Solve $\frac{dy}{dx} + 2xy = y$.

$y = Ce^{x - x^2}$

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

Given that, $\frac{dy}{dx} + 2xy = y$

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

$\Rightarrow \frac{dy}{dx} + (2x - 1)y = 0$

which is a linear differential equation.

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

$P = (2x - 1), Q = 0$

$\text{I.F.} = e^{\int P dx} = e^{\int (2x - 1) dx}$

$= e^{\left( \frac{2x^2}{2} - x \right)} = e^{x^2 - x}$

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

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

$\Rightarrow y \cdot e^{x^2 - x} = 0 + C$

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