let be
p(x)=(x^{2} -1
q(x)=3(x+1)
r(x)=1
the three coefficients of the equation
a is a singular point of the equation if lim p(x) =0
x------>a
so let's find a
lim p(x) = lim x²-1=a²-1=0
x------>a x------>a
a²-1=0 implies a=+ or -1
so the sigular points are a= -1 or a=1
case 1
for a= -1
lim (x-(-1)) q(x)/p(x)=lim (x+1) 3(x+1)/x²-1=lim3(x+1)/x-1= 0/-2=0
x------> -1 x------> -1 x------> -1
lim (x-(-1))² r(x)/p(x)= lim(x+1)²/x²-1= 0/-2=0
x------> -1 x------> -1
lim (x-(-1)) q(x)/p(x) and lim (x-(-1))² r(x)/p(x) are finite so -1 is regular
x------> -1 x------> -1
singular point
case 2
a=1
lim (x-1)) q(x)/p(x)=lim (x-1) 3(x+1)/x²-1=lim3(x+1)/x+1= 3
x------> 1 x------> 1 x------> 1
lim (x-1))² r(x)/p(x)= lim(x-1)²/x²-1= =0
x------> 1 x------> 1
1 is also a regular singular point
