# Thread: finding terms of a power series solution of an IVP

1. ## finding terms of a power series solution of an IVP

Find the first six non-zero terms of the power series solution at
x = 0 of the following IVP.

$\displaystyle (x^2 + 1)y'' + xy' + 2xy = 0; y(0) = 2; y'(0) = 3$

So far I have:

supposing that $\displaystyle y=C_0+C_1x+C_2x^2+....$is the power series solution I have solved for $\displaystyle C_0=2$.

and from $\displaystyle y'=C_1+2C_2x+3C_3x^2+....$I have solved for $\displaystyle C_1=3$

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

I have got first and second derivative of the sum and have plugged into the original equation and got:

$\displaystyle y= \displaystyle\sum_{n=2}^{\infty}\ {n(n-1)C_nX^n}+ \displaystyle\sum_{n=2}^{\infty}\ {n(n-1)C_nX^(n-2)}+ \displaystyle\sum_{n=1}^{\infty}\ {nC_nX^n} - 2\displaystyle\sum_{n=0}^{\infty}\ {C_nX^(n+1)}=0$

But now I get confused as to what to do next to have the same power of x as what my sum begins with.

Can anyone tell me how I could break down my sums to solve for my 6 terms.

Thank you!!!

2. Now you need to COMBINE the sums into one big sum. Be careful though, because it's easy to make mistakes with the coefficients! If you are not sure, the way to do it is to make a change on the index so that the X has the same power in each series. After that, choose the lowest common index (say n = 2 or something like that) for all the series and convert each series to that index by taking out the terms with lower indices, i.e.:

$\displaystyle$\displaystyle \sum_{n=0}^{\infty}a_nx^n = \sum_{n=3}^{\infty}a_nx^n +a_0+a_1x+a_2x^2

Once you combine the series you will get some big series and since it's equal to 0, it must be equal to 0 for all x. This can only happen if all the coefficients are 0. Then set the big coefficient of the combined series to 0 and now you have a recurrence relation, which you can use to calculate the coefficients!