Originally Posted by
HallsofIvy Step 1: take a Calculus class and learn to integrate!
You say yourself this equation is "separable"- so separate it. $\displaystyle \frac{dy}{dx}= \frac{y^3+ 2y}{x^2+ 3x}$ becomes $\displaystyle \frac{dy}{y^3+ 2y}= \frac{dx}{x^2+ 3x}$.
$\displaystyle \frac{dy}{y(y^2+ 2)}= \frac{dx}{x(x+ 3)}$
Now use "partial fractions" as you learned in Calculus to integrate those. That is what SlipEternal did.
$\displaystyle \frac{1}{y(y^2+ 2)}= \frac{A}{y}+ \frac{By+ C}{y^2+ 2}$
Multiply both sides by $\displaystyle y(y^2+ 2)$ to get $\displaystyle 1= A(y^2+ 2)+ (By+ C)y$. You need three equations to determine the three unknowns, A, B, and C, so substitute three different values for y to get them. If y= 0, 1= 2A. If y= 1, 1= 3A+ B+ C. If y= -1, 1= 3A+ B- C.
From 1= 2A, A= 1/2. The other two equations become 1= (3/2)+ B+ C or B+ C= 1- 3/2= -1/2 and 1= (3/2)+ B- C or B- C= 1- 3/2= -1/2. Adding those two equations 2B= -1 so B= -1/2. Subtracting 2C= 0 so C= 0 as slip eternal said.
$\displaystyle \frac{1}{y(y^2+ 2)}= \frac{1/2}{y}- \frac{1/2}{y^2+ 2}$.
On the other side you have $\displaystyle \frac{1}{x(x+ 3)}= \frac{A}{x}+ \frac{B}{x+ 3}$. (Of course, these are different "A" and "B" than before.)
Multiply on both sides by $\displaystyle x(x+ 3)$ to get $\displaystyle 1= A(x+ 3)+ Bx$. If x= 0, that becomes $\displaystyle 1= 3A$ so $\displaystyle A= \frac{1}{3}$. If x= -3, that becomes $\displaystyle 1= -3B$ so $\displaystyle B= -\frac{1}{3}$
$\displaystyle \frac{1}{x(x+ 3)}= \frac{1/3}{x}- \frac{1/3}{x+ 3}$.
So your equation finally reduces to
$\displaystyle \frac{1}{2}\frac{dy}{y}- \frac{1}{2}\frac{dy}{y^2+ 2}= \frac{1}{3}\frac{dx}{x}- \frac{1}{3}\frac{dx}{x+ 3}$.
Can you integrate that?