请教一道关于常微分方程题

Question

求解 X' = X^2 - 1, X(0) = 1. [提示：(X^2 - 1)^-1 = (1/2)/(X-1) - (1/2)/(X+1)]

谢谢大神！
求详细点的过程！

Answer 1

分离变量
dx/dt=x^2-1
dx/(x^2-1)=dt
∫[(1/2)/(X-1) - (1/2)/(X+1)]dx=∫dt

(1/2)ln|x-1|-(1/2)ln|x+1|=t+C
(1/2)ln|(x-1)/(x+1)|=t+C
ln|(x-1)/(x+1)|=2t+C
|(x-1)/(x+1)|=e^(2t+C)
(x-1)/(x+1)=Ce^(2t)
x=[1+Ce^(2t)]/[1-Ce^(2t)]
代入x(0)=1
1=(1+C)/(1-C)
C=0
所以
x=1

Answer 2