How to solve these two ODE's? For each one of them $y=y\left(x\right)$
$\text{(1) }\left(2x-y^{2}\right)\cdot y'=2y$
$\text{(2) }\left(x^{3}+e^{y}\right)\cdot y'=3x^{2}$
I suspect they might be "of the same kind".
What I tried is looking on this
$\left(y\cdot\left(2x-y^{2}\right)\right)'=y'\cdot\left(2x-y^{2}\right)+y\left(2-2y\cdot y'\right)=\underset{\text{=0 for a sol}}{\underbrace{y'\cdot\left(2x-y^{2}\right)-2y}}+2y\left(2-y\cdot y'\right)$
and this
$\left(y\cdot\left(x^{3}+e^{y}\right)\right)'=y'\cdot\left(x^{3}+e^{y}\right)+y\cdot\left(3x^{2}+e^{y}\cdot y'\right)$
But i don't really know if that helps
Any suggestions?
Thanks in advance
 A: For the first equation
$$\text{(1) }\left(2x-y^{2}\right)\cdot y'=2y$$
It's far easier to use $x'=\frac {dx}{dy}$ instead
$$\left(2x-y^{2}\right)\frac {dy}{dx}=2y  \implies \left(2x-y^{2}\right)=2y\frac {dx}{dy}$$
$$(2x-y^{2})=2yx'$$
$$yx'-x=-\frac 12y^2$$
$$\frac {yx'-x}{y^2}=-\frac 12$$
$$(\frac xy)'=-\frac 12$$
Integrate simply
$$(\frac xy)=-\frac 12\int dy$$
$$\frac xy=-\frac 12y+K$$
$$\boxed {x(y)=-\frac 12y^2+Ky}$$

First equation can also be solved easily this way :
$$(2x-y^{2})y'=2y$$
$$2(xy'-y)=y^{2}y' \implies y-xy'=-\frac 12y^{2}y'$$
$$ \left (\frac xy\right )'=-\frac 12y'$$
Integrate
$$ \left (\frac xy\right )=-\frac 12y+K$$
Therefore
$$ x(y)=-\frac 12y^2+Ky$$

For the second equation
$$\text{(2) }\left(x^{3}+e^{y}\right)\cdot y'=3x^{2}$$
$$\left(x^{3}+e^{y}\right)\frac {dy}{dx}=3x^{2} \implies \left(x^{3}+e^{y}\right)=3x^{2}\frac {dx}{dy}$$
$$x^{3}+e^{y}=3x^{2}x'$$
$$x^{3}+e^{y}=(x^3)'$$
Substitute $z=x^3$
$$z+e^{y}=z' \implies z'-z=e^y$$
Multiply both sides by $e^{-y}$
$$(ze^{-y})'=1$$
$$\boxed {x^3e^{-y}-y=K}$$
A: Wr6titing your first equation in the form
$$-2y(x)+(2x-y(x)^2)y'(x)=0$$
we need an integrating factor $$\mu(y)$$
and we get
$$2y\frac{d\mu(y)}{dx}-2\mu(y)=2\mu(y)$$
from here we get  $$\mu(y)=\frac{1}{y^2}$$
multiplying our equation by $\mu(y)$
$$-\frac{2}{y(x)}+\left(\frac{2x}{\mu(x)^2}-1\right)\frac{dy(x)}{dx}=0$$
so
$$f(x,y)=\int \frac{2}{y}dx=-\frac{2x}{y}+g(y)$$
to compute $g(y)$
$$\frac{\partial f(x,y)}{\partial y}=\frac{2x}{y(x)^2}+\frac{dg(y)}{dy}$$
so $$g(y)=-\int dy=-y$$
and we get
$$y=\frac{1}{2}\left(-\sqrt{-8x+C_1^2}-C_1\right)$$
or
$$y=\frac{1}{2}\left(\sqrt{-8x+C_1^2}-C_1\right)$$
to Your second equation 
writing
$$-3x^2+(e^{y(x)}+x^3)\frac{dy}{dx}=0$$
computing an integrating factor
$$-3\frac{d\mu(y)}{dy}x^2=3\mu(y)x^2$$
so $$\mu(y)=e^{\mu(y)}$$ and we get
$$-3e^{-y(x)}x^2+(e^{y(x)}x^3+1)\frac{dy(x)}{dx}=0$$
the equation is exact:
$$f(x,y)=\int -3e^{-y}x^2dx=e^{-y}x^3+g(y)$$
computing $g(y)$
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial}{\partial y}(-e^{-y}x^3+g(y))=e^{-y}x^3+\frac{dg(y)}{dy}$$
so
$$g(y)=y$$
and
$$f(x,y)=-e^{-y}x^3+y$$
and the solution is given by
$$-e^{-y}x^3+y=C_1$$
