y✀✀✀=√(1+y✀✀눀)的通解怎么求

∵y''+y'²=1 ==>dy'/dx=1-y'²
==>dy'/(1-y'²)=dx
==>[1/(1+y')+1/(1-y')]dy'=2dx
==>ln│(1+y')/(1-y')│=2x+ln│C1│ (C1是积分常数)
==>(1+y')/(1-y')=C1e^(2x)
==>y'=[C1e^(2x)-1]/[C1e^(2x)+1]
∴y=∫{[C1e^(2x)-1]/[C1e^(2x)+1]}dx
=∫{1-2/[C1e^(2x)+1]}dx
=x+∫{e^(-2x)/[C1+e^(-2x)]}d(-2x)
=x+∫{1/[C1+e^(-2x)]}d[C1+e^(-2x)]
=x+ln│C1+e^(-2x)│+C2 (C2是积分常数)
故原方程的通解是y=x+ln│C1+e^(-2x)│+C2 (C1,C2是积分常数).

设y''=p(x),则y'''=p'(x),方程变为
p'(x)=√(1+p^2),
∴dp/√(1+p^2)=dx,
积分得ln[p+√（1+p^2)]=x+c,
∴p+√(1+p^2)=ce^x,
解得p=[ce^x-1/(ce^x)]/2,
积分得y'=[ce^x+1/(ce^x)]/2+c2,
y=[ce^x-1/(ce^x)]/2+c2x+c3.