<s>[INST] Solve the differential equation $ 2xy' = y+ \frac{3}{2}x^2 $ Question: Solve the differential equation below 

$$ 2xy' = y+ \frac{3}{2}x^2 $$


Normal $y'= dy/x$ doest not work I think. 
Should I start it with substitution? $$ y' =u'x+u $$ $$y=\frac{u}{x} $$
But that method leads me to nothing, how I should do it?
 [/INST] $$y'-\frac{1}{2x}y=\frac{3}{4}x$$
$$\frac{y'}{\sqrt{x}}-\frac{1}{2x^{3/2}}y=\frac{3}{4}\sqrt{x}$$
$$\frac{d}{dx}\frac{y}{\sqrt{x}}=\frac{3}{4}\sqrt{x}$$
$$\frac{y}{\sqrt{x}}=\int\frac{3}{4}\sqrt{x}dx=\frac{3}{4}\frac{x^{3/2}}{3/2}+C=\frac{1}{2}x^{3/2}+C$$
$$y=\frac{1}{2}x^2+C\sqrt{x}$$
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