Construct Two Linearly Independent Power Series Solutions to $(1+z^2)u''+3zu'+u=0$

Question

I am trying to construct two linearly independent, power series solutions to the ODE $$(1+z^2)u''+3zu'+u=0$$

My attempt:

Let $$u=\sum_{k=0}^{\infty}A_kz^k\implies u'=\sum_{k=1}^{\infty}kA_kz^{k-1}\implies u''=\sum_{k=2}^{\infty}k(k-1)A_kz^{k-2}$$

Substituting this into the ODE, I find that $$\sum_{k=0}^{\infty}\left((k+2)(k+1)A_{k+2}+(k+1)^2A_k\right)z^k=0$$ Hence $u$ is a solution iff $$A_{k+2}=-\frac{k+1}{k+2}A_k \ \ \ k\geq 0$$ I do not see how to proceed, especially considering there are no intial conditions to find $A_0$ and $A_1$. Any advice would be greatly appreciated.

Answer 1

Take $A_0=0$ and $A_1=1$ and compute one solution. Then take $A_0=1$ and $A_1=0$ and compute another. You will get two linearly independent solutions.

Answer 2

It seems there is a mistake in your reccurent relation $$A_{k+2}=-\frac{1}{k+2}A_k \ \ \ k\geq 0$$ I get from your last equality

$$A_{2n}=(\frac {-1}2)^n\frac {A_0}{n!}$$ $$\implies u_1(z)=\sum_{n=0}^\infty (\frac {-z^2}2)^n\frac {A_0}{n!}$$ Which is not exact...Check again your last equality

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Edit from this $$A_{k+2}=-\frac{k+1}{k+2}A_k \ \ \ k\geq 0$$ you evaluate $A_2,A_4,A_6$ $$A_2=-\frac 12A_0$$ $$A_4=-\frac 34A_1=\frac {1\times 3}{2 \times 4}A_0$$ $$A_6=-\frac {1\times 3 \times 5}{2 \times 4 \times 6}A_0$$ Can you deduce the pattern ?

In the denominator you have that $$2\times 4 \times 6.... =2 \times 2 \times 2 \times ...... \times 1 \times 2  \times 3..=2^nn!$$

So that $$A_{2n}=(-1)^n \frac {1 \times 3 \times 5 ...}{2^n(n!)}A_0$$ Do the same for the numerator you have $$1\times 3 \times 5 \times ...$$ Wehave that $$1\times 3 \times 5 \times ..= \frac {1 \times 2 \times 3 \times 4 \times 5 \times 6} {2\times 4 \times 6}=\frac {(2n)!}{2^n (n!)}$$ Finally $$\boxed {A_{2n}=(-1)^n \frac {(2n)!}{2^{2n}(n!)^2}A_0}$$ $$u_1=\sum_ {n=0}^\infty(-1)^n \frac {(2n)!}{(n!)^2}(\frac {z}{2})^{2n}$$

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Note that you can check the solution by direct integration

$$(1+z^2)u''+3zu'+u=0$$ $$(1+z^2)u''+2zu'+zu'+u=0$$ $$((1+z^2)u')'+(zu)'=0$$ integrate $$(1+z^2)u'+zu=K_1$$