<s>[INST] Basic Homogenous Differential Equation Solve the following differential equation
$$\frac{dy}{dx} = \frac{2x^2y+x^3}{y^3+2x^3}$$
I have identified that the differential equation is homogeneous because $f(rx,ry)$ = $r^3f(x,y)$ . I then used the substitution $u = y/x$ and $\frac{dy}{dx} = x\frac{du}{dx} + u$ reducing the equation to $$x\frac{du}{dx} = \frac{1}{2u^3} + \frac{1}{u^2}$$ I believe i have made a mistake somewhere in the middle of this and if not how would I go about solving this differential equation. (The only methods that have been presented in class are 1st order linear using mu substitution, separable, exact, homogeneous and Bernoulli).  If it helps my profesor wrote down a solution $y = \int \frac{v+1}{v^3+2}$dv (He did not have the dv but i believe he just forgot it so I added it)
 [/INST] On substituting $u=\frac{y}{x}$ in the differential equation $$\frac{dy}{dx} = \frac{2x^2y+x^3}{y^3+2x^3}$$
we get 
$$x\frac{du}{dx} + u = \frac{2u+1}{u^3+2}$$
or $$x\frac{du}{dx} = \frac{2u+1-u^4-2u}{u^3+2}$$
or $$\frac{u^3+2}{1-u^4}du = \frac{dx}{x}$$
or $$-\frac{1}{4}\cdot \frac{-4u^3}{1-u^4}du + \frac{2}{(1-u^2)(1+u^2)}du= \frac{dx}{x}$$
or $$-\frac{1}{4}\cdot \frac{d(1-u^4)}{(1-u^4)} + \frac{(1-u^2)+(1+u^2)}{(1-u^2)(1+u^2)}du= \frac{dx}{x}$$
or $$-\frac{1}{4}\cdot \frac{d(1-u^4)}{(1-u^4)} + \frac{du}{1+u^2} + \frac{du}{1-u^2}= \frac{dx}{x}$$
Integrating both sides, we get that
$$-\frac{1}{4}\cdot \ln|1-u^4| + \arctan u + \frac{1}{2}\ln|\frac{1+u}{1-u}|=\ln |x| + c$$
Hope this helps.
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