Generating Function for weird Fibonacci sequence

Question

I couldn't solve the following excercise from my course of Differential Equations. The question is to find a power series solution for the differential equation: \begin{equation} y''=y'+y, y(0)=0, y'(0)=1 \text{ of the form: } y(x)=\sum_{n=0}^{\infty} \frac{F_n}{n!}x^n \end{equation}

With $F_n$ the n-th number in the sequence of Fibonacci.

I was able to write this power series with a recursive relation: \begin{equation} (n)(n-1)a_n=(n-1)a_{n-1}+a_n \text{ or }a_n=\frac{a_{n-1}}{n}+\frac{a_n}{n(n-1)} \end{equation}

I did try to write $a_{n-1}$ as the sum of his antecedents, which gives \begin{equation} \frac{2a_{n-2}(n-2)+a_{n-3}}{n(n-1)(n-2)} \end{equation} Continuation does indeed bring the Fibonacci series, since: \begin{equation} F_n=F_{n-1}+F_{n-2}=2\cdot F_{n-2}+F_{n-3} \end{equation} And continuing of the recursive relation all the way would give the same result as $F_n/n!$ since the factor $(n-2)$ would become one , but how would I go about proving this more accurately?

Answer 1

Start with your (corrected) $(n)(n-1)a_n=(n-1)a_{n-1}+a_{n-2} $ for $y(x) =\sum_{n=0}^{\infty} a_nx^n $. Multiplying by $(n-2)!$ gives $n!a_n=(n-1)!a_{n-1}+(n-2)!a_{n-2} $. Finally, setting $n!a_n = b_n$ gives $b_n = b_{n-1}+b_{n-2} $, the Fibonacci recurrence.

If $y(0) = 0$ then $a_0 = 0$ so $b_0 = 0$.

If $y'(0) = 1$ then $a_1 = 1$ so $b_1 = 1$ so that $b_n = F_n$ (with $F_{0..3} = 0, 1, 1, 2$) and $a_n =\dfrac{b_n}{n!} =\dfrac{F_n}{n!} $.

Note that we need to have $y'(0)$ specified.