Answer to Question #102570 in Differential Equations for Ajay

Question

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

Answer 1

We want to solve

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

The usual thing is to replace $y'=v$ and therefore we get

$v'=1+v^2.$

This ODE can be solved by the usual methods, just use what you know. For example separation of variables i.e. solve for v

$\int \frac{dv}{1+v^2} = \int dx$

$arctan(v)=x+C.$

We get $v(x)=\tan(x+C)$ where C is some constant. Now resubstitute i.e.

$y'=v=\tan(x+C)$

Just integrate both sides with respect to x i.e.

$y(x)=\int \tan(x+C) \ dx$

This can be solved easily, for example use substitution $s=\cos(x+C)$

$\int \tan(x+C) \ dx=-\int \frac{ds}{s}=-\log(s)=-\log(\cos(x+C))$