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?
 A: $$2xy'=y+\frac { 3 }{ 2 } x^{ 2 }\\ 2xy'-y=\frac { 3 }{ 2 } x^{ 2 }\\ 2xy'-y=0\\ 2x\frac { dy }{ dx } =y\\ 2\int { \frac { dy }{ y }  } =\int { \frac { dx }{ x }  } \\ \ln { \left| y \right| =\ln { \left| Cx \right|  }  } \\ { y }^{ 2 }=Cx\\ y=C\sqrt { x } \\ y=C\left( x \right) \sqrt { x } \\ { y }^{ \prime  }={ C }^{ \prime  }\left( x \right) \sqrt { x } +\frac { C\left( x \right)  }{ 2\sqrt { x }  } \\ 2x\left( { C }^{ \prime  }\left( x \right) \sqrt { x } +\frac { C\left( x \right)  }{ 2\sqrt { x }  }  \right) =C\left( x \right) \sqrt { x } +\frac { 3 }{ 2 } { x }^{ 2 }\\ 2x\sqrt { x } { C }^{ \prime  }\left( x \right) =\frac { 3 }{ 2 } { x }^{ 2 }\\ { C }^{ \prime  }\left( x \right) =\frac { 3\sqrt { x }  }{ 4 } \\ { C }\left( x \right) =\frac { 1 }{ 2 } x\sqrt { x } +{ C }_{ 1 }\\ y=C\left( x \right) \sqrt { x } =\sqrt { x } \left( \frac { 1 }{ 2 } x\sqrt { x } +{ C }_{ 1 } \right) =\frac { { x }^{ 2 } }{ 2 } +{ C }_{ 1 }\sqrt { x } \\ \\ $$
A: $$y'-\frac{1}{2x}y=\frac{3}{4}x$$
$$\frac{y'}{\sqrt{x}}-\frac{1}{2x^{3/2}}y=\frac{3}{4}\sqrt{x}$$
$$\frac{d}{dx}\frac{y}{\sqrt{x}}=\frac{3}{4}\sqrt{x}$$
$$\frac{y}{\sqrt{x}}=\int\frac{3}{4}\sqrt{x}dx=\frac{3}{4}\frac{x^{3/2}}{3/2}+C=\frac{1}{2}x^{3/2}+C$$
$$y=\frac{1}{2}x^2+C\sqrt{x}$$
