# Series solution, second order diff. eq.

by Telemachus
Tags: diff, order, series, solution
 P: 533 Hi there. I have this differential equation: $$x^4y''+2x^3y'-y=0$$ And I have to find one solution of the form: $$\sum_0^{\infty}a_nx^{-n},x>0$$ So I have: $$y(x)=\sum_0^{\infty}a_n x^{-n}$$ $$y'(x)=\sum_1^{\infty}(-n) a_n x^{-n-1}$$ $$y''(x)=\sum_2^{\infty}(-n)(-n-1) a_n x^{-n-2}$$ Then, replacing in the diff. eq. $$x^4\sum_2^{\infty}(-n)(-n-1) a_n x^{-n-2}+2x^3\sum_1^{\infty}(-n) a_n x^{-n-1}-\sum_0^{\infty}a_n x^{-n}=0$$ $$\sum_2^{\infty}(-n)(-n-1) a_n x^{-n+2}+2\sum_1^{\infty}(-n) a_n x^{-n+2}-\sum_0^{\infty}a_n x^{-n}=0$$ Expanding the first term for the second summation, and using k=n-2->n=k+2 for the first and the second sum: $$-\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{-k}-2a_1x-2\sum_0^{\infty}(k+2) a_n x^{-k}-\sum_0^{\infty}a_k x^{-k}=0$$ $$-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+4) a_{k+2} +a_k \right ]=0$$ Then: $$-2a_1=0\rightarrow a_1=0$$ $$a_{k+2}=\frac{-a_k}{(k+2)(k+4)}$$ After trying some terms I get for the recurrence relation: $$a_{2n}=\frac{(-1)^n a_0}{2^{2n}(n+1)!n!}$$ And $$a_{2n+1}=0\forall n$$ So then I have one solution: $$y(x)=\sum_{n=0}^{\infty}\frac{(-1)^n a_0}{2^{2n}(n+1)!n!}x^{-2n}$$ Now, I think this is wrong, but I don't know where I've committed the mistake. I hoped to find a series expansion for $$cosh(1/x)$$ or $$sinh(1/x)$$ because wolframalpha gives the solution: $$y(x)=c_1 cosh(1/x)-ic_2 sinh(1/x)$$ For the differential equation (you can check it here) Would you help me to find the mistake in here?
 P: 533 Ok. I've found a mistake there. $$-\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{-k}-2a_1x \color{red} - \color{red} 2\sum_0^{\infty}(k+2) a_n x^{-k} -\sum_0^{\infty}a_k x^{-k}=0$$ So this gives: $$-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+5) a_{k+2} +a_k \right ]=0$$ And the recurrence formula: $$a_{k+2}=\frac{-a_k}{(k+2)(k+5)}$$ It's worse now, because I couldn't even find the recurrence relation.
HW Helper
Thanks
P: 26,160
Hi Telemachus!

Always check the minuses first …

 Quote by Telemachus Expanding the first term for the second summation, and using k=n-2->n=k+2 for the first and the second sum: $$-\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{-k}-2a_1x-2\sum_0^{\infty}(k+2) a_n x^{-k}-\sum_0^{\infty}a_k x^{-k}=0$$ $$-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+4) a_{k+2} +a_k \right ]=0$$
should be …
$$\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{-k}-2a_1x-2\sum_0^{\infty}(k+2) a_n x^{-k}-\sum_0^{\infty}a_k x^{-k}=0$$
$$-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+1) a_{k+2} +a_k \right ]=0$$

 P: 533 Series solution, second order diff. eq. Aw, you're right! thank you very much. Didn't noticed that I had (-1)^2 :p

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