Question

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

Answer 1

∵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.