Question

Solve the differential equation y ′′ =1+(y')^2

Answer 1

Let us substitute $z=y' \Rightarrow y''=z', \; z'=1+z^2.$

Therefore, $\dfrac{dz} {1+z^2} =dx.$

We should integrate it with respect to x,

$\int\dfrac{dz} {1+z^2} =\int dx, \\ \tan^{-1}z=x+C_1, \\ z=\tan(x+C_1).$

Next, we solve an equation

$y'=\tan (x+C_1), \\ y=\int \tan (x+C_1)dx = - \ln(\cos(x+C_1)) +C_2.$