# Thread: General solution in the form of a power series? HELP

1. ## General solution in the form of a power series? HELP

y'' - 3xy' - y = 0

The question asks me to find the general solution of the equation in the form of a power series around x = 0.

Ive been at this for a good while now and I cant seem to finish it.

I end up with a series split into a0's and a1's, the questions ive done before this all had the numerator equal to one so turning the result into a power series was easier. Im left with horrible numerators that i cant find a pattern to transform it into a power series. Im dont know how to write the maths symbols, heres my best effort, this is what i end up with

y = a0 + a1x + a0/2!x2 + 4a1/3!x3 + 7a0/4!x4 + 40a1/5!x5 + 91a0/6!x6 + ........

Have I gone wrong somewhere?

I think im on the right track, its just the numerator thats ruining my life currently, How to turn to power series? Any help would be greatly appreciated.

Nappy

2. Originally Posted by Nappy
y'' - 3xy' - y = 0

The question asks me to find the general solution of the equation in the form of a power series around x = 0.

Ive been at this for a good while now and I cant seem to finish it.

I end up with a series split into a0's and a1's, the questions ive done before this all had the numerator equal to one so turning the result into a power series was easier. Im left with horrible numerators that i cant find a pattern to transform it into a power series. Im dont know how to write the maths symbols, heres my best effort, this is what i end up with

y = a0 + a1x + a0/2!x2 + 4a1/3!x3 + 7a0/4!x4 + 40a1/5!x5 + 91a0/6!x6 + ........

Have I gone wrong somewhere?

I think im on the right track, its just the numerator thats ruining my life currently, How to turn to power series? Any help would be greatly appreciated.

Nappy
Here is how I did it.

$\displaystyle y=\sum_{n=0}^{\infty}C_nx^n$

$\displaystyle y'=\sum_{n=1}^{\infty}nC_nx^{n-1}$

$\displaystyle y'=\sum_{n=2}^{\infty}n(n-1)C_nx^{n-2}$

Plug into $\displaystyle y''-3xy'-y=0$

$\displaystyle \sum_{n=2}^{\infty}n(n-1)C_nx^{n-2}-3x\sum_{n=1}^{\infty}nC_nx^{n-1}-\sum_{n=0}^{\infty}C_nx^n\rightarrow$

$\displaystyle \sum_{n=2}^{\infty}n(n-1)C_nx^{n-2}-3\sum_{n=1}^{\infty}nC_nx^{n}-\sum_{n=0}^{\infty}C_nx^n$

k=n-2 and k=n so now we substitute

$\displaystyle \sum_{k=0}^{\infty}(k+2)(k+1)C_{k+2}x^{k}-3\sum_{k=1}^{\infty}kC_kx^{k}-\sum_{k=0}^{\infty}C_kx^k$

Since the ks start at different values, we need to up the indexing on k=0

$\displaystyle 2C_2+\sum_{k=1}^{\infty}(k+2)(k+1)C_{k+2}x^{k}-3\sum_{k=1}^{\infty}kC_kx^{k}-C_0-\sum_{k=1}^{\infty}C_kx^k$

Combine summations now.

$\displaystyle 2C_2-C_0+\sum_{k=1}^{\infty}(k+2)(k+1)C_{k+2}x^{k}-3kC_kx^k-C_kx^k\rightarrow$

$\displaystyle 2C_2-C_0+\sum_{k=1}^{\infty}x^k[(k+2)(k+1)C_{k+2}-C_k(3k+1)]$

$\displaystyle C_2=\frac{C_0}{2}$

$\displaystyle C_{k+2}=\frac{C_k(3k+1)}{(k+2)(k+1)}$

$\displaystyle k=1\rightarrow C_3=\frac{4C_1}{3*2}$

$\displaystyle k=2\rightarrow C_4=\frac{7C_2}{4*3}=\frac{7C_0}{2*3*4}$

$\displaystyle k=3\rightarrow C_5=\frac{10C_3}{5*4}=\frac{40C_1}{2*3*4*5}$

You should probably solve up to at least k=8.....

$\displaystyle y_1=C_0\bigg[1+\frac{x^2}{2!}+\frac{7x^4}{4!}+....\bigg]$

$\displaystyle y_2=C_1\bigg[???\bigg]$