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  • If x = x(y) is the solution of the differential equation, <img alt="\mathrm{ydx}-\left(\mathrm{x}+2 \mathrm{y}^{2}\right) \mathrm{d} \mathrm{y}=0" src="https://entrancecorner.oncodecogs.com/gif.lat

If x = x(y) is the solution of the differential equation, \mathrm{ydx}-\left(\mathrm{x}+2 \mathrm{y}^{2}\right) \mathrm{d} \mathrm{y}=0 , with \mathrm{x}(-\pi)=\pi^{2} , then x is equal to:

Option: 1

x = y^{2}-2\pi^{2}


Option: 2

x = y^{2}+2\pi^{2}


Option: 3

x =2 y^{2}-\pi^{2}


Option: 4

x=2y^{2}+\pi


Answers (1)

best_answer

 

 

Exact Differential Equation -

A differential equation of the type P(x, y) dx + Q (x, y) dy = 0 is called an exact differential equation if there exists a function of two variables u(x, y) with continuous partial derivatives such that 

\mathrm{d u\;(x, y)=P(x, y) \;d x+Q(x, y) \;d y}

The general solution of the exact equation is given by 

u (x, y) = C

Where C is an arbitrary constant. 

In some cases the integrating factor is found by inspection. Using the following exact differentials, it is easy to find the integrating factors :

\\\mathbf{1.\;\;\;}\mathit{x d y+y d x=d(x y)}\\\\\mathbf{2.\;\;\;}\mathit{x d x+y d y=\frac{1}{2} d\left(x^{2}+y^{2}\right)}\\\\\mathbf{3.\;\;\;}\mathit{\frac{x d y-y d x}{x^{2}}=d\left(\frac{y}{x}\right)}\\\\\mathbf{4.\;\;\;}\mathit{\frac{y d x-x d y}{y^{2}}=d\left(\frac{x}{y}\right)}\\\\\mathbf{5.\;\;\;}\mathit{\frac{x d y-y d x}{x y}=\frac{d y}{y}-\frac{d x}{x}=d\left[\log \left(\frac{y}{x}\right)\right]}

\\\mathbf{6.\;\;\;}\mathit{\frac{y d x-x d y}{x y}=d\left[\log \left(\frac{x}{y}\right)\right]}\\\\\mathbf{7.\;\;\;}\mathit{\frac{x d y-y d x}{x^{2}+y^{2}}=\frac{\frac{x d y-y d x}{x^{2}}}{1+\frac{y^{2}}{x^{2}}}=\frac{d\left(\frac{y}{x}\right)}{1+\left(\frac{y}{x}\right)^{2}}=d\left[\tan ^{-1}\left(\frac{y}{x}\right)\right]}\\\\\mathbf{8.\;\;\;}\mathit{\frac{d x+d y}{x+y}=d[\ln (x+y)]}\\\\\mathbf{9.\;\;\;}\mathit{d(\ln (x y))=\frac{x d y+y d x}{x y}}\\\\\mathbf{10.\;\;\;}\mathit{d\left(\frac{1}{2} \ln \left(x^{2}+y^{2}\right)\right)=\frac{x d x+y d y}{x^{2}+y^{2}}}

\\\mathbf{11.\;\;\;}\mathit{d\left(\frac{e^{y}}{x}\right)=\frac{x e^{y} d y-e^{y} d x}{x^{2}}}\\\\\mathbf{12.\;\;\;}\mathit{d\left(\frac{e^{x}}{y}\right)=\frac{y e^{x} d x-e^{x} d y}{y^{2}}}

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\text { yd } x-\left(x+2 y^{2}\right) d y =0 \\ \frac{y d x-x d y}{y^{2}} =2 d y \\\\ \int d\left(\frac{x}{y}\right) =\int 2 d y \\\\ \frac{x}{y}=2y+C \\ x= 2 y^{2}+Cy=f(y) \\ C =\pi \\ x =2 y^{2}+\pi

Posted by

Sanket Gandhi

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