Important Note

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Definitions and Formulas

Dependent variable:- Dependent variable is a variable whose value depends on another variable (Generally it is denoted by y).

Independent variable:- Independent variable is a variable whose value does not depend on another variable (Generally it is denoted by x).

For example:- x + y = 7

In this equation, y is dependent and x is the independent variable.

Differential Equation:- Differential equation is an equation that contains dependent, independent variables and derivative of the dependent variable with respect to the independent variable x. For example:-

{"font":{"family":"Arial","size":11,"color":"#000000"},"id":"1-0","code":"$x\\diff{y}{x}+y=0$","type":"$","ts":1603520677982,"cs":"VRKpBJnD5NhKXavIdFh27Q==","size":{"width":96,"height":24}}

In general, an equation involving derivatives of the dependent variable with respect to the independent variable x is called a differential equation.

NOTE:- Notations for derivatives:

{"code":"$\\diff{y}{x}=y^{\\prime }$","id":"2-0","font":{"family":"Arial","size":11,"color":"#000000"},"type":"$","ts":1603521004049,"cs":"l5jfsu7ipyFEhtY7iLd0Eg==","size":{"width":60,"height":24}},{"type":"$","id":"2-1","code":"$\\frac{d^{2}y}{dx^{2}}=y^{\\prime \\prime }$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1603521076451,"cs":"mdTQFjMCK43K3COu+zwuEw==","size":{"width":68,"height":25}},{"type":"$","font":{"color":"#000000","family":"Arial","size":11},"id":"2-2","code":"$\\frac{d^{3}y}{dx^{3}}=y^{\\prime \\prime \\prime }$","ts":1603521134804,"cs":"a81LRsOdPBcD5IUktIBJLg==","size":{"width":72,"height":25}}

Order of differential equation:- The order of the differential equation is the order of the highest derivative present in the given equation.

OR

Order of differential equation is the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation. For example:-

{"id":"1-1-0","code":"$\\diff{y}{x}=e^{x}....\\left(1\\right)$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1603521931644,"cs":"7xgqsJnAKmNElTub5FlJ8g==","size":{"width":104,"height":20}}

{"id":"1-1-1","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","code":"$\\frac{d^{2}y}{dx^{2}}+y=0....\\left(2\\right)$","ts":1603521979772,"cs":"x8vudKpOG8jFEMCfIjCeqw==","size":{"width":125,"height":22}}

{"type":"$","code":"$\\left(\\frac{d^{3}y}{dx^{3}}\\right)+x^{2}\\left(\\frac{d^{2}y}{dx^{2}}\\right)^{3}=0....\\left(3\\right)$","id":"1-1-2-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1603522049743,"cs":"rzvtp2wxfc627cRs7UEJmg==","size":{"width":200,"height":32}}

Order of differential eq. 1,2 and 3 respectively are 1, 2, 3.

Degree of a differential equation:- For the study of the degree of a differential equation it is necessary that the equation must be polynomial of derivatives then the degree of that differential equation is the power of highest order derivative. 

For example:- Degree of the eq. 1, 2 and 3 respectively are 1, 1, 1.

{"type":"$","code":"$\\diff{y}{x}+\\sin\\left(\\diff{y}{x}\\right)=0....\\left(4\\right)$","font":{"family":"Arial","color":"#000000","size":10},"id":"1-1-2-1","ts":1603523218995,"cs":"rpsq5wGYmidN60qO1ync+A==","size":{"width":173,"height":28}}

Order of differential eq. 4 is 1 but the degree of this differential eq. is not defined because eq. 4 is not a polynomial of derivatives.

General solutions of a differential equation:- The solution which contains arbitrary constants is called the general(primitive) solution of the differential equation.

Particular solutions of a differential equation:- The solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.

Formation of differential equation whose general equation is given:- 

The required differential equation is obtained by eliminating constants from equations.

NOTE:- The order of differential equation is the same as the number of arbitrary constants present in the given equation.

