Help to know what kind of differential equation is this I've to solve this differential equation:
$
(2y^3-2y^2+3x)dx+(3xy^2-2xy)dy=0
$
But I even don't know what kind of differential equation is it.
Please, any help?
Thanks in advance.
 A: As I've mentioned in a comment above, this is a differential equation reducible to an exact differential equation.
A differential equation of the form $M(x,y)dx +N(x,y)dy=0$ is said to be an exact D.E. if it's left hand member is the exact differential of some function $U(x,y)$
Or, $$du=M(x,y)dx +N(x,y)dy=0 $$Thus, it's solution is $U(x,y)=c$
Ex. Consider the D.E.  $xdy +ydx=0$
Note that, $d(xy)= xdy + ydx$
Therefore, the solution of the given D.E. is $xy=c$

*

*A necessary condition for a D.E. $M(x,y)dx +N(x,y)dy=0$ to be exact is $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

*The solution of an exact differential equation $M(x,y)dx +N(x,y)dy=0$ is given by:
$$\int_{y=constant} Mdx + \int N(y)dy=c$$ where in the first part of LHS, we are integrating the expression $M(x,y)$ treating y as a constant, and $N(y)$ are the terms of $N(x,y)$ which contain only y.

I will not be providing the proof for the points 1 and 2, but you can check  it out here
Even if it may be a bit hard to follow, it will definitely make life easier with respect to solving most differential equations.
Sometimes a differential equation which is not exact may become so, on multiplication by a suitable function known as the integrating factor ().
The IF can be obtained as follows:
1.If  $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} }{N}=g(x)$ (function of x alone), then IF=$e^{\int g(x)dx}$
2.If  $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} }{M}=h(y)$ (function of y alone), then IF=$e^{\int -h(y)dy}$
There are more ways to obtain the IF, but for your question, the first one will do.
SOLUTION:
$$(2y^3-2y^2+3x)dx+(3xy^2-2xy)dy=0$$
$$M(x,y)=2y^3-2y^2+3x$$
$$\frac{\partial M}{\partial y}= 6y^2-4y$$
$$N(x,y)=3xy^2 -2xy$$
$$\frac{\partial N}{\partial x}= 3y^2-2y$$
$$\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}= 3y^2 - 2y$$
$$\frac{\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}}{N}= \frac{3y^2 - 2y}{3xy^2-2xy}= \frac{1}{x}$$
As$\frac{1}{x}$ is a function of x alone, the IF would be:
$$I.F.= e^{\int \frac{1}{x}dx}=x$$
Multiplying the equation throughout by x, we get:
$$(2xy^3-2xy^2+3x^2)dx + (3x^2 y^2 -2x^2y)dy=0$$
This will definitely satisfy the first condition I've mentioned(in the beginning)
Thus, it's solution will be:
$$\int_{y=constant} (2xy^3-2xy^2+3x^2)dx=c$$
As you can see, I haven't taken the second integral, as there is no term of y, independent of x.
$$x^2y^3-x^2y^2+x^3=c$$
This is the general solution of the given differential equation.
A: $$(2y^3-2y^2+3x)dx+(3xy^2-2xy)dy=0$$
Rewrite it as:
$$2(y^3-y^2)dx+3xdx+xd(y^3-y^2)=0$$
$$2wdx+3xdx+xdw=0$$
Where $w=y^3-y^2$. It's easier to solve now. The integrating factor is obvious:
$$\mu(x)=x$$
$$(wdx^2+x^2dw)+3x^2dx=0$$
$$dwx^2+3x^2dx=0$$
