hint on solving $1+y'^2-y\cdot y''=0$? I have the following ODE: $$1+y'^2-y\cdot y''=0$$
I've never solved an ODE where two "versions" (don't know the term) of $y$ are multiplied with eachother, in this case $y$ and $y''$.
Can I have a hint how to approach this, without too much of an answer?
 A: Substitute $v(y)=\frac{dy}{dx}$, treating $y$ as the independent variable. Therefore, we evaluate $\frac{d^2 y}{dx^2}$ to be:
$$\frac{d^2 y(x)}{dx^2}=\frac{d}{dx}\left(\frac{dy(x)}{dx}\right)=\frac{dv(y)}{dx}=\frac{dv(y)}{dy}\cdot \frac{dy}{dx}=\frac{dv(y)}{dy}\cdot v(y)$$
Hence, applying the substitution gives:
$$1+v^2-y\cdot v\cdot \frac{dv}{dy}=0$$
This is clearly a separable differential equation:
$$\frac{dv}{dy}=\frac{1+v^2}{v\cdot y}$$
Solving this for $v(y)$ and substituting back for $v(y)=\frac{dy}{dx}$ should not be a problem.
A: Another approach. Treat this differential equation as an equation where $x$ is a function of $y$. The differential equation becomes:
$$1 + \frac{1}{x'^2} + \frac{yx''}{x'^3}=0$$
or equivalently:
$$x'^3 + x' + yx''=0$$
Now, substitute $z(y) = x'$
The differential equation becomes:
$$z^3 + z + yz' = 0$$
which is a seperable differential equation.
Note that you will have to solve another differential equation once you solved for $z$, but this shouldn't be a problem.
A: I'll elaborate a bit on the substitution $v=y'$ in @projectilemotion's answer.
In general, differential equations with nonlinear dependence on lower derivatives are quite difficult and often not possible to solve analytically, so people use different tricks or try to guess a form of solution. In this equation
$$
y\cdot y''-(y')^2=1\tag{Eq}
$$
the LHS looks like the numerator in the quotient rule for
$$
\left(\frac{y}{y'}\right)'=\frac{(y')^2-y\cdot y''}{(y')^2}=-\frac{1}{(y')^2},
$$
however, in this form it does not give much. Let's try to replace the denominator with something more general - a function of $y'$ - and see what we can get out of that
$$
\left(\frac{y}{f(y')}\right)'=\frac{y'\cdot f(y')-y\cdot f'(y')y''}{f^2(y')}\stackrel{(\rm Eq)}{=}\frac{y'\cdot f(y')-(1+y'^2) f'(y')}{f^2(y')}.
$$
If we manage to find $f$ such that the numerator is zero then we are (almost) done. It looks like a differential equation for $f$ if we treat $y'$ as a variable $v$, i.e. if $f$ solves
$$
vf(v)-(1+v^2)f'(v)=0
$$
then the numerator is zero if we set $y'$ as $v$. Solving it for $f$ we get one solution to be $f(v)=\sqrt{1+v^2}$, hence, the original solutions can be found among 
$$
\left(\frac{y}{\sqrt{1+y'^2}}\right)'=0.
$$
P.S.
If you get this equation as the Euler-Lagrange equation within calculus of variations context then you can use the standard way to get around the difficulty you've faced here.
