Investigating some problem in optics I am faced with a nonlinear differential equation of the form $$ - y(x)\frac{{{d^2}}}{{d{x^2}}}\left( {\frac{1}{{y(x)}}} \right) + {y^2}(x) = f(x)$$ with initial conditions $y(0) = 0$ and $y'(0) = \varepsilon $ where $\varepsilon > 0$ is a small parameter and ${f(x)}$ is a function for which we can analytically solve (explicitly provide two independent solutions) of the linear equation $p''(x) + f(x)p(x) = 0$.
It can be seen that (because $y(x) \to 0$ as $x \to 0$) $\frac{1}{{y(x)}}$ should satisfy in the limit $x \to 0$ $$\frac{{{d^2}}}{{d{x^2}}}\left( {\frac{1}{{y(x)}}} \right) + \frac{{f(x)}}{{y(x)}} = 0$$ and we know a solution of this equation, but the first problem arises here. In some cases, it is not uniquely determined by the initial conditions. For example, if $f(x) = 1 - 2{x^{ - 2}}$ we get $$y(x) \to \frac{{\varepsilon x}}{{\left( {1 + Cx} \right)\cos x + \left( {x - C} \right)\sin x}}$$ as $x \to 0$ so I don't even understand how to solve it numerically.
I tried to use this asymptotic ${y_0}(x)$ to get $$\frac{{{d^2}\psi }}{{d{x^2}}} + f(x)\psi = \frac{1}{{{y_0}(x) + \psi (x)}}$$ with $\psi (0) = \psi '(0) = 0$ but failed to find any good solutions using an expansion $\psi = \sum\limits_{k = 1}^{ + \infty } {{\psi _k}{\varepsilon ^k}} $ (in the first order they seem to have a singularity).
So I didn't expect this to have any exact/explicit solutions but tried to find that it is possible to find a "wide-range" asymptotic using a small parameter $\varepsilon $ or just even solve it numerically -- I try to use ${y_0}(x)$ to "shift" the initial conditions from 0 to some small number, but the numerical solution from that point has a singularity that strongly depends on the initial point. I don't have much experience with nonlinear ODEs, so any help would be appreciated.