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Series solution, second order diff. eq.

by Telemachus
Tags: diff, order, series, solution
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Telemachus
#1
Mar17-12, 10:36 AM
P: 552
Hi there. I have this differential equation: [tex]x^4y''+2x^3y'-y=0[/tex]
And I have to find one solution of the form: [tex]\sum_0^{\infty}a_nx^{-n},x>0[/tex]
So I have:
[tex]y(x)=\sum_0^{\infty}a_n x^{-n}[/tex]
[tex]y'(x)=\sum_1^{\infty}(-n) a_n x^{-n-1}[/tex]
[tex]y''(x)=\sum_2^{\infty}(-n)(-n-1) a_n x^{-n-2}[/tex]

Then, replacing in the diff. eq.
[tex]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[/tex]
[tex]\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[/tex]
Expanding the first term for the second summation, and using k=n-2->n=k+2 for the first and the second sum:
[tex]-\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[/tex]
[tex]-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+4) a_{k+2} +a_k \right ]=0[/tex]
Then:
[tex]-2a_1=0\rightarrow a_1=0[/tex]
[tex]a_{k+2}=\frac{-a_k}{(k+2)(k+4)}[/tex]

After trying some terms I get for the recurrence relation:
[tex]a_{2n}=\frac{(-1)^n a_0}{2^{2n}(n+1)!n!}[/tex]
And
[tex]a_{2n+1}=0\forall n[/tex]

So then I have one solution:
[tex]y(x)=\sum_{n=0}^{\infty}\frac{(-1)^n a_0}{2^{2n}(n+1)!n!}x^{-2n}[/tex]

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 [tex]cosh(1/x)[/tex] or [tex]sinh(1/x)[/tex] because wolframalpha gives the solution:
[tex]y(x)=c_1 cosh(1/x)-ic_2 sinh(1/x)[/tex]
For the differential equation (you can check it here)
Would you help me to find the mistake in here?
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Telemachus
#2
Mar17-12, 01:14 PM
P: 552
Ok. I've found a mistake there.

[tex]-\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[/tex]

So this gives:
[tex]-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+5) a_{k+2} +a_k \right ]=0[/tex]

And the recurrence formula:
[tex]a_{k+2}=\frac{-a_k}{(k+2)(k+5)}[/tex]

It's worse now, because I couldn't even find the recurrence relation.
tiny-tim
#3
Mar17-12, 01:22 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
Hi Telemachus!

Always check the minuses first

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

Telemachus
#4
Mar17-12, 02:00 PM
P: 552
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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