I can't find similar worked questions anywhere, can someone help me with this Power Series Method for Differential Equations Question?

Question

I am trying to figure out the Power Series Method for DE's with Initial Value Problems. In my Textbook, all of the Worked Examples have the form:

y'' + p(x)y' + g(x)y = 0; y(0) = c1, y'(0) = c2 (just an example)

The Questions I am struggling with are the ones that are of the Form:

p(x)y'' + q(x)y' + r(x)y = 0; y(0) = c1, y'(0) = c2

Nowhere can I find a work question of similar form to the second function, not in my Textbook, not on the Internet and I can't talk to my Lecturers because they have only pointed me to worked examples of the first equation.

This is the Question I am working on with instructions:

Use the Power Series Method to solve the Initial Value Problem:

(1-x^2)y'' - 6xy' - 4y = 0; y(0)=1, y'(0) = 2

I'd appreciate it if this exact Question were to NOT be answered as I wish to answer it myself. Instead, could someone provide me a worked example (with steps) for a Initial Value Problem of similar Form Please.

Answer 1

The Ansatz is $$y(x)=\sum_{k=0}^\infty a_k x^k\tag{1}$$ with unknown $a_k$. The initial conditions give $$a_0=1,\qquad a_1=2\ .$$ Now introduce $(1)$ into the given ODE: $$(1-x^2)\sum_{k=2}^\infty k(k-1)a_k x^{k-2}-6x \sum_{k=1}^\infty ka_k x^{k-1}-4\sum_{k=0}^\infty a_kx^k=0\ .$$ Arrange the indexing of the sums such that "all is in terms of $x^k\>$": $$\sum_{k=0}^\infty (k+2)(k+1)a_{k+2}x^k-\sum_{k=2}^\infty k(k-1)a_k x^k-\sum_{k=1}^\infty 6k a_k x^k-4\sum_{k=0}a_k x^k=0\ .$$ Now compare coefficients of $x^k$ for each exponent $k\geq0$. For $k=0$ and $k=1$ you obtain special equations, but for all $k\geq2$ you obtain $$(k+2)(k+1)a_{k+2}-k(k-1)a_k-6 ka_k-4a_k=0\ ,$$ or $$(k+2)(k+1)a_{k+2}=(k^2+5k+4)a_k\qquad(k\geq2)\ ,$$ and this can still be simplified. In this way you obtain a recursion for all $a_k$. Compute the first $a_k$ by hand, until you see the general law for them.