I am attempting to solve the differential equation: $(x+1)y''-(2-x)y'+y=0$ using power series, but I run into a problem when arriving at the recurrence to compute the coefficients of the power series solution.
Here is my work leading up to the mistake:
$$(x+1)y''-(2-x)y'+y=0$$
$$y=\sum_{n=0}^{\infty}C_nx^n, y'=\sum_{n=1}^{\infty}nC_nx^{n-1},y''=\sum_{n=2}^{\infty}n(n-1)C_nx^{n-2}$$
$$(x+1)\sum_{n=2}^{\infty}n(n-1)C_nx^{n-2}-(2-x)\sum_{n=1}^{\infty}nC_nx^{n-1}+\sum_{n=0}^{\infty}C_nx^n=0$$
$$\sum_{n=2}^{\infty}n(n-1)C_nx^{n-1}+\sum_{n=2}^{\infty}n(n-1)C_nx^{n-2}-2\sum_{n=1}^{\infty}nC_nx^{n-1}+\sum_{n=1}^{\infty}nC_nx^{n}+\sum_{n=0}^{\infty}C_nx^n=0$$
$$\sum_{k=1}^{\infty}k(k+1)C_{k+1}x^k+\sum_{k=0}^{\infty}(k+1)(k+2)C_{k+2}x^k-2\sum_{k=0}^{\infty}(k+1)C_{k+1}x^k+\sum_{k=1}^{\infty}kC_kx^k+\sum_{k=0}^{\infty}C_kx^k=0$$
$$2C_2-2C_1+C_0+\sum_{k=1}^{\infty}\left[ k(k+1)C_{k+1}+(k+1)(k+2)C_{k+2}-2(k+1)C_{k+1}+kC_k+C_k \right]x^k=0$$
$$2C_2-2C_1+C_0+\sum_{k=1}^{\infty}\left[ [k(k+1)-2(k+1)]C_{k+1}+[(k+1)(k+2)]C_{k+2}+[(k+1)]C_k \right]x^k=0$$
$$2C_2-2C_1+C_0+\sum_{k=1}^{\infty}\left[ [(k+1)(k-2)]C_{k+1}+[(k+1)(k+2)]C_{k+2}+[(k+1)]C_k \right]x^k=0$$
$$2C_2-2C_1+C_0=0,C_2=C_1-\frac{1}{2}C_0$$
$$[(k+1)(k-2)]C_{k+1}+[(k+1)(k+2)]C_{k+2}+[(k+1)]C_k=0$$
My Recurrence Relation is incorrect. It is apparently supposed to be:
$$[-(k+1)]C_{k+1}+[(k+1)(k+2)]C_{k+2}+[(k+1)]C_k=0$$
I am unable to find where I made a mistake but I suspect it happened when I combined the series into a single summation. If anyone could point out where I made a mistake I would be grateful. This has been bothering me all day.
Best Answer
Every time I see a question related to solving ODEs with varying coefficients via power series expansions I have a gut feeling it is cheating. The coefficients of the ODE are usually chosen in such a way that the power series ansatz leads to a simple recursion i.e. such that involves two terms only and can easily be solved in close form. This example is however different since as we can see the recursion involves three terms and we do not know how to solve it. Therefore we need to resort to a different method which I will outline right now.
As we know every linear second order ODE of the form: \begin{equation} y^{''}(x) + a_1(x) y^{'}(x) + a_2(x) y(x) = 0 \end{equation} can be reduced to the one without a term involving the first derivative by the following assumption $y(x) := \exp(-1/2 \int a_1(x) dx) v(x)$ with the function $v$ satisfying the following ODE: \begin{equation} v^{''}(x) + Q(x) v(x) =0 \end{equation} where $Q(x) := \left(-\frac{1}{4} a_1(x)^2 +a_2(x) -\frac{1}{2} a_1^{'}(x) \right)$.
In our case $a_1(x):= (x-2)/(x+1)$ and $a_2(x) := 1/(x+1)$ and $y(x):= \exp(-x/2) (1+x)^{3/2} v(x)$ and therefore $Q(x) = (-15+10 u-u^2)/(4 u^2)$ with $u:=(x+1)$.
This however resembles the confluent hypergeometric equation . Indeed an equation of the form: \begin{equation} v^{''}(x) + \left( \frac{-b^2+2 b+x(-4 a+2 b)-x^2}{4 x^2}\right)v(x)=0 \end{equation} is solved by $x^{b/2} \exp(-x/2)( C_1 U(a,b,x)+ C_2 M(a,b,x) )$ .
We therefore solve the equations: \begin{eqnarray} -15 &=& -b^2+2 b\\ +10&=& -4 a+2 b \end{eqnarray} which gives us $(a,b)=(-4,-3)$ or $(a,b)=(0,5)$. The first solution is spurious and the second leads to a result. We have:
\begin{equation} y(x) = \exp(-x) \cdot (1+x)^4 \end{equation}
The second independent solution needs to be found still. It can be found using the Wronskian though.