This is an initial value problem.(adsbygoogle = window.adsbygoogle || []).push({});

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.

[tex]y=\sum_{n=1}^\infty\frac{F_{n}}{n!}x^n[/tex]

[tex]y'=\sum_{n=2}^\infty\\n\frac{F_{n}}{n(n-1)!}x^{n-1}[/tex]

[tex]y''=\sum_{n=3}^\infty\\n(n-1)\frac{F_{n}}{n(n-2)(n-1)!}x^{n-2}[/tex]

[tex]\sum_{n=0}^\infty\\(n+3)(n+2)\frac{F_{n+3}}{(n+3)(n+2)!}x^{n+1}[/tex][tex]=\sum_{n=2}^\infty\\n\frac{F_{n}}{n(n-1)!}x^{n-1}[/tex][tex]+\sum_{n=1}^\infty\frac{F_{n}}{n!}x^n[/tex]

[tex]\sum_{n=0}^\infty\\(n+3)(n+2)\frac{F_{n+3}}{(n+3)(n+2)!}x^{n+1}[/tex][tex]=\sum_{n=0}^\infty\\(n+2)(n+1)\frac{F_{n}}{(n+2)!}x^{n+1}[/tex][tex]+\sum_{n=0}^\infty\frac{F_{n}}{(n+1)!}x^{n+1}}[/tex]

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?

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# Power series differential equation

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