Series solution of a second order ordinary DE

[Lord Anoobis]

Homework Statement

Use the power series method to solve the initial value problem:
$(x^2 +1)y'' - 6xy' + 12y = 0, y(0) = 1, y'(0) = 1$

Homework Equations

The Attempt at a Solution

The trouble here is that after the process above I end up with $c_{k+2} = - \frac{(k-3)(k-4)}{(k+2)(k+1)}c_k$, giving nothing but zeroes for $k = 3$ onward, which cannot be correct. Where have I gone wrong here?

 

[Lord Anoobis]

They would not look any different, but is it not necessary in order to express the whole works as a single summation?

 

[Orodruin]

I am sorry, but your uploaded image is barely readable. Since it is LaTeXed it should not be much of an effort to copy paste the source here and changing the delimiters for the forum appropriate ones.

Also, I suggest that you try to avoid posting replies to your own threads. It makes it seem as if the thread has gotten replies and decreases the probability that it will be read.

 

[Lord Anoobis]

I am sorry, but your uploaded image is barely readable. Since it is LaTeXed it should not be much of an effort to copy paste the source here and changing the delimiters for the forum appropriate ones.

Also, I suggest that you try to avoid posting replies to your own threads. It makes it seem as if the thread has gotten replies and decreases the probability that it will be read.

Let me try that. Also, the reply button under Kuruman's response was giving an error message earlier for some reason.

 

[Lord Anoobis]

I am sorry, but your uploaded image is barely readable. Since it is LaTeXed it should not be much of an effort to copy paste the source here and changing the delimiters for the forum appropriate ones.

Also, I suggest that you try to avoid posting replies to your own threads. It makes it seem as if the thread has gotten replies and decreases the probability that it will be read.

For a solution centred about the ordinary point

,

Then

and

Substitute into the equation:

After extracting the first two terms from the second series, the first term from the third and the first two terms from the fourth in the equation above we have:

Now for each series respectively, let

, then

Then

And

 

[kuruman]

Also, I suggest that you try to avoid posting replies to your own threads. It makes it seem as if the thread has gotten replies and decreases the probability that it will be read.

Let me try that. Also, the reply button under Kuruman's response was giving an error message earlier for some reason.

To be honest, OP's #2 was a response to a post that I deleted soon after I posted but not before OP posted #2. My initial response was incomplete and I needed time to complete, but now it's probably no longer necessary.

 

[Orodruin]

So your conclusion with $y(0) = c_0 = 1$ and $y'(0) = c_1 = 1$ would be that $c_2 = -6$, $c_3 = -1$, and $c_4 = 1$, i.e.,
$$
y(x) = 1 + x - 6x^2 - x^3 + x^4.
$$
Why do you think that is wrong?

Also, there is no reason for separating the equations for $n = 2$ and $n = 3$. They are the same functional expression as you can add zeros to the sums without affecting them, for example
$$
\sum_{n=2}^\infty c_n n(n-1) x^n = \sum_{n=0}^\infty c_n n(n-1) x^n
$$
since $n(n-1) = 0$ for both $n = 0$ and $n = 1$.

 

[Lord Anoobis]

Actually, it appears to be correct. I thought it a bit odd that every term after that is zero. A bit of a sneaky question, I guess.

 

[Orodruin]

A series expansion that terminates is still a series expansion ...

In fact, this is essentially how I introduce the Legendre polynomials in my textbook, i.e., by looking at Legendre's differential equation and a series expansion of the solution. As it turns out, requiring that solution is regular at $x = 1$ forces the series expansion to terminate and leads to a relation between the eigenvalue and the parameter $\ell$ in Legendre's differential equation, $\mu = \ell(\ell + 1)$.

 

[Lord Anoobis]

A series expansion that terminates is still a series expansion ...

In fact, this is essentially how I introduce the Legendre polynomials in my textbook, i.e., by looking at Legendre's differential equation and a series expansion of the solution. As it turns out, requiring that solution is regular at $x = 1$ forces the series expansion to terminate and leads to a relation between the eigenvalue and the parameter $\ell$ in Legendre's differential equation, $\mu = \ell(\ell + 1)$.

So it is. One tends to forget such things when extremely tired. It's approaching midnight here so it's time to say thank you and goodnight.