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
<p>问题：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<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">房立存的回答：<div class="content-b">网友采纳 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
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