Series solution up to a term, power series

[dp182]

Homework Statement

consider the initial value problem (1-x)y^,,+xy^,-2y=0 find the series solution up to the term with x⁶

Homework Equations

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

The Attempt at a Solution

assuming the answer has the form $\Sigma$a_nxⁿ
that gives y^,,=$\Sigma$na_nx^n-1 and y^,,=$\Sigma$n(n-1)a_nx^n-2 then plugging these back in and getting rid of the x^n-2 and x^n-1 you get the equation $\Sigma$(n+2)(n+1)a_n+2xⁿ-$\Sigma$n(n+1)a_n+1xⁿ+$\Sigma$na_nxⁿ-2$\Sigma$a_nxⁿ so what I'm wondering is how do you replace the a_n+1 with an a_n so you can solve for a_n+2

 

[tiny-tim]

hi dp182!

… how do you replace the a_n+1 with an a_n so you can solve for a_n+2

just subtract 1 from everything

(eg ∑a_n+1xⁿ = ∑a_nx^n-1)

 

[vela]

how do you replace the a_n+1 with an a_n so you can solve for a_n+2

You don't. The recurrence relation involves a_n, a_n+1, and a_n+2. There's nothing wrong with that.

 

[id the sloth]

It makes sense that the recurrence relation for a_n+2 involves a_n+1 and a_n. It would require you to define a_0 and a_1 and Second order differential equations require two boundary/initial conditions for a solution.

 

[dp182]

So If I solve for a_n+2 I get a_n+2=n(n+1)a_n+1+a_n(2-n)/(n+2)(n+1) so to get my terms up to x6 I will just input values of n, so my series will be in terms of a₀and a₁?

 

[vela]

Yup, but you made an algebra error solving for a_n+2 (or omitted very needed parentheses).