ODE Substitution Problem Have some practice exam questions and answers to work through, I'm almost there but get stuck.
what we have is:

$$
\begin{cases}
\dfrac{dx}{dt}=\dfrac{x^{3}-xt^{2}}{t^{3}}  \\ 
x(1)=2 \\
\end{cases}
$$

and the (general solution) answer is:
$x(t) = \pm t\sqrt{\frac{2}{1-Dt^{4}}}$
now I figured that this is a substitution problem, using $y=\frac{x}{t}$ so that
$\frac{dx}{dt}=y^{3}-y$
and
$\frac{dy}{dt}=y+t\frac{dy}{dy}$
giving:
$y^{3}-2y = t\frac{dy}{dt}$
$\int{\frac{1}{y^3-2y}dy}=\int{\frac{1}{t}dt}$
$\ln{y^{3}-2y}=\ln{t}+C$
$y^{3}-2y=Dt$
and therefore:
$\frac{x^{3}}{t^{3}}-2\frac{x}{t}=Dt$
$x^3-2xt^{2}=Dt^{4}$
>
Here is where I get stuck... so I reverse engineered the solution to find that:
$x^{3}-2xt^{2}=x^{3}Dt^{4}$
and am now completely lost... where did that extra $x^{3}$ on the RHS of the equation come from??
 A: Assume that $\frac{dy}{y^3-2y}=\frac{dt}{t}$ is done alright. The anti derivative of $\frac{1}{t}dt$ is indeed $ln|t|$, however the anti derivative of $\frac{dy}{y^3-2y}$ cannot be $ln|y^3-2y|$ for the very simple reason, the derivative of $ln|y^3-2y|$ is not $\frac{1}{y^3-2y}$, in fact it is $\frac{3y^2-2}{y^3-2y}$ by Chain Rule!. So how to do $\int\frac{dy}{y^3-2y}$? The generic answer is using partial fraction decomposition (although this one could be done in a different way as well, see end of post).
So setting up the fractions (through factoring): $\frac{1}{y^3-2y}=\frac{A}{y}+\frac{B}{y-\sqrt{2}}+\frac{C}{y+\sqrt{2}}$. Multiply through by $y^3-2y$ and equating each equation with corresponding coefficients (which is 0,0,1 respectively from the numerator) gives $A+B+C=0,B\sqrt{2}-C\sqrt{2}=0,-2A=1$ results in $A=-0.5,B=1/4,C=1/4$ and you can now integrate $\int\frac{-0.5}{y}$+$\int\frac{0.25}{y-\sqrt{2}}$+$\int\frac{0.25}{y+\sqrt{2}}$$dy$. And done.   Now in THIS (I repeat THIS!) particular case, you could have subbed in $\int\frac{dy}{y^3-2y}$the substitution  $y=\frac{1}{v}$ and your integral would have become $\int\frac{-vdv}{1-2v^2}$ (verify!) which straight away would have become an $ln$ term. I will now leave it up to you to work out the rudiments. Good luck!
