# Find the general solution of $xy' - 2y = x^2$

$(A)\;y = x \;(\log | x| - c)\\ (B)\;y = x \;(\log | x| + c) \\(C)\; y = x^2\;(\log | x| + c) \\ (D)\; y = x^2\;(\log | x| - c)$

The standard form of this equation is $y' + P (x)y = Q(x)$
$y' - \large\frac{2}{x}$$y=x \Rightarrow P (x) = \large\frac{-2}{x} \rightarrow \large\int$$ P(x) dx = \large \int \frac{2}{x}$$dx = - \log x^2 e$$^{ \int \normalsize P(x) dx}$ $= e^{-\log x^2} = \large\frac{1}{x^2}$
Multipling the standard form equation by $\large\frac{1}{x^2}$, we get $\large\frac{y'}{x^2}$$- \large\frac{2y}{x^3}$$ = \large\frac{1}{x}$
$\Rightarrow \large\frac{d}{dx}$$\bigg (\large\frac{y}{x^2} \bigg) =\large\frac{1}{x} \Rightarrow \large\frac{y}{x^2} = \large\int \frac{1}{x}$$dx = \log | x| + c$
$\Rightarrow y = x^2\;(\log | x| + c$)