Question

看这个微分怎么求：x^2*dy+(y-2xy-x^2)*dx=0

是不是题错了？

Answer 1

解：∵x²dy+(y-2xy)dx=0 ==>x²dy/dx+(1-2x)y=0
==>dy/y+(1/x²-2/x)dx=0
==>ln|y|-1/x-2ln|x|=ln|C| (C是积分常数)
==>ln|y|=ln|Cx²|+1/x
==>y=Cx²e^(1/x)
∴设原方程的通解为y=C(x)x²e^(1/x) (C(x)是关于x的函数)
代入原方程整理得C'(x)=e^(-1/x)/x²
==>C(x)=∫e^(-1/x)d(-1/x)
==>C(x)=e^(-1/x)+C (C是积分常数)
∴y=[e^(-1/x)+C]x²e^(1/x)=x²+Cx²e^(1/x)
故原方程的通解是y=x²+Cx²e^(1/x) (C是积分常数)

Answer 2

x^2*dy = -(y - 2 x y - x^2)*dx
dy/dx = -(y - 2 x y - x^2)/x^2