From the Question $y''=1+(y')^2$
I used Reduction of Order by taking $y'=z$ to get,
$$z\left(\frac{dz}{dy}\right)=1+z^2$$
Solve the equation by Separation of Variable, to get
$$\frac{1}{2} \ln(1+z^2) = y+c $$
I tried to substitute $z=y'$ back to the equation above and solve the equation by Integration, but that was far very complex, so I got struck here. Please help
2 Answers
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 $v$
$$\int \frac{dv}{1+v^2} = \int dx$$
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 $s=\cos(x+C)$
$$\int \tan(x+C) \ dx=-\int \frac{ds}{s}=-\log(s)=-\log(\cos(x+C))$$
One thing to recognize is that the derivative of $\tan^{-1}$:
$$ \frac{d}{dx} \tan^{-1}(x) \;\; =\;\; \frac{1}{1+x^2}. $$
We see here that we can rewrite our differential equation as
$$ \frac{dy'}{1 + (y')^2} \;\; =\;\; dx $$
which can be solved as
$$ \tan^{-1}\left (y' \right ) \;\; =\;\; x + c. $$
The tricky part now integrating $y'(x) = \tan(x+c)$. Are you allowed to look this up?
