<s>[INST] Solve the differential equation IVP 
Solve $\displaystyle x \frac{dy}{dx} = 2y + x^2, y(1) = 2$

We can use the homogeneous form, $y = vx$, so $dy/dx = xdv/dx + v$
$x^2dv/dx + vx = 2x(v + x) \implies x^2 dv/dx = vx + x \implies x dv/dx = v + 1 \implies (v+1) dv = xdx \implies v^2/2 + v = x^2 + C$ 
Then we would just substitute $v = y/x$
But is there a simpler method?
 [/INST] \begin{eqnarray}
xy'&=&2y+x^2\\
x^2y'-2xy&=&x^3\\
\dfrac{x^2y'-2xy}{x^4}&=&\dfrac{1}{x}\\
\left(\dfrac{y}{x^2}\right)'&=&\dfrac{1}{x}\\
\int\left(\dfrac{y}{x^2}\right)'&=&\int\dfrac{1}{x}\\
\dfrac{y}{x^2}&=&\ln x+C\\
y&=&x^2\ln x+2x^2
\end{eqnarray}
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