Advice on second order non-linear ordinary differential equation

Question

I'm currently working on some problems concerning the calculus of variations and I have come up with the following differential equation that I now want to solve: $$1 + y'(x)^2 - y''(x)(y(x)-\lambda) = 0.$$ I'm only used to solving linear differential equations so I would appreciate some advice on where to start.

Answer 1

let $u = y - \lambda, v = u' = y'.$ then we can rewrite the differential equation as $$1 + v^2 - v'u = 0 \to u \frac{dv}{dx} = v' u = 1 + v^2, \frac{du}{dx} = u' = v $$ which implies $$u\frac{ dv}{du} = \frac{1+v^2}{v} \to \frac{2vdv}{1+v^2} = \frac{2du}u$$ on integration, we get $$(1 + v^2) = cu^2 \to v = \pm \sqrt{1-cu^2} \to \frac{du}{\sqrt{1-cu^2}} = \pm dx$$ now the integration will depend on the sign of $c$: if $c> 0,$ you get an inverse sine, $\sin^{-1}$ and if $c < 0,$ then $\ln().$

Answer 2

As abel noted, with $u = y - \lambda$ we get $1 + (u')^2 - u'' u = 0$. Now if you recall that $\cosh^2 - \sinh^2 = 1$ and $\cosh'' = \cosh$, you are led to consider something of the form $u = a \cosh(b x + c)$, and we find that this works if $b = 1/a$. Thus we get the solution

$$ y = \lambda + a \cosh(x/a + c)$$

where $a \ne 0$ and $c$ are arbitrary parameters.