Power Series Solution to a Diff EQ

1. Mar 24, 2008

[SOLVED] !!!Power Series Solution to a Diff EQ!!!

1. The problem statement, all variables and given/known data
Find the first 5 term of a Power series solution of

$$y'+2xy=0$$ (1)

Missed this class, so please bear with my attempt here.

3. The attempt at a solution

Assuming that y takes the form

$$y=\sum_{n=0}^{\infty}c_nx^n$$

Then (1) can be written:

$$\sum_{n=1}^{\infty}nc_nx^{n-1}+2x\sum_{n=0}^{\infty}c_nx^n=0$$

Re-written 'in phase' and with the same indices (in terms of k):

$$c_1+\sum_{k=1}^{\infty}(k+1)c_{k+1}x^k+\sum_{k=1}^{\infty}2c_{k-1}x^k=0$$

$$\Rightarrow c_1+\sum_{k=1}^{\infty}[(k+1)c_{k+1}+2c_{k-1}]x^k=0$$

Now invoking the identity property, I can say that all coefficients of powers of x are equal to zero (including $c_1*x^0$)

So I can write:

$c_1=0$ and

$$c_{k+1}=-\frac{2c_{k-1}}{k+1}$$

Now I am stuck (I know I am almost there though!)

Should I just start plugging in numbers for k=1,2,3,4,5 ? Will this generate enough 'recursiveness' to solve for the 1st five terms?

Is that the correct approach?

Thanks!!

2. Mar 24, 2008

Don't worry guys, I got it. And for those who might make future use of this thread, my approach was correct. Solving

$$c_{k+1}=-\frac{2c_{k-1}}{k+1}$$

For k=1,2....,9 generates enough coefficients to write out the first five nonzero terms of the solution by plugging them back into

$$y=\sum_{n=0}^{\infty}c_nx^n$$

Last edited: Mar 24, 2008
3. Mar 24, 2008

Kreizhn

I think you need one more initial condition on your series (example $y(0)=x_0$). You need to define $c_0$ (or you can just leave $c_0$ as the 'integration' constant of the ODE). However, the recursion formula becomes pretty obvious once you have that. If you can't see it right away, try plugging in a few values. Since $c_1=0$ what can we say about all the odd labelled coefficients? You should get something that looks like the series of an exponential function. In fact, the series will have a closed form solution if you can see how your answer relates to the exponential series.