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
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![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](/uploads/image/z/10760334-6-4.jpg?t=%E6%B1%82%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8B%E7%9A%84%E5%88%9D%E5%80%BC%E9%97%AE%E9%A2%982y%28d%5E2y%2Fdx%5E2%29%3D%28dy%2Fdx%29%5E2%2By%5E2%2Cy%7C%28x%3D0%29%3D1%2Cdy%2Fdx%7C%28x%3D0%29%3D-1)
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
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
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
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
上课听什么呢?