Question

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

Answer 1

Given that, $ydx-xdy=x^{2}ydx$
$\Rightarrow \frac{1}{x^2}-\frac{1}{xy}.\frac{dy}{dx}=1$ [dividing throught by $x^{2}ydx$]
$\Rightarrow -\frac{1}{xy}.\frac{dy}{dx}+\frac{1}{x^2}-1=0\Rightarrow \frac{dy}{dx}-\frac{xy}{x^2}+xy=0\Rightarrow \frac{dy}{dx}-\frac{y}{x}+xy=0\Rightarrow \frac{dy}{dx}+(x-\frac{1}{x})y=0$
which is a linear differential equation.
On comparing it with $\frac{dy}{dx}+Py=Q,$ we get
$P=(x-\frac{1}{x}),Q=0IF=e^{\int Pdx}=e^{\int (x-\frac{1}{x})dx}=e^{\frac{x^2}{2}-logx}=e^{\frac{x^2}{x}},e^{-logx}=\frac{1}{x}{e}^{\frac{x^2}{2}}$
The general solution is
$y.\frac{1}{x}{e}^{x^2/2}=\int 0.\frac{1}{x}{e}^{x^2/2}dx+C\Rightarrow y.\frac{1}{x}{e}^{x^2/2}=C\Rightarrow y=Cx{e}^{-x^2/2}$