Question

solve the D.E. ydouble prime =1+( y prime)^2

Answer 1

$y''=1+(y')^2$

$y'=dy/dx=t$

$y''=dt/dx=t'$

$dt/dx=1+t^2$

$\int dt/(1+t^2)=\int dx$

$arctan(t)=x+C_1$

$t=tan(x+C_1)$

$dy/dx=tan(x+C_1)$

$\int dy=\int tan(x+C_1)dx$

substitution for the integral: $z=cos(x+C_1), dz=-sin(x+C_1)dx$

$y=\int -dz/z=-log(z)+C_2=-log(cos(x+C_1))+C_2$

answer: $y=-log(cos(x+C_1))+C_2$