Solve This Ordinary Differential Equation How to solve this ordinary differential equation:
$$
(y^{3}+4e^{x}y)dx + (2e^{x}+3y^2)dy = 0
$$
This is from the book Fundamentals of Differential Equations by Nagle, Saff and Sniders. The eight edition on the page 77 problem 12.
I appreciate any hints.
 A: Hint:
$(y^3+4e^xy)~dx+(2e^x+3y^2)~dy=0$
$(2e^x+3y^2)~dy=-(y^3+4e^xy)~dx$
$(3y^2+2e^x)\dfrac{dy}{dx}=-y^3-4e^xy$
Let $u=y^2$ ,
Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$
$\therefore\dfrac{3y^2+2e^x}{2y}\dfrac{du}{dx}=-y^3-4e^xy$
$(3y^2+2e^x)\dfrac{du}{dx}=-2y^4-8e^xy^2$
$(3u+2e^x)\dfrac{du}{dx}=-2u^2-8e^xu$
This belongs to an Abel equation of the second kind.
A: $$(y^{3}+4e^{x}y)dx + (2e^{x}+3y^2)dy = 0\to\\
(y^{3}+4e^{x}y)+ (2e^{x}+3y^2)y^{'} = 0$$
Consider substitution $y=te^{Ax}$, where $t-$function of $x$, $A-$some constant.
$$
\left(t^3e^{3Ax}+4te^{(1+A)x}\right)+(2e^x+3t^2e^{2Ax})(t^{'}e^{Ax}+Ate^{Ax})=0\to\\
e^{3Ax}\left(t^3+4te^{(1-2A)x}\right)+e^{3Ax}(3t^2+2e^{(1-2A)x})(t^{'}+At)=0
$$
Notice that equation has the simplest form if we choose $A=\frac{1}{2}$. Also, as far as $A$ is finite, we can omit multiplicator $e^{3Ax}$. 
$$
\left(t^3+4t\right)+(3t^2+2)\left(t^{'}+\frac{t}{2}\right)=0\to t^{'}=-\frac{t(t^2+4)}{3t^2+2}-\frac{t}{2}\to\\
\frac{2(3t^2+2)}{5t(t^2+2)}dt=-dx
$$
In terms of partial fractions:
$$
\frac{2}{5}\left(\frac{1}{t}+\frac{2t}{t^2+2}\right)dt=-dx
$$
Now it is easy to integrate
$$
\frac{2}{5}\left(\ln(t)+\ln(t^2+2)\right)=-x+C\to\frac{2}{5}\ln(t(t^2+2))=C-x\to\\
(t^3+2t)^{\frac{2}{5}}=e^{C-x}
$$
Back to the initial variables $t=ye^{-\frac{1}{2}x}$
$$
(y^3e^{-\frac{3}{2}x}+2ye^{-\frac{1}{2}x})^{\frac{2}{5}}=e^{C-x}\to\\
(y^3e^{-x}+2y)^{\frac{2}{5}}=e^{C-\frac{4x}{5}}
$$
A: You need to find the intergration factor, which in this case comes out to be e^x . Multiply whole equation with it and you see that you have perfect differentials 
