Question

求方程(y^2+xy^2)dx+(x^2-yx^2)dy=0的通解

Answer 1

∵(y^2+xy^2)dx+(x^2-yx^2)dy=0
==>y²(1+x)dx+x²(1-y)dy=0
==>[(y-1)/y²]dy=[(1+x)/x²]dx
==>(1/y-1/y²)dy=(1/x+1/x²)dx
==>ln|y|+i/y=ln|x|-1/x+C1 (C1是积分常数)
==>ln|y|-ln|x|=-1/x-1/y+C1
==>ln|y/x|=-1/x-1/y+C1
==>y/x=e^(-1/x-1/y)e^C1
==>y=Cxe^(-1/x-1/y) (令C=e^C1)
∴方程(y^2+xy^2)dx+(x^2-yx^2)dy=0的通解是：
y=Cxe^(-1/x-1/y) (C是积分常数)