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

Given that, $ydx - xdy = {x^2}ydx$

$\Rightarrow$ $\frac{1}{{{x^2}}} - \frac{1}{{xy}} \cdot \frac{{dy}}{{dx}} = 1$

[dividing throughout by ${x^2}ydx$]

$\Rightarrow$ $- \frac{1}{{xy}} \cdot \frac{{dy}}{{dx}} + \frac{1}{{{x^2}}} - 1 = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} - \frac{{xy}}{{{x^2}}} + xy = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} - \frac{y}{x} + xy = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} + \left( {x - \frac{1}{x}} \right)y = 0$

which is a linear differential equation.
On comparing it with

$\frac{{dy}}{{dx}} + Py = Q,$

we get
$P = \left( {x - \frac{1}{x}} \right),Q = 0$

${\rm{IF}} = {e^{\int P dx}}$
$= {e^{\int {\left( {x - \frac{1}{x}} \right)} dx}}$

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

$= {e^{\frac{{{x^2}}}{x}}},{e^{ - \log x}}$

$= \frac{1}{x}{e^{\frac{{{x^2}}}{2}}}$

The general solution is

$y \cdot \frac{1}{x}{e^{{x^2}/2}} = \int 0 \cdot \frac{1}{x}{e^{{x^2}/2}}dx + C$

$\Rightarrow$ $y \cdot \frac{1}{x}{e^{{x^2}/2}} = C$

$\Rightarrow$ $y = Cx{e^{ - {x^2}/2}}$