Hi there. I have this differential equation: [tex]x^4y''+2x^3y'-y=0[/tex](adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums - The Fusion of Science and Community**

# Series solution, second order diff. eq.

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Series solution, second order diff. eq.

Loading...

**Physics Forums - The Fusion of Science and Community**