MHB How do you solve the Euler ODE $x^2y'' + xy' - n^2y = 0$ using $y=x^p$?

[ognik]

Given $ x^2y'' + xy' - n^2y = 0 $

I think this is an Euler ODE, so I try $y=x^p, \therefore y'=p x^{p-1}, \therefore y''= p (p-1) x^{p-2}$

Substituting: $x^p p(p-1) + x^p p - n^2 x^p = 0, \therefore p^2 = n^2, \therefore p= \pm n$

$ \therefore y=C_1 x^n + C_2 x^{-n} $?

 

[MarkFL]

I will check your result using another approach to Cauchy-Euler equations, and that is to make the substitution:

$$x=e^t\implies\d{y}{t}=x\d{y}{x}\implies x^2\d{^2y}{x^2}=\d{^2y}{t^2}-\d{y}{t}$$

And so the given ODE becomes:

$$\d{^2y}{t^2}-n^2y=0$$

The characteristic equation is:

$$r^2-n^2=(r+n)(r-n)=0$$

And so the general solution is:

$$y(t)=c_1e^{-nt}+c_2e^{nt}$$

Back-substituting for $x$, we then obtain:

$$y(x)=c_1x^{-n}+c_2x^{n}$$

And this agrees with your result. :)