The question from my notes involves solving the ODE in the form of: $$x^2y''+xpy'+qy=r$$
To solve this I set $y=e^t$ and I stated that: $$x^2\frac{\mathrm{d}^2y}{\mathrm{d} x^2}=\frac{\mathrm{d}^2y}{\mathrm{d} t^2}-\frac{\mathrm{d}y}{\mathrm{d} t}$$
But when I do this now I get: $$x^2\frac{\mathrm{d}^2y}{\mathrm{d} x^2}=\frac{\mathrm{d}^2y}{\mathrm{d} t^2}$$
I can't see where I got the extra term from?
Hint: solve first $$x^2y''+xpy'+qy=0$$
by setting $y=e^t$ and derive with respect to $t$. This will give you the $homogeneous$ solution $y_h$. Find after that the $particular$ solution by setting your $y_h$ in $$x^2y''+xpy'+qy=r$$
You will obtain the $particular$ solution $y_p$.
The solution of the differential equation will be
$$y=y_h+y_p$$
I think a much more easier way to solve this Euler equation is to try potential solutions for the homogeneous part of the solution:
$$y_h(x) = K x^n.$$
When substituted this in your original ODE with $r=0$ it yields:
$$ n(n-1) + p n + q = 0,$$
which is a quadratic equation for $n$. Solving for $n$ you will obtain (if $n_1 \neq n_2$):
$$y(x) = C x^{n_1} + D x^{n_2} + y_p(x),$$
where $C$ and $D$ are constants of integration and $y_p$ is the partircular solution. Since the non-homogenous term, $r$, seems to be constant, $y_p$ is then constant too. Prove that the particular solution is then:
$$y_p = r/q.$$
Cheers!
