A physical problem brought me to the following boundary-value problem
$y''(x)= \kappa^2 \left(y(x) + \frac{y(x)^2}{2} \right)$
with $y(0)=0$ and $y(C)=-58$ for some $C>0.$
If there was no quadratic term, we would just have the equation of a harmonic oscillator (up to the incorrect sign, due to $\kappa^2>0$, leading to $\cosh$ rather than $\cos$ type solutions.)
Does anybody know if this equation admits closed-form solutions? Or differently, if you know that this equation can only be treated numerically or approximately with perturbative methods, this would also answer my question.
If anything is unclear, please let me know.
Well, the system is Hamiltonian (hardly surprising if it's derived from a physical problem) with Hamiltonian $$H = \frac{1}{2}(y')^2 - \frac{1}{2}\kappa^2\left(y^2+\frac{y^3}{3}\right)$$ and can be reduced to the quadrature $$\frac{dx}{dy} = \pm\frac{1}{\kappa\sqrt{\left(y^2+\frac{y^3}{3}\right) + 2H}}$$ for constant $H$. I don't know if this is solvable in terms of ordinary functions, but do you really need to know the explicit $x$-$y$ relation? In dynamical systems, it's often enough to know the character of the system or it's long term behaviour. Since the system is Hamiltonian the phase portrait is easy to sketch, you can find the stationary points etc... so you can tell a lot about it without solving that ugly integral. Depends what's useful.