求微分方程的初值问题2y(d^2y/dx^2)=(dy/dx)^2+y^2,y|(x=0)=1,dy/dx|(x=0)=-1
问题描述：
求微分方程的初值问题2y(d^2y/dx^2)=(dy/dx)^2+y^2,y|(x=0)=1,dy/dx|(x=0)=-1
答
令y'=p,则y''=dy'/dx=dy'/dy*dy/dx=pdp/dx
所以2pydp/dy=p^2+y^2
p(0)≠0,所以p不恒等于0
2p/y*dp/dy=(p/y)^2+1
令u=p/y,则dp/dy=u+y*du/dy
2u(u+y*du/dy)=u^2+1
y*du/dy=1-u^2
du/(1-u^2)=dy/y
1/2*(1/(1+u)+1/(1-u))du=dy/y
1/2*(ln|1+u|-ln|1-u|)=ln|y|+C1
(1+u)/(1-u)=C1y^2
令x=0：0=C1
所以u=p/y=-1
dy/y=-dx
ln|y|=-x+C2
y=C2e^(-x)
令x=0：1=C2
y=e^(-x)
经检验符合题意