- #1

pinkbabe02

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## Homework Statement

(1+2x^2)y''+6xy'+2y=0

1. find the power series solutions of the equation near x0=0...show the recurrence relation for an, derive a formula for an in terms of a0 and a1, and show the solution in the form y=a0y1(x)+a1y2(x)

2.what is the lower bound for the radii convergence of the power series solutions in part 1

3.what is the lower bound for the power series solutions of the same equation near x0=-1

## Homework Equations

## The Attempt at a Solution

I attached the word doc I typed it up in because that will probably be easier to read.

(1+〖2x〗^2 ) y^''+6xy^'+2y=0

Let…

y=∑_(n=0)^∞▒〖a_n x^n 〗

y^'=∑_(n=1)^∞▒〖〖na〗_n x^(n-1) 〗

y^''=∑_(n=2)^∞▒〖n(n-1)a_n x^(n-2) 〗

(1+〖2x〗^2 ) ∑_(n=2)^∞▒〖n(n-1)a_n x^(n-2) 〗+6x∑_(n=1)^∞▒〖〖na〗_n x^(n-1) 〗+2∑_(n=0)^∞▒〖a_n x^n 〗=0

∑_(n=2)^∞▒〖2n(n-1) a_n x^n+∑_(n=2)^∞▒〖n(n-1)a_n x^(n-2) 〗〗+∑_(n=1)^∞▒〖〖6na〗_n x^n+∑_(n=0)^∞▒〖〖2a〗_n x^n 〗〗=0

∑_(n=2)^∞▒〖2n(n-1) a_n x^n+∑_(n=0)^∞▒〖(n+2)(n+1)a_(n+2) x^n 〗〗+∑_(n=1)^∞▒〖〖6na〗_n x^n+∑_(n=0)^∞▒〖〖2a〗_n x^n 〗〗=0

Since the first term is 0 for both n=0, 1, we can start the summation at n=0.

Since the third term is 0 for n=0, we can start the summation at n=0, also.

∑_(n=0)^∞▒〖2n(n-1) a_n x^n+∑_(n=0)^∞▒〖(n+2)(n+1)a_(n+2) x^n 〗〗+∑_(n=0)^∞▒〖〖6na〗_n x^n+∑_(n=0)^∞▒〖〖2a〗_n x^n 〗〗=0

∑_(n=0)^∞▒〖[2n(n+1) a_n+(n+2)(n+1) a_(n+2)+6na_n+2a_n ] x^n 〗=0

∑_(n=0)^∞▒〖[2(n^2+4n+1) a_n+(n+2)(n+1) a_(n+2) ] x^n 〗=0

2(n^2+4n+1) a_n+(n+2)(n+1) a_(n+2)=0 n=0,1,2,…

〖(a)recurrence relation→a〗_(n+2)=(-2(n^2+4n+1))/((n+2)(n+1)) a_n n=0,1,2,…

n=0 a_2=(-2)/2 a_0 a_2=-a_0

n=1 a_3=(-12)/6 a_1 a_3=-〖2a〗_1

n=2 a_4=(-13)/6 a_2 a_4=13/6 a_0

n=3 a_5=(-11)/5 a_3 a_5=22/5 a_1

I am having trouble creating a formula in terms of a_0 and a_1. More specifically, I don’t know how to create a general summation for this case because of the weird numbers in the numerator. I am pretty sure that the denominator for the odd case is (2k+1)! and the even case would be (2k)!

For part(c), I have y(x)=a_0 (-1+13/6 x^2+⋯)+a_1 (-2x+22/5 x^3+⋯)

As for the lower bound for radii convergence of the power series solutions from the last part, I tried to solve this but I`m not sure if I am approaching it correctly since I didn’t know how to find the power series from the last part. I found p(x) and q(x)…

p(x)=6x/(1+〖2x〗^2 ) and q(x)=2/(1+〖2x〗^2 )

My problem with this is that 1+〖2x〗^2 does not have any zeroes. Does that mean that the radius of convergence is infinity?

For the final part which asks for the lower bound for the power series solutions of the same equation near x_0=-1, I used the same p(x) and q(x) as I did in the last part. But since the denominator has no zeroes I feel like the lower bound is infinity also. The questions about radii of convergence really confuse me so any help would be greatly appreciated!