Solve ydx - xdy = x^2 y dx.

The correct answer is Option (2) → $y = Cx e^{-x^2/2}$ ##

Given that, $ydx - xdy = x^2 y dx$

$\Rightarrow \frac{1}{x^2} - \frac{1}{xy} \frac{dy}{dx} = 1 \quad \text{[dividing throughout by } x^2 y dx \text{]}$

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

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

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

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

which is a linear differential equation.

On comparing it with $\frac{dy}{dx} + Py = Q$, we get

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

$\text{I.F.} = e^{\int P dx} = e^{\int \left( x - \frac{1}{x} \right) dx}$

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

$= e^{\frac{x^2}{2}} \cdot e^{-\log x} \quad \left[ ∵-\log x = \log \left( \frac{1}{x} \right) \right]$

$\Rightarrow = e^{\frac{x^2}{2}} \cdot e^{\log \left( \frac{1}{x} \right)}$

$= \frac{1}{x} e^{\frac{x^2}{2}} \quad \left[ ∵e^{\log \left( \frac{1}{x} \right)} = \frac{1}{x} \text{ as } e^{\log x} = x \right]$

The general solution is $y \cdot \text{I.F.} = \int Q \cdot \text{I.F.} dx + C$

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

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

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