Question

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

Answer 1

Given $y''=1+y'^2.$

Put $z = y'.$ Then the given equation becomes,

$z' = 1+ z^{2}$ .

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

Integrating both sides, we get

$\arctan z = x + c_{1}\\ z = \tan (x+c_{1})\\ y'=\tan (x+c_{1})~~\text{(Using the substitution we have taken)}\\ \dfrac{dy}{dx}=\tan (x+c_{1})\\ \text{Integrating both sides with respect to $x$,}\\ y = \ln\sec (x+c_{1})+c_{2}$