# Homework Help: Solve Series and Find General Solution

1. Apr 5, 2006

### Zem

These are differential equations problems. They are now "redo's", so I have hints from the grader that I don't understand.

First problem:
(x^2 + 1)y'' + 6xy' + 4y = 0
After isolating C_n+2, I have this.

(n+2)(n+1)C_n+2 * X^n = -n(n-1)C_n * X^n - 6nC_n * X^n + 4C_n
C_n+2 = [-n(n-1)C_n - 6nC_n + 4C_n] / (n+2)(n+1)

Since the grader is saying my solution for C_n+2 is equal to
[-(n+4) / (n+2)]C_n, it appears that I am right up until this point.

I have simplified the numerator, and it still doesn't look right.
C_n+2 = C_n[-n(n-1) - 6n + 4] / (n+2)(n+1)
It looks like it should be
C_n+2 = C_n[(-n)^2 - 5n + 4]
C_n+2 = C_n[(n-4)(n-1)] / (n+2)(n+1)
Is there a (n+1) in the numerator that cancels?

Second problem:
y'' + xy' + y = 0
y'' = Sigma_n=2 n(n-1)C_nX^n-2
y' = Sigma_n=1 nC_nX^n-1
y = Sigma_n=0 C_nX^n
Sigma_n=2 n(n-1)C_nX^n-2 + Sigma_n=1 nC_nX^n-1 + Sigma_n=0 C_nX^n = 0

Sigma_n=0 (n+2)(n+1)C_n+2X^n + Sigma_n=0 nC_nX^n + Sigma_n=0C_nX^n = 0

X^n[(n+2)(n+1)C_n+2 + nC_nX^n + C_nX^n = 0

X^n(n+2)(n+1)C_n+2 = -X^n(n)C_nX^n - X^nC_n
C_n+2 = -X^n(n)C_n - X^nC_n
C_n+2 = (-nC_n - C_n) / (n+2)(n+1)
Is this incorrect? The grader says this should be
C_n+2 = -C_n / (n+2)

Last edited: Apr 5, 2006
2. Apr 5, 2006

### dextercioby

You got

$$C_{n+2}=\frac{-nC_{n}-C_{n}}{(n+1)(n+2)} =-\frac{C_{n}(n+1)}{(n+1)(n+2)} =-\frac{C_{n}}{n+2}}$$

which is what you were supposed to get, right...?

Daniel.

3. Apr 5, 2006

### Zem

Yes, so I'll make my patterns...

$$C_{n+2}=\frac{-C_{n}}{(n+2)}$$
$$n=0: C_{2} = \frac{-C_{0}}{2}$$
$$n=1: C_{3} = \frac{-C_{1}}{3}$$
$$n=2: C_{4} = \frac{(-1)C_{0}}{2*4}$$
$$n=3: C_{5} = \frac{(-1)C_{1}}{3*5}$$
$$n=4: C_{6} = \frac{(-1)C_{0}}{4*6}$$

The even ones look like $$C_{2n} = \frac{(-1)^nC_{0}}{2^{n}n!}$$

But the series for the odd ones is strange. Here is what is in the denominator when n=1, n=3, n=5:
1*3*5*7
This looks like $$(2n+1)(2n-1)(2n-3)$$
How do I represent that in the series?

4. Apr 6, 2006

### dextercioby

To me it looks like

$$(2k \pm 1) !!$$

Daniel.

5. Apr 8, 2006

### Zem

Thanks! I have found the radius of convergence and general solution of the second problem.
Back to the first problem. Note:

Code (Text):
$$(x^2 + 1)y'' + 6xy' + 4y = 0$$
$$(\sum_{n = 2}^{\infty} \{n(n-1)C_n*X^n}) + (\sum_{n = 2}^{\infty} \{n(n-1)C_n*X^{{n-2}}) + 6(\sum_{n=1}^{\infty} \{n}{C_n}*X^n}) + 4(\sum_{n=0}^{\infty} \{C_n*X^n}) = 0$$
$$(\sum_{n = 2}^{\infty} \{n(n-1)C_n*X^n}) + (\sum_{n = 0}^{\infty} \{(n+2)(n+1)C_{n+2}*X^n}) + 6(\sum_{n=0}^{\infty} \{n}{C_n}*X^n}) + 4(\sum_{n=0}^{\infty} \{C_n*X^n}) = 0$$
$$(\sum_{n = 0}^{\infty} \{n(n-1)C_n*X^n}) + (\sum_{n = 0}^{\infty} \{(n+2)(n+1)C_{n+2}*X^n}) + 6(\sum_{n=0}^{\infty} \{n}{C_n}*X^n}) + 4(\sum_{n=0}^{\infty} \{C_n*X^n}) = 0$$
$$(n+2)(n+1)C_{n+2}*X^n = -n(n-1)C_n * X^n - 6nC_n * X^n + 4C_n * X^n$$

$$(n+2)(n+1)C_{n+2} = C_n[-n(n-1) - 6n + 4]/(n+2)(n+1)$$
$$C_{n+2} = C_n[(-n)^2 - 5n + 4]/(n+2)(n+1)$$
$$C_{n+2} = C_n[(n-4)(n-1)]/(n+2)(n+1)$$

The simplified equation for C_n+2 is
Code (Text):
$$C_{n+2} = C_n(n-4)/(n+2)$$
but I don't see how because (n+1) does not cancel (n-1). Have I made a mistake somewhere?

Last edited: Apr 8, 2006