为自己加油

求微分方程dy/dx=(1+y²)/(x²-1)y的通解．

将方程分离变量，得[y/(1+y²)]dy=[1/(x²-1)]dx两边取不定积分，得∫[y/(1+y²)]dy=∫[1/(x²-1)]dx积分得1/2ln(1+y²)=(1/2)ln[(x-1)/(x+1)]+(1/2)lnC，变形得y²=C(x-1)/(x+1)-1，故方程的通解为y²=C(x-1)/(x+1)-1．