题目

救命：怎么解这个微分方程
sinx*y''+(y'^2-sin^2x)^(1/2)*y'^2-(cosx)*y'=0, y(0)=0, y'(0)=1
猜答案不给分（答案：y=x）

提问时间：2021-03-26

答案

∵sinx*y''+(y'²-sin²x)^(1/2)*y'²-(cosx)*y'=0
==>(cosx)*y'-sinx*y''=(y'²-sin²x)^(1/2)*y'²
==>[(cosx)*y'-sinx*y'']/y'²=(y'²-sin²x)^(1/2)
==>(sinx/y')'=y'[1-(sinx/y')²]^(1/2)
==>d(sinx/y')/[1-(sinx/y')²]^(1/2)=dy
∴arcsin(sinx/y')=y+C1 (C1是任意常数)
∵y(0)=0,y'(0)=1,代入上式得 C1=0
∴arcsin(sinx/y')=y
==>sinx/y'=siny
==>y'siny=sinx
==>siny*dy=sinx*dx
==>cosy=cosx+C2 (C2是任意常数)
∵y(0)=0,y'(0)=1,代入上式得 C2=0
∴cosy=cosx ==>y=x
故原方程在初始条件（y(0)=0,y'(0)=1）下的特解是 y=x.