Question

求微分方程y′′+（y′）²=1满足y=|x=0=0,y′|x=0=0的特解.

Answer 1

令z=y'

原方程变为

z'+z²=1

dz/dx=1-z²

dz/(1-z²)=dx

(1/2)[dz/(1-z)+dz/(1+z)]=dx

(1/2)ln|(1+z)/(1-z)|=x+C

代入x=0,z(0)=y'(0)=0

C=0

ln|(1+z)/(1-z)|=2x

(1+z)/(1-z)=e^(2x)

1+z=e^2x-e^2x z

z=(e^2x-1)/(e^2x+1)

y'=(e^2x-1)/(e^2x+1)=1-2/(e^2x +1)

积分

y=C+x-2积分 e^x dx/[e^x(e^2x+1)]

t=e^x,dt=e^xdx

y=C+x-2积分 dt/[t(t^2+1)]

=C+x-2积分 [dt/t-t dt /(t^2+1)]

=C+x-2ln |t|+ln|t^2+1|

=C+x-2x+ln(e^2x +1)

=C-x+ln(e^2x+1)

代入x=0,y(0)=0

C=-ln2

y=-ln2-x+ln(e^(2x)+1)