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?
 A: $\bf hint:$  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 $A = 3, B = -1.$
A: Use partial fractions
$$\begin{align}
\frac{2u^{3} - 1}{u^{4} + u} &= \frac{2u^{3} - 1}{u(u^{3} + 1)} \\
&= \frac{A}{u} + \frac{Bu^{2} + Cu + D}{u^{3} + 1} \\
\implies 2u^{3} - 1 &= (A + B)u^{3} + Cu^{2} + Du + A \\
\end{align}$$
Equating coefficients
$$\begin{align}
A &= -1 \\
B &= 3 \\
\end{align}$$
Hence
$$\begin{align}
\frac{2u^{3} - 1}{u^{4} + u} &= \frac{3u^{2}}{u^{3} + 1} - \frac{1}{u}
\end{align}$$
Integrating
$$\begin{align}
\int \bigg(\frac{3u^{2}}{u^{3} + 1} - \frac{1}{u} \bigg)du &= - \int \frac{1}{x} dx \\
\implies \ln(u^{3} + 1) - \ln(u) &= -\ln(x) + K_1 \\
\implies \ln \bigg( \frac{u^{3} + 1}{u} \bigg) &= - \ln(x) + K_1 \\
\implies \frac{u^{3} + 1}{u} &= \frac{K_2}{x} \\
\end{align}$$
With 
$$\begin{align}
u &= \frac{y}{x} \\
\implies \frac{y^{2}}{x^{2}} + \frac{x}{y} &= \frac{K_2}{x} \\
\implies y^{3} + x^{3} &= K_2xy
\end{align}$$
A: Of course, it is possible to solve the ODE, but it is not what is wanted.
The question is : Show that the solution is on the given form. 
Proving that the given solution is correct is not the same as finding the solution : It is a waste of time to try to solve the ODE since it is already solved and the solution known.
It is sufficient to bring back the known solution : $x^3+y^3=3Cxy$ into the ODE $\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$  and show that it agrees.
The differentiation of  $x^3+y^3=3Cxy$ leads to :
$$3x^2 dx+3y^2dy=3Cy dx+3Cx dy$$
$$\frac{dy}{dx}=-\frac{3x^2+3Cy}{3y^2-3Cx}$$
From $x^3+y^3=3Cxy$ we have $3C=\frac{x^3+y^3}{xy}$. Use it to eliminate $C$. 
A: here is a cheap way of doing this knowing that $x^3 + y^3 = Cxy$ is a solution. we will make the change of variables $$u = x^3+y^3, t = xy$$ 
the deifferential equation satisfied by $u$ and $t$ are:
$$\frac{du}{dx} = 3x^2 + 3y^2 \frac{dy}{dx} = \frac{3(y^6-x^6)}{x(2y^3 - x^3}, \frac{dt}{dx} = x\frac{dy}{dx} + y = \frac{3y(y^3-x^3)}{2y^3 - x^3}  $$ 
we can divide it out and get a separable equation $$\frac{du}{dt} = \frac{u}{t} $$ which has solution of the form $$u = Ct, x^3 + y^3 = Cxy$$
A: Left hand side simplifies by integration:
$ \int \dfrac {2 u^3 - 1} {u(u^3 +1)} du  = log  \dfrac{(1+u^3)}{u} + const. $ 
