<s>[INST] exercise: ordinary differential equations I am struggling with an exercise. Can you please give me a hint?
Exercise:
Show that the solution curves of the differential equation:
$\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$, are of the form $x^3+y^3=3Cxy$.
I tried the substitution $u=y/x \rightarrow y=xu, \frac{dy}{dx}=u+x\frac{du}{dx}$.
Hence I get:
$u+x\frac{du}{dx}=-\frac{u(2-u^3)}{2u^3-1}$
This gives:
$2u^4-u+(2u^3-1)x\frac{du}{dx}=-2u+u^4$
$(2u^3-1)x\frac{du}{dx}=-(u^4+u)$
$\frac{(2u^3-1)}{(u^4+u)}\frac{du}{dx}=-\frac{1}{x}$
So I can atleast reduce the problem to a seperable differential equation, but I am not able to integrate the left side up. Do you have any tips?
 [/INST] $\bf hint:$  so you need to integrate $\int \frac{2u^3 - 1}{u^4 + u}$ we can do this by using partial fraction. here is how it goes 
$$ \frac{2u^3 - 1}{u^4 + u} = 
\frac{1}{2}\left(\frac{4u^3 +1}{u^4 + u}  - \frac{3}{u(u+1)(u^2-u+1)}\right)$$
now you need to find the constants $A, B, C$ and $D$ so that $$ \frac{3}{u(u+1)(u^2-u+1)} = \frac{A}{u} + \frac{B}{u+1} + \frac{C(2u-1)}{u^2-u+1} + \frac{D}{u^2 - u + 1} $$ is an identity. 
finding $A, B$ are easier. just by looking at the behaviour near $u = 0, u = -1.$ they turn out to be $A = 3, B = -1.$
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