align*}\n{\\log_{}\\left|4\\right|}&={\\log_{}\\left|\\frac{11}{10}\\right|t}\t\n\\end{align*}","ts":1603880714551,"cs":"o4Wy49HBGI/HhXgI/HYqiw==","size":{"width":124,"height":34}}

{"type":"align*","code":"\\begin{align*}\n{t}&={\\frac{\\log_{}\\left|4\\right|}{\\log_{}\\left|\\frac{11}{10}\\right|}}\\\\\n{t}&={\\frac{2\\log_{}\\left|2\\right|}{\\log_{}\\left|\\frac{11}{10}\\right|}}\t\n\\end{align*}","id":"41-2-1-1-1","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603880809556,"cs":"mKPn3At7NShAoP99OFQKIw==","size":{"width":85,"height":88}}

23. The general solution of the differential equation {"id":"36","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={e^{x+y}}\t\n\\end{align*}","ts":1603881041011,"cs":"ciNoD1A7tKuyP2bE118hXQ==","size":{"width":74,"height":32}}

Solun:- Given equation is

{"id":"36","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={e^{x+y}}\t\n\\end{align*}","ts":1603881041011,"cs":"ciNoD1A7tKuyP2bE118hXQ==","size":{"width":74,"height":32}}

{"type":"align*","id":"36","code":"\\begin{align*}\n{\\diff{y}{x}}&={e^{x}.e^{y}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1603881360031,"cs":"OhDuRP1Ommzz6yd6KoUw5A==","size":{"width":80,"height":32}}........(1)

Separate variables of eq. 1:-

{"id":"33-2-1-0-1","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\frac{dy}{e^{y}}}&={e^{x}.dx}\t\n\\end{align*}","ts":1603881409878,"cs":"oDe20d4wawf9JM6tJEDRBA==","size":{"width":82,"height":32}}

⇒ e-y.dy = ex.dx

Integrating both sides:-

{"type":"align*","id":"42-0","code":"\\begin{align*}\n{\\int_{}^{}e^{-y}.dy}&={\\int_{}^{}e^{x}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603881502792,"cs":"zwakR3dvUZuqlnxC/yR8Pw==","size":{"width":144,"height":36}}

We know that:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}e^{x}.dx}&={e^{x}+C}\t\n\\end{align*}","id":"25-1-1-1-1-3","type":"align*","ts":1603881534410,"cs":"toz+8MkY9sbN/OLKlaIfVg==","size":{"width":124,"height":36}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"42-1","code":"\\begin{align*}\n{-e^{-y}}&={e^{x}+c}\t\n\\end{align*}","ts":1603881762170,"cs":"SuAacI3lHmoxrxM+K3x4Lw==","size":{"width":96,"height":16}}

{"type":"align*","id":"43","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{e^{x}+e^{-y}}&={C}\t\n\\end{align*}","ts":1603881733106,"cs":"tyq6UYNkXWoOx/YXVs2zMw==","size":{"width":88,"height":16}}

The correct answer is A.



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