Solve the following differential equation: $xy''-y'+k^2y=0$

Question

It is not an Euler equation. I tried power series, but I can't develop them around $x=0$. I also tried dividing Bessel's equation by $x$, but it doesn't seem to be possible to arrive at my equation. Any hints would be much appreciated.

Answer 1

As the analytic solution of this ODE involves Bessel functions of first and second kind, a power series approach is the only way out. Let's transform our initial expression :

$$xy'' - y' + c^2y = 0 \Rightarrow y'' - \frac{y'}{x}+\frac{c^2}{x}y=0$$

Since for $x=0$ we have an undeterminate form, we recall the Frobenius Method, which is exactly suited for such cases. For this particular example, $q(x) = k^2$.

We now seek a power series solution of the form $$y(x) = \sum_{k=0}^\infty A_kz^{k+r}, \quad A_0 \neq 0$$ By differentiating, we yield : $$y'(x) = \sum_{k=0}^\infty (k+r)A_kz^{(k+r)-1} \quad \text{and} \quad y''(x) = \sum_{k=0}^\infty [(k+r)-1](k+r)A_kz^{(k+r)-2}$$

By substituting now in the ODE, you should be able to yield a reccurence relation involving $\{A_k\}$.

Answer 2

If I have not made substitution errors we should get by substituting (I'm not writing the details, just the odes I arrived to on paper)

$y(x)=u\left(2k\sqrt{x}\right)\iff u(x)=y\left(\left(\dfrac{x}{2k}\right)^2\right)$

the new ode written below : $$x(u''+u)=(2k^2+1)u'$$

Now set $n=k^2+1$ and $u(x)=x^n\,v(x)$

You get this ode : $$x^2v''+xv'+(x^2-n^2)v=0$$

Which is the Bessel differential equation:
http://mathworld.wolfram.com/BesselDifferentialEquation.html

Whose solution is $v(x)=c_1J_n(x)+c_2H_n(x)$

Explaining the result you get in the other answer comments by Rebellos.