Question

(x²+2xy－y²)dx+(y²+2xy－x²)dy=0，求通解

1

Answer 1

简单计算一下即可，答案如图所示

Answer 2

设y=tx,则dy=xdt+tdx,方程变为
(1+2t-t^2)dx+(t^2+2t-1)(xdt+tdx)=0,
整理得(t^3+t^2+t+1)dx+x(t^2+2t-1)dt=0,
分离变量得dx/x=-(t^2+2t-1)dt/[(t+1)(t^2+1)]
=[2/(t+1)-(3t-1)/(t^2+1)]dt,
积分得lnx=2ln(t+1)-(3/2)ln(t^2+1)+arctant+lnc,
所以x=c(t+1)^2/(t^2+1)^(3/2)*e^arctant,
即x=c(y/x+1)^2/(y^2/x^2+1)^(3/2)*e^arctan(y/x),为所求。