Solve the following differential equation: $xy''-y'+k^2y=0$ 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. 
 A: 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 $\{A_k\}$.
A: 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.
