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V0ODO0CH1LD
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Homework Statement
Find the power series solution of the differential equation
[tex] y''-\frac{2}{(1-x)^2}y=0 [/tex]
around the point ##x=0##.
Homework Equations
[tex] y=\sum_{n=0}^\infty{}c_nx^n [/tex]
[tex] y'=\sum_{n=0}^\infty{}c_{n+1}(n+1)x^n [/tex]
[tex] y''=\sum_{n=0}^\infty{}c_{n+2}(n+2)(n+1)x^n [/tex]
The Attempt at a Solution
If I substitute the power series for ##y## in the differential equation (and mess around with it a bit) I get:
[tex] \sum_{n=0}^\infty{}\left[c_{n+2}(n+2)(n+1)-\frac{2}{(1-x)^2}c_n\right]x^n=0. [/tex]
Okay, so I replaced the problem of solving a differential equation with the problem of finding the coefficients of an infinite power series that satisfy the equation above, right? So what is the condition that the coefficients have to satisfy in order that the equation above is true for (at least) every ##x\in\mathbb{R}-{1}##. Usually the next step here is to say that
[tex] c_{n+2}(n+2)(n+1)-\frac{2}{(1-x)^2}c_n=0\,\Longrightarrow\, c_{n+2}=\frac{2}{(n+2)(n+1)(1-x)^2}c_n, [/tex]
but why is this the "general" condition that the ##c_i##'s have to satisfy? Are the terms inside the brackets always bigger than or equal to zero?
Also, what is the meaning of "around the point ##x=0##"? I assume I should take the limit as ##x## goes to zero at some point, but when?