Methods of Solving First Order, First Degree Differential Equations

(1) Differential equations with variable separable:- 

(i) Separate both independent and dependent variables.

(ii) Integrate both sides then calculating the general solution of the given differential equation.

Example:- Find the general solution of the differential equation {"type":"$","id":"3","code":"$\\diff{y}{x}=\\frac{x+1}{2-y}$","font":{"family":"Arial","color":"#000000","size":10},"ts":1603530170418,"cs":"RJXXyGqJ1bcPYM4gFmE+0g==","size":{"width":68,"height":22}}, (y≠2).

Solun:- Given

{"type":"$","id":"3","code":"$\\diff{y}{x}=\\frac{x+1}{2-y}$","font":{"family":"Arial","color":"#000000","size":10},"ts":1603530170418,"cs":"RJXXyGqJ1bcPYM4gFmE+0g==","size":{"width":68,"height":22}}

Separating the variables:-

{"id":"4-0-0","type":"$","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\left(2-y\\right).dy=\\left(x+1\\right).dx$","ts":1603530356974,"cs":"wOdpvbHtaY3LRaeORfOL8A==","size":{"width":164,"height":16}}

Integrating both sides:-

{"id":"4-1","font":{"family":"Arial","color":"#000000","size":10},"type":"$","code":"$\\int_{}^{}\\left(2-y\\right).dy=\\int_{}^{}\\left(x+1\\right).dx$","ts":1603530398905,"cs":"cWvO/ndSGxevHX9HDOjAvg==","size":{"width":186,"height":17}}

{"id":"5","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{2y-\\frac{y^{2}}{2}}&={\\frac{x^{2}}{2}+x+C}\\\\\n{4y\\,-y^{2}\\,}&={x^{2}+2x+c}\t\n\\end{align*}","type":"align*","ts":1603530554644,"cs":"uDF+4SuhvFooVwfTo41OWA==","size":{"width":160,"height":56}}

x2 + y2 + 2x - 4y + c = 0

(2) Homogeneous differential equations:-

If x and y are replaced by λx and λy respectively and then F(λx, λy) = λnF(x, y) then the given equation is called a homogeneous equation if it is a homogeneous function of degree zero where λ is a non-zero constant.

Solving Homogeneous Differential Equation:- If the given equation is a homogeneous equation then

Type-1: {"font":{"size":10,"color":"#222222","family":"Arial"},"type":"$","code":"$\\diff{y}{x}=F\\left(x,y\\right)=g\\left(\\frac{y}{x}\\right)$","id":"6-0","ts":1603531547795,"cs":"kReRzfrfl0/2F48DBehS4g==","size":{"width":144,"height":20}}......(1)

(i) Substitute y = vx and then differentiate with respect to x

{"id":"7","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":"X6CDF7con0L/vHFj8QNAMA==","size":{"width":132,"height":69}}..........(2)

(ii) Replace dy/dx in given equation 1 with eq. 2

(iii) Separate variables and then integrating both sides and put the value of v in integrating result

(iv) Calculate the general solution of the given equation

Type-2: {"font":{"family":"Arial","size":10,"color":"#222222"},"code":"$\\diff{x}{y}=F\\left(x,y\\right)=g\\left(\\frac{x}{y}\\right)$","type":"$","id":"6-1","ts":1603532049263,"cs":"a6gz8y0eK4oYJJNFQbLaWg==","size":{"width":149,"height":28}}......(1)

(i) Substitute x = vy and then differentiate with respect to y

{"font":{"family":"Arial","color":"#222222","size":10},"type":"align*","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*}","id":"7","ts":1603532117719,"cs":"2wp8NbxJI4RhH4qXQK/D2Q==","size":{"width":128,"height":76}}..........(2)

(ii) Replace dx/dx in given equation 1 with eq. 2

(iii) Separate variables and then integrating both sides and put the value of v in integrating result

(iv) Calculate the general solution of the given equation

Example:- Show that the differential equation {"font":{"color":"#000000","family":"Arial","size":10},"type":"$","id":"3","code":"$\\left(x-y\\right)\\diff{y}{x}=x+2y$","ts":1603532784055,"cs":"cbKSMNUmo9iS427Ixcvo1w==","size":{"width":132,"height":20}} is a homogeneous equation and solve it.

