Show that every equation of the form $$ax'' + b(x^2 - 1) x' + cx = 0$$ where $a, b, c > 0$ can be transformed into a van der Pol equation by a change in the independent variable.
I am unable to find this replacement. If anyone could help me or give a hint I would be grateful.
By dividing through by $c$ and pulling a negative out of the $x^2$ term we get the equation:
$$\frac acx''-\frac bc(1-x^2)x'+x=0 $$
Now, let the dependent variable (usually $t$) be: $t=\sqrt{\frac ca} u$.
So now the derivatives via the chain rule will be:
$$ \sqrt{\frac ca}x' $$ And $$ \frac cax'' $$
So plug this in and we see that the $a/c$ term cancels and we are left with:
$$ x''-\frac{b\sqrt c}{c\sqrt a}(1-x^2)x'+x=0 $$
Then let $\mu=\frac{b\sqrt c}{c\sqrt a} $.
$$ x''-\mu(1-x^2)x'+x=0 $$
EDIT
From the variable change our functions are now $x\left(\sqrt{\frac ca} u\right)$.
The primed notation can be a little confusing sometimes, so what we really have is:
$$ \frac{d}{du}x\left(\sqrt{\frac ca} u\right)=x'\left(\sqrt{\frac ca} u\right)\frac{d}{du}\left(\sqrt{\frac ca} u\right)=\sqrt{\frac ca}x' $$
Then just repeat this process for the second derivative to obtain the $c/a$.