Question

solve as a (x^2-1)y''+2xy=0 as a series solution can someone help me.
so far I got c_n = ((2-n)(n-1)c_(n+2)-2c_(n-1))/(n(n-1))
found for the C_n's
am I on the right track?
but when I solved I c_6 = 1/15
c_4=1/4 these numbers do not sound right to me

Answer 1

suppose \(y(x)=\Sigma_{n=0}^{\infty}a_x(x-x_0)^n\)
now find it's first and second derivatives:
\(y'(x)=\Sigma_{n=1}^{\infty}na_x(x-x_0)^{n-1}\)
\(y''(x)=\Sigma_{n=2}^{\infty}n(n-1)a_x(x-x_0)^{n-2}\)
now substitute into your equation:
\((x^2-1)\Sigma_{n=2}^{\infty}n(n-1)a_x(x-x_0)^{n-2}+2x\Sigma_{n=0}^{\infty}a_x(x-x_0)^n=0\)
now try to put it all under same sumation by ajusting the coeficients.

Answer 2

ok I get \[\sum_{n=2}^{\infty}n(n-1)c_nx^n+2c_2+2c_3x+\sum_{n=2}^{\infty}(2-n)(n+1)c_n+1x^n+2c_0x^1+\sum_{n=2}^{\infty}2c_n-1x^n\]

Answer 3

What are the initial conditions?

Answer 4

The solution for y(0)=1 and y'(0)=1 is
\[
1+ x+\frac{x^3}3+\frac{x^4}6+\frac{x^5}{10} \cdots
\]

Answer 5

initial condition is that the series is centered at 0

Answer 6

the series I got is \[y=-x^3-\frac{ 3 }{ 2 }x^5+\frac{ 1 }{ 5 }x^6+\frac{ 5 }{ 3 }x^7+....\]
I do not think this is right