求微积分方程（1+y）dx-（1-x）dy=0的通解

(1 + y) dx - (1 - x) dy = 0(1 + y) dx = (1 - x) dy[1/(1 - x)]dx =[1 /(1 + y)]dyd(ln(1 - x))=d(ln(1 + y))ln(1 - x) + C1 = ln(1 + y) （C1为任意实数）1 + y = e^[ln(1 - x) + C1]y = C*(1 - x) - 1 (C为任意...