# Homework Help: Find the power series in x for the general solution of?

1. Jul 27, 2016

### Math10

1. The problem statement, all variables and given/known data
Find the power series in x for the general solution of (1+x^2)y"+6xy'+6y=0.

2. Relevant equations
None.

3. The attempt at a solution
I got up to an+2=-an(n+3)/(n+1)
for n=1, 2, 3, 4, 5, 6...
a3=-2a1
a4=0
a5=3a1
a6=0
a7=-4a1
a8=0
The answer in the book says y=a0sigma from m=0 to infinity of (-1)^m * (2m+1)x2m+a1sigma from m=0 to infinity of (-1)^m * (m+1) x^(2m+1). How do I get the correct answer?

2. Jul 28, 2016

### Stephen Tashi

You didn't explain how you got those results. One way to solve this kind of problem is to do manipulations with summations using the $\Sigma$ notation. Another way is to work out cases for specific powers of $x$ until you see how to express the result as a pattern using the $\Sigma$ notation. Which method do your class materials expect you to use ?

Assume $y = A_0 + A_1 x + A_2 x^2 + A_3 x^3 + .... = \Sigma_{i=0}^\infty A_i x^i$. To start things off, what is the summation that represents $y'$ ?

3. Jul 30, 2016

### Math10

Just like you mentioned, I got those results by finding y, y' and y'' and then plugging those into the differential equation and simplify. But at the end after finding all the values of a3, a4, a5, etc. I don't know how to get to the correct answer in the textbook.

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4. Jul 30, 2016

### Stephen Tashi

The picture of your work is upside down.

5. Jul 31, 2016

### Math10

Now I believe you can take a look at my work. Sorry about that.

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6. Aug 1, 2016

### Stephen Tashi

After the step:

$\sum_{n=0}^\infty {(n+2)(n+1)A_{n+2}x^n} + \sum_{n=2}^\infty {(n)(n-1)A_nx^n}+ 6 \sum_{n=1}^\infty {(n)A_n x^n }+ 6 \sum_{n=0}^\infty A_n x^n = 0$

I don't see how you concluded $A_0 = 0$

$(2)(1) A_2 x^0 + (3)(2)A_3x^1 + \sum_{n=2}^\infty {(n+2)(n+1)A_{n+2}x^n}$
$+ \sum_{n=2}^\infty {(n)(n-1)A_nx^n}$
$+ (6)(1)A_1x^1 + 6 \sum_{n=2}^\infty {(n)A_n x^n }$
$+ 6A_0 x^0 + 6A_1 x^1 + 6 \sum_{n=2}^\infty A_n x^n = 0$

$( 2 A_2 + 6A_0)x^0$
$+ (6A_3 + 12A_1)x^1$
$+ \sum_{n=2}^{\infty} {[ (n+2)(n+1)A_{n+2} + (n)(n-1)A_n + 6nA_n + 6A_n] x^n} = 0$

$( 2 A_2 + 6A_0)x^0$
$+ (6A_3 + 12A_1)x^1$
$+ \sum_{n=2}^{\infty} {[ n+2)(n+1)A_{n+2} + (n)(n-1)A_n + 6nA_n + 6A_n] x^n} = 0$

$( 2 A_2 + 6A_0)$
$+ (6A_3 + 12A_1)x^1$
$+ \sum_{n=2}^{\infty} {[ n+2)(n+1)A_{n+2} + ( (n)(n-1) + 6n + 6)A_n] x^n} = 0$

7. Aug 2, 2016

### Math10

Okay, so from where you left off above,
2a2+6a0=0
that means a2=0 and so does a0=0.
And (6a3+12a1)x=0,
that means a3=-2a1.
Now what? How to find the answer from here?

8. Aug 2, 2016

### Staff: Mentor

Look at what you have so far:

$$y = a_0(1-3x^2...)+a_1(x-2x^3...)$$

9. Aug 2, 2016

### Math10

Okay, so I got y=a0(1-3x^2-5/3x^4+7/3x^6-3x^8)+a1(x-2x^3+3x^5-4x^7).

10. Aug 2, 2016

### Staff: Mentor

The $a_0$ sum does not match the book answer, but the $a_1$ sum does.

11. Aug 2, 2016

### Math10

But how do I get to the answer in the book?

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12. Aug 2, 2016

### Staff: Mentor

Does the answer in the book satisfy the differential equation?

13. Aug 2, 2016

### Math10

To be honest, I'm not too sure...

14. Aug 2, 2016

### Staff: Mentor

What do you get if you substitute their answer term for term?

15. Aug 2, 2016

### Math10

I still don't get it.

16. Aug 2, 2016

### Staff: Mentor

Your methodology looks correct. That's the important part. If their answer is correct and you don't match it, you must have made a simple "arithmetic" error somewhere.

17. Aug 2, 2016

### Math10

After checking my work, I think I have made a stupid mistake.
Now I got y=a0(1-3x^2-5x^4+7x^6-9x^8)+a1(x-2x^3+3x^5-4x^7). Does this match the answer in the book?

18. Aug 2, 2016

### Stephen Tashi

No, it doesn't imply a definite value for $A_0$ or $A_2$. (After all, the solution to a differential equation should contain some arbitrary constants that are determined only when you are given some initial conditions.)

19. Aug 2, 2016

### Stephen Tashi

$A_2 = - 3 A_0$

$A_3 = - 2 A_1$

$A_{n+2} = -\frac{n^2 + 5n + 6}{(n+2)(n+1)} A_n = -\frac{(n+2)(n+3)}{(n+2)(n+1) } A_n$

So we can take $A_0$ and $A_1$ as the arbitrary constants and express all the other $A_i$ in terms of $A_0$ and $A_1$.

20. Aug 3, 2016

### Staff: Mentor

You should be able to determine this yourself. The answer does match.