Question

求微分方程y(dy/dx)=x(1-y^2)的通解

Answer 1

令u=x-3,v=y+2,那么x=u+3,y=v-2,dy/dx=d(v-2)/d(u+3)=dv/du
dv/du=2(((v-2)+2)/((u+3)+(v-2)-1))^2=2(v/(u+v))^2
du/dv=(1/2)*(u/v + 1)^2
令z=u/v,u=zv,u'=z+z'v
z+z'v=(1/2)*(z+1)^2
1/(z^2+z+1)dz=(1/2v)dv
(2/√3)/{[(2z/√3)+(1/√3)]^2+1} d[(2z/√3)+(1/√3)]=(1/2v)dv
(2/√3)arctan[(2z/√3)+(1/√3)]=(ln|v|)/2+C
(2/√3)arctan[(2u/v√3)+(1/√3)]=(ln|v|)/2+C
(2/√3)arctan[(2(x-3)/√3(y+2))+(1/√3)]=(ln|y+2|)/2+C

Answer 2

整理得ydy/(1-y²)=xdx
积分,∫ydy/(1-y²)=∫xdx
-1/2*ln|1-y²|=x²/2+C
ln|1-y²|=-x²+C

1-y²=Ce^(-x²)
y²=1-Ce^(-x²)为通解