How to separate this differential equation? I am having trouble separating this differential equation
$$xy' + y = x^2 \sqrt{y}$$
I've gotten as far as
$$\frac{1}{\sqrt{y}} dy - \sqrt{y} \frac{dx}{x} = x \;dx$$
but I can't finish it. It obviously isn't an exact equation, either.
 A: Observe that $(\sqrt{y})' = \frac12 y' / \sqrt{y}$. So you can re-write your equation, assuming $y\neq 0$, as
$$ 2 x (\sqrt{y})' + \sqrt{y} = x^2 $$
Now observe that
$$ (\sqrt{x}\sqrt{y})' = \frac12 \frac{1}{\sqrt{x}} \sqrt{y} + \sqrt{x} (\sqrt{y})'$$
this means that your equation can be re-written as
$$ 2\sqrt{x} \left( \sqrt{xy}\right)' = x^2 $$
which you can solve by directly integrating. 
A: Let us rewrite differential equation in the form :
$\frac{dy}{dx}+\frac{1}{x} \cdot y =x \cdot y^{\frac{1}{2}}$ 
Now, if we make substitution $v=y^{\frac{1}{2}}$ we can write :
$v'= \frac{1}{2}\cdot y^{\frac{-1}{2}}\cdot y'\Rightarrow \frac{dv}{dx}+\frac{1}{2} \cdot \frac{1}{x} \cdot v=\frac{1}{2} \cdot x$
The last equation is linear differential equation in terms of variable $v$ .
A: I'd put $y = u^2$ to get
$$2xu' + u = x^2$$
Then investigate a solution of the form $u = Ax^2$
$$4Ax^2 + Ax^2 = x^2$$
Thus you have $A = \frac{1}{5}$ and finally $y = \frac{x^4}{25}$
Do you need the general solution too?
