# Solving Recurrence Relation for Series Solution of an ODE

I am trying to solve the below problem:

Assume $y = \sum_{n=0}^{\infty}a_nx^n$ is a solution to $(x-1)y''-(x-3)y'-y=0$. Find $a_n$.

I took both derivatives of $y$, plug them into the equation, modify the indices until each series has the same $x^n$, and take terms out of the series such that they have the same index to bring all the terms under the same summation. I arrive at

$\sum_{n=0}^{\infty}[(n+1)(n)a_n-(n+1)(n+2)a_{n+2} - na_n+3(n+1)a_{n+1}-a_n]x^n + (-2a_2+3a_1-a_0)=0$

Therefore $$a_2 = \frac{3a_1}{2}-\frac{a_0}{2}$$ and $$a_{n+2} = \frac{(n+3)a_{n+1}-a_n}{(n+2)}, n = 1,2,3...$$

I also find some terms as $$a_3 = \frac{4a_2}{3}-a_1 = \left(\frac{4\cdot 3}{3\cdot 2} - 1 \right) a_1-\frac{4}{3\cdot 2}a_0$$ $$a_4 = \left(\frac{5\cdot 4\cdot 3}{4\cdot 3\cdot 2} - \frac{13}{8} \right) a_1-\left(\frac{5\cdot 4}{4\cdot 3\cdot 2} - \frac{1}{8} \right)a_0$$ $$a_5 = \left(\frac{6\cdot 5\cdot 4\cdot 3}{5\cdot 4\cdot 3\cdot 2} - \frac{53}{40} \right) a_1-\left(\frac{6\cdot 5\cdot 4}{5\cdot 4\cdot 3\cdot 2} - \frac{17}{60} \right)a_0$$

I can see a pattern for the first set of coefficients, but not the next. For example the first part for $a_1$ is $a_n = \frac{n+1}{2}$

Any help would be greatly appreciated. Thanks!

Following along up until the difference equation for the coefficients leads one to $$a_{n+2} = \frac{(n+3) \, a_{n+1} - a_{n}}{n+2} \hspace{5mm} n \geq 0.$$ Now, for a few values of $n$ it is discovered that \begin{align} a_{2} &= \frac{3 a_{1} - a_{0}}{2!} \\ a_{3} &= \frac{4 a_{2} - a_{1}}{3} = \frac{10 a_{1} - 4 a_{0}}{3!} \\ a_{4} &= \frac{5 a_{3} - a_{2}}{4} = \frac{41 a_{1} - 17 a_{0}}{4!}. \end{align} The general form follows as $$a_{n} = \frac{\alpha_{n} \, a_{1} - \beta_{n} \, a_{0}}{n!} \hspace{5mm} n \geq 0.$$ Further values of $n$ will lead to the discovery of $\alpha_{n} \in \{ 0, 1, 3, 10, 41, 206, 1237, \cdots \}$ which can be seen as $\lfloor (e-1) \, n! \rfloor$, or sequence A002627 of the Oeis. Similarly $\beta_{n} = \lfloor (e-2) \, n! \rfloor$ and $\beta_{n} \in \{ 1, 0, 1, 4, 17, 86, 517, \cdots \}$ and is sequence A056542.
From this the solution can then be seen as $$f(x) = a_{1} \, \sum_{n=0}^{\infty} \lfloor (e-1) \, n! \rfloor \, \frac{x^{n}}{n!} - a_{0} \, \sum_{n=0}^{\infty} \lfloor (e-2) \, n! \rfloor \, \frac{x^{n}}{n!}.$$
This could also be seen in the form: $$f(x) = a_{0} - a_{0} \, \sum_{n=2}^{\infty} \lfloor (e-2) \, n! \rfloor \, \frac{x^{n}}{n!} + a_{1} \, \sum_{n=1}^{\infty} \lfloor (e-1) \, n! \rfloor \, \frac{x^{n}}{n!}.$$
• Thanks for the reply. I would never had thought of this form. But to achieve a non-zero $a_0$ term, don't we need to start the sums at $n=1$ and append the $a_0$ constant? Then the final sum would look like $$f(x) = a_0 + a_{1} \, \sum_{n=1}^{\infty} \lfloor (e-1) \, n! \rfloor \, \frac{x^{n}}{n!} - a_{0} \, \sum_{n=1}^{\infty} \lfloor (e-2) \, n! \rfloor \, \frac{x^{n}}{n!}.$$ – Titan May 22 '18 at 1:12
Since Leucippus identified the proper sequences, if $$a_{n+2} = \frac{(n+3) \, a_{n+1} - a_{n}}{n+2}$$ the coefficients can write $$a_n=(2 a_0 -a_1) +e \,(a_1 -a_0 )\, \frac{\Gamma (n+1,1)}{n!}$$ where appears the incomplete gamma function.