Find the power series in x for the general solution of?

[Math10]

Homework Statement

Find the power series in x for the general solution of (1+x^2)y"+6xy'+6y=0.

Homework Equations

None.

The Attempt at a Solution

I got up to a_n+2=-a_n(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)x^2m+a₁sigma from m=0 to infinity of (-1)^m * (m+1) x^(2m+1). How do I get the correct answer?

 

[Stephen Tashi]

I got up to a_n+2=-a_n(n+3)/(n+1)
for n=1, 2, 3, 4, 5, 6...
a₃=-2a₁

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 $y'$ ?

 

[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.

 

[Stephen Tashi]

The picture of your work is upside down.

 

[Math10]

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

 

[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 $A_0 = 0$

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

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

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

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

 

[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?

 

[Chestermiller]

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?

Look at what you have so far:

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

 

[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).

 

[Chestermiller]

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).

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

 

[Math10]

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

 

[Chestermiller]

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

Does the answer in the book satisfy the differential equation?

 

[Math10]

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

 

[Chestermiller]

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

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

 

[Math10]

I still don't get it.

 

[Chestermiller]

I still don't get it.

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.

 

[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?

 

[Stephen Tashi]

2a₂+6a₀=0
that means a₂=0 and so does a₀=0.

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.)

 

[Stephen Tashi]

Now what? How to find the answer from here?

$A_2 = - 3 A_0$

$A_3 = - 2 A_1$

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

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 $A_1$.

 

[Chestermiller]

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?

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

 

[Math10]

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

Thank you so much for the great help! You've made this type of problem much easier to solve now.