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1. The problem statement, all variables and given/known data

(1+2x-x^2)y''-6xy'-6y=0 find power series solution of the equation near x0=1

(a)show the recurrence relation for an,

(b)derive a formula for an in terms of a0 and a1, and

(c)show the solution in the form y=a0y1(x)+a1y2(x)

2. Relevant equations

3. The attempt at a solution

I`ll attach the word file I typed it up in if that is easier to read...

(1+2x-x^2 ) y^''-6xy^'-6y=0, x_0=1

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

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

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

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

I can turn the second term into n=0 since when n=0, the value equals 0 anyhow…the same goes for the third term (when n=0, 1 the value is 0), so we have…

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

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

This is where I am stuck. I don’t know how to find the recurrence relation since I have a_n,a_(n+1),and a_(n+2). If someone could give me a hint that would be extremely helpful.

[STRIKE][STRIKE][/STRIKE][/STRIKE]

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# Differential equation help

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