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

  1. #1
    s3a
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    Finding coefficients for a power series solution for a differential equation (Trig)

    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!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Finding coefficients for a power series solution for a differential equation (Tri

    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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