Question

Solve $ydx-xdy=x^{2}ydx$

Answer 1

We have,

$ydx-xdy=x^{2}ydx$

$ydx-x^{2}ydx=xdy$

$y(1-x^2)ydx=xdy$

$\left(\frac{1-x^2}{x}\right)dx=\frac{dy}{y}$

$(\frac{1}{x}-\frac{x^2}{x})dx=\frac{dy}{y}$

$\frac{1}{x}dx-\frac{x^2}{x}dx=\frac{dy}{y}$

$\frac{1}{x}dx-xdx=\frac{dy}{y}$

on integrating we get,

$\log x-\frac{x^2}{2}=\log y+\log C$

$\log x-\frac{x^2}{2}=\log yC$

$\log yC-\log x=-\frac{x^2}{2}$

$\log \frac{yC}{x}=-\frac{x^2}{2}$

$\frac{yC}{x}=e^{-\frac{x^2}{2}}$

$yC=x{e}^{-\frac{x^2}{2}}$

$y=\frac{e^{-\frac{x^2}{2}}}{C}$

$y=C_1{e}^{-\frac{x^2}{2}}\therefore C_1=\frac{1}{C}$

Hence this is the answer.