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Find a series solution to $(x^2-2)y''+6xy'+4y=0$.

A. Find the recurrence relation to $a_n$: My answer is $a_{n+2}=a_n\cdot \frac{n+4}{2(n+2)}$ which is correct.

B. Using A, write two independent solutions to the ODE.

So, for the odd n's: $$a_3=\frac{1}{2} \frac{5}{3}a_1 \\ a_5=\frac{1}{2}\frac{7}{5}a_3=\frac{1}{2}^2\frac{7}{3}a_1 \\ a_7=\frac{1}{2}\frac{9}{7}a_5=(\frac{1}{2})^3\cdot \frac{9}{3}a_1$$ So that if I iterate right, the $2m+1$'th term is $$a_{2m+1}=(\frac{1}{2})^{m+1}\cdot (m+5)\cdot \frac{a_1}{3}$$ Which means the a solution is $$y_1(x)=\sum_{m=0}^\infty (\frac{1}{2})^{m+1}\cdot (m+5)\cdot \frac{a_1}{3}x^{2m+1}$$But, the answer in the textbook is slightly different: $$y(x)=\sum_{m=0}^\infty \frac{2m+3}{2^m}x^{2m+1}$$ Same with even terms:

I came up with $$y_2(x)=\sum_{m=0}^\infty \frac{m+2}{2^{m+1}}x^{2m}$$ and the solution I have is $$y(x)=\sum_{n=0}^\infty \frac{m+1}{2^m}x^{2m}$$ So what's is wrong in my solution?

C. Find the Solution to the ODE when $y(0)=1, \ y'(0)=0$ as a series solution around $x=0$. So how do I know that $a_0=1, \ a_1=0$ and not vise versa?

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    $\begingroup$ Does it have to be a series solution? It is fairly easy to write down the closed form solution. Are you allowed to deduce the series from that? $\endgroup$
    – almagest
    Sep 14, 2014 at 16:59
  • $\begingroup$ Unfortunately not. This one came from a final exam where they ask to do it the way I have written above. $\endgroup$
    – E Be
    Sep 14, 2014 at 20:10

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