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

 

[blue_raver22]

Hello! Thank you for sharing your work on solving this second order differential equation using a series solution. Your approach looks correct, but there are a few things that may have caused some confusion.

Firstly, in your initial solution for y(x), you have used x^-n as the exponent for each term. However, in the differential equation, the exponent should be -n+2. So your solution should be y(x) = \sum_{n=0}^{\infty} a_n x^{-n+2}.

Secondly, in your recurrence relation, you have used a_{k+2} and a_k, but these should actually be a_{n+2} and a_n, since you are using n as the index for the summation. This will change the values you get for a_{2n} and a_{2n+1}.

Finally, the solution given by WolframAlpha uses hyperbolic functions, while your solution is using a combination of powers and factorials. This is because the differential equation given has two distinct roots, which is why there are two solutions. Your solution is the one for the root x=0, while the other solution is for the root x=-1. This is why you are getting a different result from WolframAlpha.

To find the other solution, you can use a similar approach, but with a different initial value for a_0. Alternatively, you can use the Wronskian method to find the second solution.

I hope this helps clarify things and good luck with your work!