Series solution, second order diff. eq.

[Telemachus]

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?

 

[Telemachus]

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.

 

[tiny-tim]

Hi Telemachus!

Always check the minuses first …

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 …
$$\sum_0^{\infty}(k+2)(k+3) a_{k+2} x^{-k}-2a_1x-2\sum_0^{\infty}(k+2) a_n x^{-k}-\sum_0^{\infty}a_k x^{-k}=0$$
$$-\sum_0^{\infty}x^{-k} \left [ (k+2)(k+1) a_{k+2} +a_k \right ]=0$$

 

[Telemachus]

Aw, you're right! thank you very much. Didn't noticed that I had (-1)^2 :p