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Series solution, second order diff. eq. 
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#1
Mar1712, 10:36 AM

P: 535

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^{n1}[/tex] [tex]y''(x)=\sum_2^{\infty}(n)(n1) a_n x^{n2}[/tex] Then, replacing in the diff. eq. [tex]x^4\sum_2^{\infty}(n)(n1) a_n x^{n2}+2x^3\sum_1^{\infty}(n) a_n x^{n1}\sum_0^{\infty}a_n x^{n}=0[/tex] [tex]\sum_2^{\infty}(n)(n1) 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=n2>n=k+2 for the first and the second sum: [tex]\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{k}2a_1x2\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? 


#2
Mar1712, 01:14 PM

P: 535

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. 


#3
Mar1712, 01:22 PM

Sci Advisor
HW Helper
Thanks
P: 26,157

Hi Telemachus!
Always check the minuses first … [tex]\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{k}2a_1x2\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] 


#4
Mar1712, 02:00 PM

P: 535

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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