Solun:- Given

{"font":{"color":"#000000","family":"Arial","size":10},"type":"$","id":"3","code":"$\\left(x-y\\right)\\diff{y}{x}=x+2y$","ts":1603532784055,"cs":"cbKSMNUmo9iS427Ixcvo1w==","size":{"width":132,"height":20}}

{"font":{"family":"Arial","color":"#000000","size":10},"type":"$","id":"8","code":"$\\diff{y}{x}=\\frac{x+2y}{x-y}$","ts":1603533006692,"cs":"7Whbi3W/Z74Mof+4nR0b/g==","size":{"width":73,"height":22}}.....(1)

Let F(x,y)={"code":"$\\frac{x+2y}{x-y}$","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","id":"9","ts":1603533092873,"cs":"pP8oQ7haSu5L3P1qxTF4sA==","size":{"width":32,"height":22}}

Replace F(x,y) with F(λx,λy):-

{"code":"\\begin{align*}\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\frac{\\lambda x+2\\lambda y}{\\lambda x-\\lambda y}}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}\\left(\\frac{x+2y}{x-y}\\right)}\\\\\n{F\\left(\\lambda x,\\lambda y\\right)}&={\\lambda^{0}.F\\left(x,y\\right)}\t\n\\end{align*}","id":"10","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","ts":1603533277819,"cs":"JuiP9gMqR4ZiRuLFw98YCA==","size":{"width":177,"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","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":"X6CDF7con0L/vHFj8QNAMA==","size":{"width":132,"height":69}}.......(2)

From eq. 1 and 2:-

{"code":"\\begin{align*}\n{v+x\\diff{v}{x}}&={\\frac{x+2vx}{x-vx}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{x\\left(1+2v\\right)}{x\\left(1-v\\right)}}\\\\\n{v+x\\diff{v}{x}}&={\\frac{1+2v}{1-v}}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"id":"11","type":"align*","ts":1603533799462,"cs":"n5X8RQbJAC8hencMdgfjzg==","size":{"width":152,"height":116}}

{"id":"12","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{x\\diff{v}{x}}&={\\frac{1+2v}{1-v}-v}\\\\\n{x\\diff{v}{x}}&={\\frac{1+2v-v\\left(1-v\\right)}{1-v}}\\\\\n{x\\diff{v}{x}}&={\\frac{1+2v-v+v^{2}}{1-v}}\\\\\n{x\\diff{v}{x}}&={\\frac{1+v+v^{2}}{1-v}}\t\n\\end{align*}","type":"align*","ts":1603535816547,"cs":"H/IrKqZJe2/DCu6m9aH+Gw==","size":{"width":178,"height":156}}

Separating the variables:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\left(\\frac{1-v}{1+v+v^{2}}\\right).dv=\\frac{1}{x}.dx$","type":"$","id":"4-0-1-0","ts":1603535893021,"cs":"FAlmxMJzEBvb3PWOpqRgzQ==","size":{"width":145,"height":28}}

Integrating both sides:-

{"id":"4-0-1-1-0","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}\\left(\\frac{1-v}{1+v+v^{2}}\\right).dv=\\int_{}^{}\\frac{1}{x}.dx$","ts":1603535933635,"cs":"0tgSfSBtqpV4Ruoy50imNQ==","size":{"width":170,"height":28}}

We know that:-

Numerator = A.d(Denominator)/dx+B

⇒ 1-v = A.d(1+v+v2)/dv + B

⇒ 1-v = A(2v+1) + B

⇒ 1-v = 2Av + A + B

Compare both sides:-

⇒ 2A = -1

⇒ A = -1/2

⇒ A + B = 1

⇒ -1/2 + B = 1

⇒ B = 3/2

Then 1-v = -1/2(2v+1)+3/2

{"id":"4-0-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\left(\\frac{\\frac{-1}{2}\\left(2v+1\\right)+\\frac{3}{2}}{1+v+v^{2}}\\right).dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","ts":1603536736771,"cs":"08LI9BTye0BPFVTpujdETQ==","size":{"width":260,"height":46}}

