We're using Kreyszig's Advanced Engineering Mathematics in the second part of my "online" Applied Analysis course. He introduces the theory, with minimal examples, then often assigns exercises that, when I learn the solution, leave me wondering how on earth he expected me to figure out the technique he had in mind.
I'm struggling with the following problem, for instance. We're asked to solve the differential equation $x^2y''+xy'+(4x^4-1)y=0$. My lecturer said to find a real number such that $z=x^{p}$ reduces the DE to a Bessel equation for some function in $z$, but I'm struggling to see how to do it. I was hoping someone could help.
When substituting a variable, use the chain rule. If we define $z=x^p$, then using the chain rule we get that (see the wiki here for more information https://en.wikipedia.org/wiki/Chain_rule#Higher_derivatives) \begin{align} \frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dz}{\mathrm dx}\frac{\mathrm dy}{\mathrm dz}=pz^{1-1/p}\frac{\mathrm dy}{\mathrm dz}, \end{align} (not sure how much you've done with ODE's, I calculate $z'$, then substitute $x=z^{1/p}$ to get $z'$ in terms of $z$, he same goes for the higher order derivatives) and that \begin{align} \frac{\mathrm d^2y}{\mathrm dx^2}&=\left(\frac{\mathrm dz}{\mathrm dx}\right)^2\frac{\mathrm d^2y}{\mathrm dz^2}+\frac{\mathrm d^2z}{\mathrm dx^2}\frac{\mathrm dy}{\mathrm dz}\\ &=p^2z^{2(1-1/p)}\frac{\mathrm d^2y}{\mathrm dz^2}+p(p-1)z^{1-2/p}\frac{\mathrm dy}{\mathrm dz}. \end{align} Substituting the two give that \begin{align} z^{2/p}\left(p^2z^{2(1-1/p)}\frac{\mathrm d^2y}{\mathrm dz^2}+p(p-1)z^{1-2/p}\frac{\mathrm dy}{\mathrm dz}\right)+z^{1/p}\cdot pz^{1-1/p}\frac{\mathrm dy}{\mathrm dz}+(4z^{4/p}-1)y=0. \end{align} Collect your terms, \begin{align} p^2z^{2}\frac{\mathrm d^2y}{\mathrm dz^2}+p^2z\frac{\mathrm dy}{\mathrm dz}+(4z^{4/p}-1)y=0. \end{align} Note that $p$ could be any number (except 0 or 1), so choose it based on what you need. Edit: $p$ could be 1 it just wouldn't do anything.