求微分方程e^xy''+(y')^2=0满足初始条件y(0)=0,y'(0)=-1/2的特解

Question

求微分方程e^xy''+(y')^2=0满足初始条件y(0)=0,y'(0)=-1/2的特解

Answer 1

令z=y'
e^x z'+z^2=0
dz/dx=-z^2/e^x
-dz/z^2=e^(-x)dx
两边积分
1/z=-e^(-x)+C
z=1/[C-e^(-x)]
z(0)=y'(0)=-1/2
1/[C-1]=-1/2
C-1=-2
C=-1
z=1/[-1-e^(-x)]
dy/dx=1/[-1-e^(-x)]
两边积分
y=ln[1+e^(-x)]+C
y(0)=C=0
y=ln[1+e^(-x)]