{"type":"align*","id":"13","code":"\\begin{align*}\n{\\int_{}^{}\\left[\\frac{-1}{2}\\frac{\\left(2v+1\\right)}{1+v+v^{2}}+\\frac{3}{2}\\frac{1}{1+v+v^{2}}\\right].dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603536825652,"cs":"2byFCytCz76ZewsXaF+J+Q==","size":{"width":349,"height":37}}

{"type":"align*","code":"\\begin{align*}\n{\\frac{-1}{2}\\int_{}^{}\\frac{\\left(2v+1\\right)}{1+v+v^{2}}.dv+\\frac{3}{2}\\int_{}^{}\\frac{1}{v^{2}+v+1}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"color":"#222222","size":10,"family":"Arial"},"id":"14","ts":1603536955584,"cs":"l6tN02x6VXVvNzGEq7zQBw==","size":{"width":382,"height":36}}

{"type":"align*","font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\frac{-1}{2}\\int_{}^{}\\frac{\\left(2v+1\\right)}{1+v+v^{2}}.dv+\\frac{3}{2}\\int_{}^{}\\frac{1}{v^{2}+v+\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}+1}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","id":"15","ts":1603537111830,"cs":"YefHy1G8iok6ASr7qyigfQ==","size":{"width":488,"height":44}}

{"id":"16-0","code":"\\begin{align*}\n{\\frac{-1}{2}\\int_{}^{}\\frac{\\left(2v+1\\right)}{1+v+v^{2}}.dv+\\frac{3}{2}\\int_{}^{}\\frac{1}{\\left(v+\\frac{1}{2}\\right)^{2}+\\left(\\frac{{\\sqrt[]{3}}}{2}\\right)^{2}}.dv}&={\\int_{}^{}\\frac{1}{x}.dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","ts":1603537148289,"cs":"1UIgtGpsXVM+7mt+h30IKw==","size":{"width":441,"height":53}}

We know that:-

{"id":"17-0","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":"DshAgnHvjtnjTUmIC8dS9w==","size":{"width":184,"height":24}}

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx}&={\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+C}\t\n\\end{align*}","id":"17-1","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603537261723,"cs":"b45uJMMB29+QX+TNNbBUAQ==","size":{"width":244,"height":36}}

{"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","ts":1603537288100,"cs":"YLrYbqTaJVm3jwf7+qUaMg==","size":{"width":153,"height":36}}

{"font":{"family":"Arial","size":10,"color":"#222222"},"id":"16-1-0","code":"\\begin{align*}\n{\\frac{-1}{2}\\log_{}\\left|1+v+v^{2}\\right|+\\frac{3}{2}\\times\\frac{2}{{\\sqrt[]{3}}}\\tan^{-1}\\left(\\frac{2\\left(v+\\frac{1}{2}\\right)}{{\\sqrt[]{3}}}\\right)}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","type":"align*","ts":1603537746829,"cs":"jAOXGVLC7ZpUj+3JhVjCEg==","size":{"width":440,"height":46}}

{"font":{"size":10,"color":"#222222","family":"Arial"},"id":"18","type":"align*","code":"\\begin{align*}\n{\\frac{-1}{2}\\log_{}\\left|1+v+v^{2}\\right|+\\frac{3}{2}\\times\\frac{2}{{\\sqrt[]{3}}}\\tan^{-1}\\left(\\frac{2v+1}{{\\sqrt[]{3}}}\\right)}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","ts":1603537771773,"cs":"EVcgMDHZG18jHNytOh7PPA==","size":{"width":420,"height":38}}

