series solution...

**hunkydory19** · January 21st 2008, 12:35 PM

Find a series solution y(x) of the IVP

y'' -xy' + 5y = 0, y(0) = 0, y'(0) = 1

Show that for the given initial conditions the series terminates and hence the solution reduces to a fifth order polynomial.

I seriously do not have a clue where to start on this problem! Can anyone please give me any help at all?!

Thanks in advance!

**ThePerfectHacker** · January 21st 2008, 12:51 PM

Originally Posted by hunkydory19

Find a series solution y(x) of the IVP

y'' -xy' + 5y = 0, y(0) = 0, y'(0) = 1

Show that for the given initial conditions the series terminates and hence the solution reduces to a fifth order polynomial.

I seriously do not have a clue where to start on this problem! Can anyone please give me any help at all?!

Thanks in advance!

Let $y=\sum_{n=0}^{\infty}a_n x^n$ be a solution.

Then, $\sum_{n=2}^{\infty}n(n-1)a_n x^{n-2} - x\sum_{n=1}^{\infty} na_nx^{n-1} + 5\sum_{n=0}^{\infty} a_n x^n = 0$

Change index on the first, $\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n - \sum_{n=1}^{\infty} na_n x^n + \sum_{n=0}^{\infty} 5a_n x^n=0$

Evaluate 1st and 3rd sums at $n=0$ and combine, $2a_2 + 5a_0 + \sum_{n=1}^{\infty} [(n+2)(n+1)a_{n+2} - (n-5)a_n]x^n = 0$ .

This means,
(i) $2a_2 + 5a_0 = 0$
(ii) $(n+2)(n+1)a_{n+2} - (n-5)a_n = 0$ for $n\geq 1$ .

Note that (i) is implied from (ii) if we let $n=0$ .
Thus, we get the relation, $(n+2)(n+1)a_{n+2} - (n-5)a_n = 0 \implies a_{n+2} = \frac{(n-5)a_n}{(n+2)(n+1)}$ for $n\geq 0$ .

**hunkydory19** · January 21st 2008, 12:55 PM

Thank you SO much, genius!!

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