Question

y’‘+（y’）^2=1的通解的求解过程

Answer 1

解：令y'=p，则y''=pdp/dy
代入原方程，化简得
p[dp/dy-2p/(y-1)]=0
==>p=0，或dp/dy-2p/(y-1)=0
显然，p=0是dp/dy-2p/(y-1)=0的解
又，由dp/dy-2p/(y-1)=0，得
dp/dy=2p/(y-1)
==>dp/p=2dy/(y-1)
==>ln│p│=2ln│y-1│+ln│c1│
==>p=c1(y-1)²
(∵p=0是一个解，∴c1是任意常数)
==>y'=c1(y-1)²
==>dy/(y-1)²=c1dx
==>1/(1-y)=c1x+c2
(c2是任意常数)
==>(c1x+c2)(1-y)=1
故原方程的通解是
(c1x+c2)(1-y)=1
(c1,c2是任意常数)。

Answer 2

解：∵y''+y'=1
==>y''=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=x+ln│C1+e^(-2x)│+C2
(C2是积分常数)
∴原方程的通解是y=x+ln│C1+e^(-2x)│+C2
(C1,C2是积分常数)。