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Thread: Confusing series solution

  1. #1
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    Confusing series solution

    Hello everyone!

    Here's a DE: $\displaystyle y' -5x^4y=3x^9$.
    Letting $\displaystyle y=\sum_{n=0}^{\infty} c_n x^n$ and after some algebra, we get:
    $\displaystyle c_1+2c_2x+3c_3x^2+4c_4x^3-3x^9$ $\displaystyle +\sum_{k=0}^{\infty}(\left (k+5)c_{k+5}-5c_k \right)\,x^{k+4}=0$.

    does this mean that $\displaystyle c_1=c_2=c_3=c_4=0$?
    what about $\displaystyle -3x^9$? And is the relation $\displaystyle (k+5)c_{k+5}-5c_k$ even valid.

    Solved the DE w\out power series, and I checked it, I got $\displaystyle y=-\frac{3}{5}(x^5+1)$

    How to do the algebra in the series solution?
    Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    The DE is...

    $\displaystyle y^{'} = 5 x^{4} y + x^{9}$ (1)

    ... and we are searching for a solution of the form...

    $\displaystyle y(x)= c_{0} + c_{1} x + c_{2} x^{2} \dots + c_{n} x^{n} + \dots$ (2)

    ... so that is...

    $\displaystyle y^{'} (x) = c_{1} + 2 c_{2} x + \dots + n c_{n} x^{n-1} + \dots$ (3)

    Now in order to proceed we suppose to know $\displaystyle y(0)=c_{0}$. Inserting (2) and (3) in (1) we obtain first...

    $\displaystyle c_{1} = c_{2} = c_{3} = c_{4} = 0$ (4)

    ... and for $\displaystyle n=5$ ...

    $\displaystyle 5 c_{5} x^{4}= 5 c_{0} x^{4} \rightarrow c_{5}=c_{0}$ (5)

    Increasing n we obtain then...

    $\displaystyle c_{6} = c_{7} = c_{8} = c_{9} = 0$ (6)

    ... and for $\displaystyle n=10$ ...

    $\displaystyle 10 c_{10} x^{9}= (5 c_{5}+1) x^{9} \rightarrow c_{10}=\frac{5 c_{0}+1}{10}$ (7)

    All other $\displaystyle c_{n}$ are 0 so that is...

    $\displaystyle y(x)= c_{0} + c_{0} x^{5} + \frac{5 c_{0}+1}{10} x^{10} $ (8)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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