Put the value of v:-

{"id":"16-1-1","type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{\\frac{-1}{2}\\log_{}\\left|1+\\frac{y}{x}+\\frac{y^{2}}{x^{2}}\\right|+{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{\\frac{2y}{x}+1}{{\\sqrt[]{3}}}\\right)\\,\\,}&={\\log_{}\\left|x\\right|+C}\\\\\n{\\frac{-1}{2}\\log_{}\\left|\\frac{x^{2}+xy+y^{2}}{x^{2}}\\right|+{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{2y+x}{x{\\sqrt[]{3}}}\\right)}&={\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","ts":1603802472803,"cs":"42TBc8aJ1bPOBK1Xzw8RsQ==","size":{"width":406,"height":92}}

{"code":"\\begin{align*}\n{-\\log_{}\\left|\\left(\\frac{x^{2}+xy+y^{2}}{x^{2}}\\right)\\right|+2{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{2y+x}{x{\\sqrt[]{3}}}\\right)}&={2\\log_{}\\left|x\\right|+C}\\\\\n{\\log_{}\\left|\\left(\\frac{x^{2}+xy+y^{2}}{x^{2}}\\right)^{-1}\\right|+2{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{2y+x}{x{\\sqrt[]{3}}}\\right)}&={2\\log_{}\\left|x\\right|+C}\\\\\n{\\log_{}\\left|\\left(\\frac{x^{2}}{x^{2}+xy+y^{2}}\\right)\\right|+2{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{2y+x}{x{\\sqrt[]{3}}}\\right)\\,\\,\\,\\,\\,}&={2\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"19","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","ts":1603802491310,"cs":"ToouCepMpkZFkCwhp7T97A==","size":{"width":436,"height":136}}

{"type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|x^{2}\\right|-\\log_{}\\left|x^{2}+xy+y^{2}\\right|+2{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{2y+x}{x{\\sqrt[]{3}}}\\right)\\,\\,\\,\\,\\,}&={2\\log_{}\\left|x\\right|+C}\\\\\n{2\\log_{}\\left|x\\right|-\\log_{}\\left|x^{2}+xy+y^{2}\\right|+2{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{2y+x}{x{\\sqrt[]{3}}}\\right)\\,\\,\\,\\,\\,}&={2\\log_{}\\left|x\\right|+C}\t\n\\end{align*}","id":"20","font":{"family":"Arial","color":"#222222","size":10},"ts":1603802514538,"cs":"56YA9IuLOHAvxccbVRKc4g==","size":{"width":474,"height":82}}

{"code":"\\begin{align*}\n{2{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{2y+x}{x{\\sqrt[]{3}}}\\right)}&={\\log_{}\\left|x^{2}+xy+y^{2}\\right|+C}\t\n\\end{align*}","id":"21","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603802530329,"cs":"iDYgrbX/51Q0N/UxKRcpbw==","size":{"width":318,"height":38}}

{"code":"\\begin{align*}\n{2{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{2y+x}{x{\\sqrt[]{3}}}\\right)-C}&={\\log_{}\\left|x^{2}+xy+y^{2}\\right|}\t\n\\end{align*}","id":"22","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603802542286,"cs":"3LskDxSdW6pMN9fSyP7pjA==","size":{"width":318,"height":38}}

Here -C = c both are constants.

{"font":{"size":10,"color":"#222222","family":"Arial"},"id":"23","type":"align*","code":"\\begin{align*}\n{\\log_{}\\left|x^{2}+xy+y^{2}\\right|}&={2{\\sqrt[]{3}}\\tan^{-1}\\left(\\frac{2y+x}{x{\\sqrt[]{3}}}\\right)+c}\t\n\\end{align*}","ts":1603802560730,"cs":"YGAzBbrx1RROMPEOTMr1sg==","size":{"width":313,"height":38}}

(3) Linear differential equations:-

A differential of the form

Type-1: {"font":{"color":"#222222","family":"Arial","size":10},"id":"6-2","type":"$","code":"$\\diff{y}{x}+Py=Q$","ts":1603608018935,"cs":"NDtClQMB5rSxk5powXvcKQ==","size":{"width":92,"height":20}}......(1)

(i) In eq. 1, P and Q are constants or functions of x only.

