# Power series differential equation

1. Mar 13, 2006

### 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?

Last edited: Mar 13, 2006
2. Mar 13, 2006

### HallsofIvy

Staff Emeritus
It's not clear to me what you are assuming. You talk about the Fn being Fibonnacci numbers but then treat the Fn 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 Fns 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 Fn 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.

Last edited: Mar 13, 2006
3. Mar 13, 2006

### 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.

Last edited: Mar 13, 2006