# Homework Help: Series Solutions of Second Order Linear Equations

1. May 5, 2010

### Jamin2112

1. The problem statement, all variables and given/known data

(1+x2)y'' - 4xy' + 6y = 0

2. Relevant equations

I'm going to assume y can be written as [n=0 to ∞] ∑anxn

3. The attempt at a solution

y = [n=0 to ∞] ∑anxn

----> y' = [n=0 to ∞] ∑(n+1)an+1xn
----> y'' = [n=0 to ∞] ∑(n+2)(n+1)an+2xn

----> (1+x2)∑(n+2)(n+1)an+2xn -4x∑(n+1)an+1xn + 6∑(n+1)an+1xn = 0.

My problem is that, after multiplying out the functions in front of y,y',y'', everything is summed from n=0 to ∞ but the the exponents on the x's don't match up. For example, one term looks like ∑(n+1)(n+2)an+2xn+2. I can't just do a transformation j=n+2, making it

[j=-2 to ∞] ∑(j-1)jajxj = (-3)(-2)a-2x-2 + (-2)(-1)a-1x-1 + 0 + 0 + (1)(2)a2x2 + (2)(3)a3x3 + ...

because then I have those darned a-2,a-1 that screw everything up. In other words, I want everything summed from n=0 to ∞ and only xn being summed. Then I can figure out the recurrence relation, the radius of convergence, etc. Suggestions are appreciated.

NEXT QUESTION:

On another problem, I was able to get to the point of

[n=0 to ∞] ∑xn{(n+2)(n+1)an+2 + nan + 2an}=0.

For this equation to be satisfied for all x, the coefficient of each power of x must be zero; hence I conclude that

(n+2)(n+1)an+2 + nan + 2an = 0.

I'm now trying to figure out the recurrence relationship.

I write out

2a0 + 2a2 = 0 -----> a2=-a0
3a1 + 6a3 = 0 ------> a3=(-1/2)a1
4a2 + 6a3 = 0 ------> a4=-(1/3)a2=(1/3)a0
5a3 + 2a5 = 0 ------> a5=-(1/4)a3=(1/8)a1
6a4 + 30a6 = 0 -----> a6=-(1/5)a4=(-1/15)a0
7a5 + 42a7 = 0 ------>a7=(-1/6)a5=(-1/48)a1

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---------> an = a2k = (-1)k * a0/((2k-1)(2k-3)) ?????????

I dunno. Does that seem right? I'm just trying to figure it out in my head. Supposing this is correct, what then do I do with the a2k+1 stuff--you know, the a1,a3,a5,...

2. May 5, 2010

### vela

Staff Emeritus
The reason you probably can't do it is because it's often impossible. The first few terms typically have to be treated separately as special cases.
Note you can rewrite this as

$$a_{n+2} = -\frac{1}{n+1} a_n$$

So the coefficient of the even powers of x will be proportional to a0 and the coefficient of the odd powers of x will be proportional to a1. The series proportional to a0 is one solution, and the series proportional to a1 is the other solution.

3. May 5, 2010

### Jamin2112

So where do I go from here?

I've been told to:

Solve the given differential equation using a power series about x0=0. Find the recursion relation and find the first four terms in each of the two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution.

4. May 5, 2010

### vela

Staff Emeritus
I think what's confusing you is you're trying to make the limits on the summations match up too early. What you want to do instead is make all the sums contain xn. For instance, the first term gives you

$$(1+x^2)y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=2}^\infty n(n-1)a_n x^n$$

Now let k=n-2, and rewrite the first sum in terms of k:

$$(1+x^2)y'' = \sum_{k=0}^\infty (k+2)(k+1)a_{k+2} x^k + \sum_{n=2}^\infty n(n-1)a_n x^n$$

The first summation will contribute x0 and x1 terms while the second does not, so separate those two out and write the rest as one summation to get:

$$(1+x^2)y'' = 2a_2 + 6a_3 x + \sum_{n=2}^\infty [(n+2)(n+1)a_{n+2}+n(n-1)a_n] x^n$$

5. May 5, 2010

### Jamin2112

Hmmm.... I'll try this and get back to you.