Power series differential equation

[Zem]

This is an initial value problem.
y''=y'+y, y(0)=0 and y(1)=1
I am solving it by deriving the Fibonnacci power series.

The sum of F_n from n=0 to infinity is 0,1,1,2,3,5,8,13 of Fibonnacci numbers defined by F_o = 0, F_1 =1. F_n = F_n-2 + F_n-1 for n>1.

Here is what I have attempted so far.
$$y=\sum_{n=1}^\infty\frac{F_{n}}{n!}x^n$$
$$y'=\sum_{n=2}^\infty\\n\frac{F_{n}}{n(n-1)!}x^{n-1}$$
$$y''=\sum_{n=3}^\infty\\n(n-1)\frac{F_{n}}{n(n-2)(n-1)!}x^{n-2}$$
$$\sum_{n=0}^\infty\\(n+3)(n+2)\frac{F_{n+3}}{(n+3)(n+2)!}x^{n+1}$$$$=\sum_{n=2}^\infty\\n\frac{F_{n}}{n(n-1)!}x^{n-1}$$$$+\sum_{n=1}^\infty\frac{F_{n}}{n!}x^n$$
$$\sum_{n=0}^\infty\\(n+3)(n+2)\frac{F_{n+3}}{(n+3)(n+2)!}x^{n+1}$$$$=\sum_{n=0}^\infty\\(n+2)(n+1)\frac{F_{n}}{(n+2)!}x^{n+1}$$$$+\sum_{n=0}^\infty\frac{F_{n}}{(n+1)!}x^{n+1}}$$

Is this correct so far? I am not sure how to derive the factorials in the denominators. If this is right, is my next step to solve for F_n+3 by dividing both sides by (n+3)(n+2), multiplying by (n+3)(n+2)!, and dividing by x^n+1?

 

[HallsofIvy]

It's not clear to me what you are assuming. You talk about the F_n being Fibonnacci numbers but then treat the F_n as unknowns. Oh, I see, you want to show that those coefficients are Fibonnacci numbers.
I also do not understand why you want to get xⁿ⁺¹ as the power rather than just n or why you have switched the index from starting at 1 to starting at 0 (Since y(0)= 0 the x⁰ coefficient is clearly 0).
Assuming you have reason for doing those things, and by "Fibonnacci series" you just mean the F_ns are recursively defined, you have have done is mostly correct. There area a few minor typos- especially there should not be a factor (n+2) in the second sum of the last equation and you didn't change the "n" on F_n when you made the last index changes. Then you can cancel terms in the first two sums:
Since (n+2)!= (n+2)(n+1)!, in the first sum
$$(n+3)(n+2)\frac{F_{n+3}{(n+3)(n+2)!= \frac{(n+3)(n+2)}{(n+3)(n+2)}\frac{F_}n+3}{(n+1)!}$$
(what I typed is not showing up here- click on it to see the correct code.)
and in the second
$$(n+1)\frac{F_{n+2}}{(n+2)!}= \frac{F_{n+2}}{(n+1)!}$$
so
$$\sum_{n=0}^\infty\\ \frac{F_{n+3}}{(n+1)!}x^{n+1}= \sum_{n=0}^\infty\\ \frac{F_{n+2}}{n!}x^{n+1}+ \sum_{n=0}^\infty\\F_{n+1} x^{n+1}$$
and all of the factorials, as well as xⁿ⁺¹, cancel.

 

[Zem]

1,000 apologies. It is nice to see how the Fibonacci series would be done. But I understand now that I can use the power series
$$\sum_{n=0}^\infty\\C_{n}x^{n}$$
because a theorem in the chapter states that it is the same series as the Fibonacci series because they represent the same function. I think this means any equation with derivated functions such as y'' = y' + y can be represented by the above power series. So I can use that simpler series, instead of the one the book provided in the problem. So I don't yet understand why the problem includes the Fib series. Here is how the problem reads.

For the initial value problem y'' = y' + y, y(0)=0, y(1)=1 derive the power series solution
$$\sum_{n=1}^\infty\frac{F_n}{n!}x^{n}$$
where [F_n] from n=0 to infinity is the sequence 0,1,1,2,3,5,8,13,... of Fibonacci numbers defined by F_o = ), F_1 = 1, F_n = F_n-2 + F_n-1 for n>1.

In answer to your questions.. I set the index for y at n=1 because that is where the Fib series starts. I thought x^(n+1) would be the correct power because it derives to x^(n-2) for y''. And the index is n=3 for y'', so to change it to n=0, I added 3 to the power of x to make it x^(n+1). It worked out to x^(n+1) for the x's on the right side, too.

I will work on the C_n power series, but also have a question about the Fibonacci one. The left side equation has (n+1)! in the denominator. Is that because it derives to (n-2)!, then add 3 to make it (n+1)! ? Does something similar happen in the denominators on the right side?

Thank you.