一阶线性微分方程 du/dt+p(t)u=Q(t) 的解是
u = e^[-∫p(t)dt]{C+∫Q(t)e^[∫p(t)dt]dt}.
本题微分方程是 dx/dy+[1/(1-y)]x=y/(1-y)
则 x = e^[-∫dy/(1-y)]{C+∫[y/(1-y)]e^[∫dy/(1-y)]dy}
= e^[ln(1-y)]{C+∫[y/(1-y)]e^[-ln(1-y)]dy}
= (1-y){C+∫[y/(1-y)^2]dy}
= (1-y){C+∫[(y-1+1)/(1-y)^2]dy}
= (1-y){C+∫[(-1/(1-y)+1/(1-y)^2]dy}
= (1-y)[C+ln|1-y|+1/(1-y)]
= (1-y)ln|1-y|+C(1-y)+1
