Question

求解可降阶的高阶微分方程(1+x^2)y′′+(y′)^2+1=0

Answer 1

令z=y'

(1+x^2)z'+z^2+1=0

(1+x^2)dz/dx=-(z^2+1)

dz/(z^2+1)=-dx/(x^2+1)

arctan z=-arctanx+C

z=tan(C-arctanx)

y'=(C-x)/(1+Cx)

积分

1.C=0

y=-x^2/2+D

2.C不等于0

y=积分-1/C+(C+1/C)/(1+Cx)

=-x/C+(1+1/C^2)ln|1+Cx|+D

C,D为任意常数