ordinary differential equations test the exactness of the O.D.E $(4xy+2x^2 y)dx+(2x^3+3y^2)dy=0$ and hence find the potential function which is the general solution.I tried to solve it and I reached ending up failing to get the integrating factor.please help me
 A: Consider $$I(x,y)dx + J(x,y)dy = 0.$$ If this is an exact ODE we should be able to find a potential function $F(x,y)$ such that $$\frac{\partial}{\partial x}F(x,y) = I(x,y) = 4xy + 2x^2y$$
and 
$$\frac{\partial}{\partial y}F(x,y) = J(x,y) = 2x^3 + 3y^2.$$
We also know by Clairaut's theorem that the mixed partial derivatives should be equal, i.e. $$\frac{\partial^2}{\partial x \partial y} F(x,y) = \frac{\partial^2}{\partial y \partial x} F(x,y).$$ 
Equivalently,
$$\frac{\partial}{\partial y} I(x,y) = \frac{\partial}{\partial x} J(x,y).$$
This is very useful, since it makes it easy to check whether or not an ODE is exact.
$$\frac{\partial}{\partial y} I(x,y) = 4x + 2x^2$$
$$\frac{\partial}{\partial x} J(x,y) = 6x^2$$
These two expressions are not equal, so we conclude that this is not an exact ODE.
A: Hint:
$(4xy+2x^2y)dx+(2x^3+3y^2)dy=0$
$(2x^3+3y^2)dy=-((2x^2+4x)y)dx$
$(2x^3+3y^2)\dfrac{dy}{dx}=-(2x^2+4x)y$
Let $u=y^2$ ,
Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$
$\therefore\dfrac{(2x^3+3y^2)}{2y}\dfrac{du}{dx}=-(2x^2+4x)y$
$(2x^3+3y^2)\dfrac{du}{dx}=-(4x^2+8x)y^2$
$(2x^3+3u)\dfrac{du}{dx}=-(4x^2+8x)u$
This belongs to an Abel equation of the second kind.
Let $v=u+\dfrac{2x^3}{3}$ ,
Then $u=v-\dfrac{2x^3}{3}$
$\dfrac{du}{dx}=\dfrac{dv}{dx}-2x^2$
$\therefore3v\left(\dfrac{dv}{dx}-2x^2\right)=-(4x^2+8x)\left(v-\dfrac{2x^3}{3}\right)$
$3v\dfrac{dv}{dx}-6x^2v=-(4x^2+8x)v+\dfrac{8x^4(x+2)}{3}$
$3v\dfrac{dv}{dx}=(2x^2-8x)v+\dfrac{8x^4(x+2)}{3}$
$v\dfrac{dv}{dx}=\dfrac{(2(x-2)^2-8)v}{3}+\dfrac{8x^4(x+2)}{9}$
Let $s=x-2$ ,
Then $\dfrac{dv}{dx}=\dfrac{dv}{ds}\dfrac{ds}{dx}=\dfrac{dv}{ds}$
$\therefore v\dfrac{dv}{ds}=\dfrac{(2s^2-8)v}{3}+\dfrac{8(s+2)^4(s+4)}{9}$