(ii) Find the Integrating factor(I.F.) = {"type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"code":"$e^{\\int_{}^{}P.dx}$","id":"24-0","ts":1603608304574,"cs":"QmbB3psp4aBLDtW55tWyGw==","size":{"width":53,"height":18}}

(iii) Then the solution of the given differential equation is:

{"type":"$","code":"$y.\\left(I.F\\right)=\\int_{}^{}\\left(Q\\times I.F\\right).dx+C$","id":"25-0","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603608400933,"cs":"clQRc01sOXiTFuR/4ornyw==","size":{"width":212,"height":17}}

Type-2: {"type":"$","font":{"family":"Arial","color":"#222222","size":10},"id":"26","code":"$\\diff{x}{y}+Px=Q$","ts":1603608432999,"cs":"0LP5tqLGsoSCDxrbX9MoPg==","size":{"width":93,"height":21}}......(2)

(i) In eq. 2, P and Q are constants or functions of y only.

(ii) Find the Integrating factor(I.F.) = {"code":"$e^{\\int_{}^{}P.dy}$","id":"24-1","type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1603608488770,"cs":"RepAP/59q1eNlheyjudLyg==","size":{"width":52,"height":18}}

(iii) Then the solution of the given differential equation is:

{"code":"$x.\\left(I.F\\right)=\\int_{}^{}\\left(Q\\times I.F\\right).dy+C$","type":"$","id":"25-1","font":{"family":"Arial","color":"#000000","size":10},"ts":1603608564793,"cs":"QYaKGSH/OybBnE70Q31W9g==","size":{"width":212,"height":17}}

Example:- Find the general solution of the differential equation {"code":"$\\diff{y}{x}-y=\\cos x$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"3","type":"$","ts":1603608690591,"cs":"1cstPpNwn2eakt1z3I/tZg==","size":{"width":100,"height":20}}.

Solun:- Given

{"code":"$\\diff{y}{x}-y=\\cos x$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"3","type":"$","ts":1603608690591,"cs":"1cstPpNwn2eakt1z3I/tZg==","size":{"width":100,"height":20}}.....(1)

Given equation is a linear differential equation of type

{"font":{"color":"#222222","family":"Arial","size":10},"id":"6-2","type":"$","code":"$\\diff{y}{x}+Py=Q$","ts":1603608018935,"cs":"NDtClQMB5rSxk5powXvcKQ==","size":{"width":92,"height":20}}......(2) then

Compare equation 1 and 2:- 

P = -1 and Q = cos x

I.F. = {"type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"code":"$e^{\\int_{}^{}P.dx}$","id":"24-2","ts":1603608304574,"cs":"KAuc/i8K8us72nWlqOxd1w==","size":{"width":53,"height":18}}

I.F. = {"type":"$","id":"24-3","code":"$e^{\\int_{}^{}\\left(-1\\right).dx}$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1603608973143,"cs":"e7BJabrICCd55wseTBfP4Q==","size":{"width":69,"height":18}}

I.F. = {"type":"$","code":"$e^{-x}$","font":{"color":"#000000","family":"Arial","size":12},"id":"24-4","ts":1603609013317,"cs":"nScj6eiy2/AoTTaIFZ5yOw==","size":{"width":29,"height":16}}

The solution of equation 1 is:- {"type":"$","code":"$y.\\left(I.F\\right)=\\int_{}^{}\\left(Q\\times I.F\\right).dx+C$","id":"25-0","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603608400933,"cs":"clQRc01sOXiTFuR/4ornyw==","size":{"width":212,"height":17}}

{"id":"27-0-0-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{y.e^{-x}}&={\\int_{}^{}\\left(\\cos x\\times e^{-x}\\right).dx+C}\t\n\\end{align*}","type":"align*","ts":1603609196061,"cs":"Z4KOYr7j7H+lOt8pPYnnoQ==","size":{"width":221,"height":36}}......(3)

