For those that are interested. This problem is from the homework. The problem was made up by the professor. It is not from the differential equations book.I just want to see what you think about this problem and if so, is it correct. The problem has an initial value which is y(0) = 1, and y'(0) = 2. The professor wants at least five terms of the solution.

$\displaystyle

y'' + x^3 y' - x y = 0

$

First, you would assume a solution in the form of:

$\displaystyle

y = \sum_{n = 0}^{ \infty} a_n x^n

$

From here you can find the first and second derivatives of this solution:

$\displaystyle

\Rightarrow y' = \sum_{n = 1}^{ \infty} n a_n x^{n - 1}

$

$\displaystyle

\Rightarrow y'' = \sum_{n = 2}^{ \infty} n(n - 1) a_n x^{n - 2}

$

Once I have found my derivatives I will plug them back to my original equation:

$\displaystyle

\sum_{n = 2}^{ \infty} n(n - 1) a_n x^{n - 2} + x^3 \sum_{n = 1}^{ \infty} n a_n x^{n - 1} - x \sum_{n = 0}^{ \infty} a_n x^n = 0

$

$\displaystyle

\Rightarrow \sum_{n = 2}^{ \infty} n(n - 1) a_n x^{n - 2} + \sum_{n = 1}^{ \infty} n a_n x^{n + 2} - \sum_{n = 0}^{ \infty} a_n x^{n + 1} = 0

$

Now we want all powers to be n.

$\displaystyle

\Rightarrow \sum_{n = 0}^{ \infty} (n + 2) (n + 1) a_{n + 2} x^n + \sum_{n = 3}^{ \infty} (n - 2) a_{n - 2} x^n - \sum_{n = 1}^{ \infty} a_{n - 1} x^n = 0

$

Now we want to get the indexes the same. The starting points for the series have to be same. The only way to do that here is to evaluate the second and third term of the series that start at n = 3 and n = 1 and write the remainder of the series as the summation from n = 3 onward, so we can get all the indexes to 1.

Now combine all the series into one:

$\displaystyle

\Rightarrow 2a_2 + 6a_3 x + 12a_4 x^2 - a_0 x - a_1 x^2 + \sum_{n = 3}^{ \infty} [(n + 2) (n + 1) a_{n + 2} + (n - 2) a_{n - 2} - a_{n - 1})] x^n = 0

$

Now we want all the coefficients to be zero.

$\displaystyle

a_0 = 1, a_1 = 2

$

$\displaystyle

2a_2 = 0 \implies a_2 = 0

$

$\displaystyle

2a_2 + x (6a_3 - a_0) + x^2 (12a_4 - a_1) = 0

$

$\displaystyle

6a_3 = a_0 \implies a_3 = \frac {1}{6}

$

$\displaystyle

12a_4 = a_1 \implies a_4 = \frac {1}{6}

$

$\displaystyle

a_5 = \frac {a_2}{20} - \frac {a_1}{20} = - \frac {1}{10}

$

$\displaystyle

a_6 = \frac {a_3}{30} - \frac {2a_2}{30} = \frac {1}{180}

$

$\displaystyle

y_G = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5 + a_6 x^6 + ....

$

$\displaystyle

y_G = 1 + 2x + \frac {x^3}{6} + \frac {x^4}{6} - \frac {x^5}{10} + \frac {x^6}{180} + ....

$

My Recursive Formula:

$\displaystyle

\boxed { a_{n + 2} = \frac {a_{n - 1} - (n - 2) a_{n - 2}}{(n + 2) (n + 1)} }

$

I have corrected my mistakes check to see if they are correct.