Finding coefficients for a power series solution for a differential equation (Trig)

**s3a** · December 3rd 2011, 03:33 PM

The question and my work are attached. I'm having trouble doing the trig parts since I am trying to get all the sums to have an x^n term so can I build my recurrence relation but I'm stuck at getting the trig stuff to have the x^n term.

Could someone please show me what to do?

Any help would be greatly appreciated!
Thanks in advance!

**chisigma** · December 7th 2011, 01:22 AM

The solution of the second order DE...

$y^{''}-\sin x\ y = \cos x,\ y(0)=-5,\ y^{'}(0)=-9$ (1)

... is supposed to be analytic in x=0 so that we can write...

$y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}=\sum_{n=0}^{\infty} a_{n}\ x^{n}$ (2)

The computation of the derivatives in x=0 is performed as follows. The initial conditions give us $a_{0}=-5$ and $a_{1}=-9$ . From (1) we obtain directly...

$y^{''}= \sin x\ y + \cos x \implies y^{''}(0)= 1 \implies a_{2}=\frac{1}{2}$ (3)

Deriving (3) we obtain...

$y^{'''}= \sin x\ y^{'} + \cos x\ y - \sin x \implies y^{'''}(0)= -5 \implies a_{3}=-\frac{5}{6}$ (4)

The computation of the successive derivatives is left to You...

Kind regards

$\chi$ $\sigma$

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