Let

{"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\left(\\cos x\\times e^{-x}\\right).dx...\\left(A\\right)}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"27-0-1","ts":1603609526339,"cs":"fdTX9wf2+Su8DBRGwIA0rA==","size":{"width":205,"height":36}}

According to ILATE:- u = cos x and v = e-x

We know that:-

{"type":"$","code":"$\\int_{}^{}\\left(u.v\\right).dx=u\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}\\int_{}^{}v.dx\\right].dx$","id":"28","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603609307767,"cs":"kczmGVSk0gCff3vdscHr1A==","size":{"width":286,"height":20}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"31","code":"\\begin{align*}\n{I}&={\\cos x\\int_{}^{}e^{-x}.dx-\\int_{}^{}\\left[\\diff{\\left(\\cos x\\right)}{x}\\int_{}^{}e^{-x}.dx\\right].dx}\\\\\n{I}&={-e^{-x}.\\cos x-\\int_{}^{}\\left[-\\sin x\\times\\left(-e^{-x}\\right)\\right].dx+C}\\\\\n{I}&={-e^{-x}.\\cos x-\\int_{}^{}\\left[e^{-x}.\\sin x\\right].dx+C}\t\n\\end{align*}","ts":1603610079256,"cs":"r7PGnJs211BoI9sNrsPRWQ==","size":{"width":340,"height":118}}

{"id":"27-1","code":"\\begin{align*}\n{I}&={-e^{-x}.\\cos x-\\left[\\sin x\\int_{}^{}e^{-x}.dx-\\int_{}^{}\\left[\\diff{\\left(\\sin x\\right)}{x}\\int_{}^{}e^{-x}.dx\\right].dx\\right]+C}\\\\\n{I}&={-e^{-x}.\\cos x-\\left[-e^{-x}.\\sin x-\\int_{}^{}\\left[\\cos x.\\left(-e^{-x}\\right)\\right].dx\\right]+C}\\\\\n{I}&={-e^{-x}.\\cos x-\\left[-e^{-x}.\\sin x+\\int_{}^{}\\left(e^{-x}.\\cos x\\right).dx\\right]+C}\\\\\n{I}&={-e^{-x}.\\cos x+e^{-x}.\\sin x-\\int_{}^{}\\left(e^{-x}.\\cos x\\right).dx+C}\\\\\n{I}&={-e^{-x}.\\cos x+e^{-x}.\\sin x-I+C}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1603610100498,"cs":"ftgccg4W0bF19PrTVuN1ng==","size":{"width":478,"height":185}}

From eq. A:-

{"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"29","code":"\\begin{align*}\n{2I}&={-e^{-x}.\\cos x+e^{-x}.\\sin x+C}\t\n\\end{align*}","ts":1603609892126,"cs":"PixwEhH5Cp2h476NLvrp5A==","size":{"width":224,"height":16}}

{"code":"\\begin{align*}\n{I}&={\\frac{e^{-x}.\\left(\\sin x-\\cos x\\right)}{2}+C}\t\n\\end{align*}","id":"30","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1603609968001,"cs":"oVof3Th8LSaHnTrl7RKsmQ==","size":{"width":193,"height":33}}

Put the value of I in equation 3:-

{"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{y.e^{-x}}&={\\frac{e^{-x}.\\left(\\sin x-\\cos x\\right)}{2}+C}\t\n\\end{align*}","id":"27-0-0-1","ts":1603610415033,"cs":"pq1TwNhELieAKa/PhqQAow==","size":{"width":224,"height":33}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\frac{y}{e^{x}}}&={\\frac{\\left(\\sin x-\\cos x\\right)}{2e^{x}}+C}\t\n\\end{align*}","id":"33","ts":1603610481084,"cs":"fVsNdw/nP/9CH9uEjLiB4A==","size":{"width":176,"height":33}}

{"id":"34","code":"\\begin{align*}\n{y}&={\\left(\\frac{\\sin x-\\cos x}{2}\\right)+Ce^{x}}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1603610543884,"cs":"vNdQk/pAF13sM/dcQtDWoQ==","size":{"width":188,"height":37}}

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