Terms of a Series (previous question with more details)?

Question

Anyone good with series?
It's basically finding the first few non-zero terms of an equation with y", y', and y using y = the series (Cn*x^n), etc.

Here's the actual question:
Find the first four non-zero terms of two linearly independent power series solutions for
(x+3)y"+4y'-xy=0
I've worked it out a long ways but I'm getting really funny answers. Just want to see what someone else gets. Use the derivatives of the series below to solve.

Additional Details:
It's a power series...and we're looking for the first four terms of the power series solutions (there will be two solutions). You have to plug in the y" and y', etc. (from the first guy that answered the previous question) into the equation, expand it all out and get it to have the same indexes, etc. Then you have to plug in "n = 0, 1,..." to get the coefficients (C1, C2, etc.) and those will be the answers. I just want to see if someone else gets the same thing I did....Thanks!

FOR PKI15!!!
I got:
A_n+3 = [A_n-(4(n+2)A_n+2)-(n+1)(n+2)A_n+2] / 3(n+2)(n+1)
instead of
a_(n+2) = (a_(n-1) - (n+1)(n+4)a_(n+1))/(3(n+2)(n+1) <>
which is obviously a big difference. Can you email me with your steps so I can see what I did. I've checked over it many times and am not getting anything different....

OH and that's for n>=0, not 1 (for me).

Answer 1

Ok, I worked this all out, it's a pretty messy problem, but power series solutions always are. I won't give you all the details, unless you'd like me to, but the recurrences I got were:

Let a_n (a sub n) be the coefficients of your power series solution.

Then:
6a_2 + 4a_1 = 0
a_(n+2) = (a_(n-1) - (n+1)(n+4)a_(n+1))/(3(n+2)(n+1))
This holds for n >= 1

Ok, now I solved this second recurrences for the first few terms to find the two independent solutions were:

#1: a_0 (1 + (1/18)x^3 - (1/36)x^4 + (7/540)x^5)
#2: a_1(x - (2/3)x^2 + (10/27)x^3 - (17/108)x^4)

Let me know if you'd like more details. Obviously there was a lot of room for error here, but I think my answers are correct.

EDIT:
Some more details:
(x+3)y'' + 4y' - xy = 0
(x+3) sum ( 2 to inf) (n(n-1) a_n x^(n-2)) + 4 sum (1 to inf) (n a_n x^(n-1) - x sum (0 to inf) a_n x^n =0
Expand, pull in x's into sums.
sum (2 to inf) n(n-1) a_n x^(n-1) + sum (2 to inf) 3n(n-1)a_n x^(n-2) + sum (1 to inf) 4na_n x^(n-1) - sum (0 to inf) a_n x^(n+1) = 0
Now get powers of x to be all n's, by shifting indices
sum(1 to inf) (n+1)na_(n+1)x^n + sum (0 to inf) 3(n+2)(n+1)a_(n+2) x^n + sum (0 to inf) 4(n+1)a_(n+1)x^n - sum (1 to inf) (a_(n-1)x^n) = 0
Now pull off terms to make every sum from 1 to Inf
sum (1 to inf) ((n+1)na_(n+1) + 3(n+2)(n+1)a_(n+2) + 4(n+1)a_(n+1) - a_(n-1))x^n + 6a_2 + 4a_1 = 0
So we get:
6a_2 + 4a_1 = 0
(n+1)na_(n+1) + 3(n+2)(n+1)a_(n+2) + 4(n+1)a_(n+1) - a_(n-1))=0 for n >= 1
Combine the a_(n+1) terms.
(n+1)(n+4)a_(n+1) + 3(n+2)(n+1)a_(n+2) - a_(n-1) = 0
a_(n+2) = (a_(n-1) - (n+1)(n+4)a_(n+1))/(3(n+2)(n+1))

Answer 2

(x+3)*y” + 4y’ – xy = 0
Assume a power series solution y = ∑ C_n * x^n

After all the tedious work, we end up with the following solution
(4C1 + 6C2) + ∑ [-C_n + (n+2)(n+5)*C_n+2 + 3(n+2)(n+3)*C_n+3 ] * x^(n+1) = 0

Thus the following recurrence relations are obtained:
C2 = -2C1 / 3
C_n+3 = [ C_n – (n+2)(n+5)*C_n+2 ] / [ 3(n+2)(n+3) ]

For the first independent solution, let C0 = 0.
For the second, let C1 = 0.
We get
y1 = C0 * [ 1 + 1/18 * x^3 – 1/36 * x^4 + 7/540 * x^5 + … ]
y2 = C1 * [x – 2/3 * x^2 + 10/27 * x^3 – 17/108 * x^4 + 101/1620 * x^5 + … ]

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But there is a much easier way of solving this problem, without going through this power series torment. Use Taylor series.
(x+3)*y”’ + 5y” – xy’ – y = 0
(x+3)*y”” + 6y”’ – xy” – 2y’ = 0
(x+3)*y””’ + 7y”” – xy”’ – 3y” = 0

Now follow the same procedures. For the first series, let y(0) = C0 and y’(0) = 0. For the second series, let y(0) = 0, y’(0) = C1. You’ll end up getting the exact same solutions with much less anguish.

Answer 3

For x=0, (x+3)y"+4y'-xy=0 implies 3y"(0)+4y'(0)=0 so
y"(0)/2!=-4y'(0)/3/2!
=-2y'(0)/3

y(x)=y(0)+y'(0)*x-2y'(0)/3*x^2
+O(x^3)