Question

I have a differential equation stating
solve: y'=(y-1)(y+1) when y(0)=3

I separated the dy/dx so I got dy/(y-1)(y+1) = dx took the integral of both sides and solved with PFD
I then got that
x= 1/2 ln(y-1) - 1/2ln(y+1) +c
I solved for C and got that C=.3466
I moved the C to the left side subtracting it from x
x-c = 1/2 ln(y-1) - 1/2ln(y+1)
I raised e to the power of both sides
e^(x-c) = (y-1)^1/2/(y+1)^1/2
Now I do not know how to finish simplifying in order to get the equation down to a form y=x

Answer 1

(y-1)^1/2/(y+1)^1/2 = e^(x-c)
(y-1)/(y+1) = e^(2(x-c))
Using a different c, that is just
(y-1)/(y+1) = ce^(2x) where the new c = e^(- old c)
y-1 = y*ce^(2x) + ce^(2x)
y(1-ce^2x) = 1+ce^2x
y = (1+ce^2x)/(1-ce^2x)