# Transforming equations of the form $ax''+b(x^2-1)x'+cx=0$ into van der Pol equations

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.

• Maybe you could include Van der Pol's equation in the post. – coffeemath Jul 5 '14 at 23:21
• Van Der Pol's equation is $x''-\mu(1-x^2)x'+x=0$ @coffeemath – ClassicStyle Jul 5 '14 at 23:51

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 $c/a$.

• Do not quite understand how you used the chain rule there. – Croos Jul 6 '14 at 1:08
• I'll update it shortly. Let me know if you still don't get it. – ClassicStyle Jul 6 '14 at 1:25
• I understand, thanks for the help – Croos Jul 6 '14 at 1:53