Question

solve this nonlinear differential equation: y''+(y'^2)+1=0
please show your work

Answer 1

y'' +(y')^2+1=0
(D^2+D^2+1)y=0
(2D^2+1)y=0
Characteristic polynomial 2m^2+1=0
2m^2=-1
m=+/-iroot1/2
ycf = c1cos root1/2x+c2sin root1/2x

Answer 2

\[y''+(y')^2+1=0\quad,\quad z=y'\]\[z'+z^2+1=0\]\[\frac{z'}{z^2+1}+1=0\]\[(\arctan{z})'+1=0\]\[(\arctan{z})'=-1\]\[\arctan{z}=-x+C_1\]\[z=\tan{(-x+C_1)}=-\tan{(x-C_1)}\]\[y'=-\tan{(x-C_1)}\]\[y=-\int\tan{(x-C_1)}\,dx=\ln{(\cos{(x-C_1)})}+C_2\]