Question

请教一道关于常微分方程题X' = X^2 - 1,X(0) = 1.[提示：(X^2 - 1)^-1 = (1/2)/(

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