求微積分方程(1+y)dx-(1-x)dy=0的通解

求微積分方程(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為任意實數)1 + y = e^[ln(1 - x)+ C1]y = C*(1 - x)- 1(C為任意…