`(1-x)^(2)(d^(2)y)/(dx^(2))-2x(dy)/(dx)+n(n+1)y=0`
Answers
How do I solve (1-x^2) d^2y/dx^2-2xdy/dx+2y=0?
“How do I solve (1-x^2) d^2y/dx^2-2xdy/dx+2y=0?”
(1−x2)y′′−2xy′+2y=0(1)
It's a second order, linear, homogeneous differential equation with variable coefficients and two singular points: x=±1 .
y′′−2xy′(1−x2)+2y(1−x2)=0
from here we see that those points are Fuchsian because the coefficient of y' has first order poles (x=±1) only and the coefficient of y too.
The indicial equation is:
r2+(p0−1)r+q0
where, for the greatest singular point, x=1 :
p0=limx→+1(x−1)⋅−2x(1−x2)=1
q0=limx→+1(x−1)2⋅2(1−x2)=0
=>r2=0=>r=0
So around x=+1 an analytic solution y(x) of the diff. equation (1) exists.
Let's use the substitution:
y(x)=x⋅u(x) :
y′=u+xu′; y′′=2u′+xu′′
The equation (1) becomes:
x(1−x2)u′′+2(1−2x2)u′=0
u′(x)=w(x) :
w′+2(1−2x2)wx(1−x2)=0
which is a first order linear equation with separable variables:
w′w=−2(1−2x2)x(1−x2)
integrating:
ln|w|=−ln|1−x2|−2ln|x|+C
w(x)=Ax2(1−x2)=u′(x)
integrating again:
u(x)=A2(−2x+ln∣∣∣1+x1−x∣∣∣)+B
y(x)=xu(x)=
=A(x2 ln∣∣∣1+x1−x∣∣∣−1)+Bx.