A problem I should solve is the following IVP: I need to find an approximation for small $\varepsilon$ of $$y'' + y + \varepsilon y^3 = 0, y(0) = 1, y'(0)=0$$
My approach so far was:
I assume that I already know a solution, call it $y(t)$. Then I have a look at the Taylor series at $t-\varepsilon$ in order to get the following:
$$y(t) = \sum_{k=0}^{\infty} \frac{y^{(k)}(t-\varepsilon)}{k!}(\varepsilon)^k$$
This gives me a dependency on epsilon, and can only have a look at the first two terms of the series and say that for sufficiently small epsilon the terms $\mathcal{O}(\varepsilon^2)$ can be neglected.
The next thing I do is using the IV, giving me:
$$y(0) = y(-\varepsilon)+ y'(-\varepsilon)\varepsilon = 1$$ and $$y'(0) = y'(-\varepsilon)+y''(-\varepsilon)\varepsilon = 0$$
Using both equations I get the ODE $$y(-\varepsilon) -\varepsilon^2y''(-\varepsilon)=1$$ and solve that Euler-type ODE.
I am confused about the meaningfulness of this last ODE and this approach in general. The solution will be dependent on $\varepsilon$ and I am unable to see how it should approximate the first given ODE.
Any help is greatly appreciated!
That is not at all the standard approach. The standard approach is to notice that the problem is well-posed when $\varepsilon=0$, indeed the solution in this case is $\cos(x)$. Now write $y=\sum_{n=0}^\infty \varepsilon^{\alpha_n} y_n(x)$, where $\alpha_n$ is an unknown increasing sequence of numbers,$\alpha_0=0$, and $y_n$ are unknown functions. (In many situations $\alpha_n=n$.) Then substitute that into the ODE, taking $y_0$ to be the solution with $\varepsilon=0$. Then use the principle of dominant balance after that. This means that you group together all terms with a given power of $\varepsilon$ and set that aggregate equal to zero. In your case the ODE looks like
$$\sum_{n=0}^\infty \varepsilon^{\alpha_n}(y_n''+y_n) + \varepsilon \left ( \sum_{n=0}^\infty \varepsilon^{\alpha_n} y_n \right )^3 = 0.$$
The lowest order term in the expansion of $\varepsilon y^3$ is $\varepsilon y_0^3$. So select $\alpha_1=1$, then solve
$$y_1''+y_1+y_0^3 = 0,y_1(0)=0,y_1'(0)=0$$
where the initial conditions are zero because all of the initial condition information is already in $y_0$. (If the initial conditions depended on $\varepsilon$ then you might have something nonzero here.) Thus you are left to solve $y_1''+y_1+\cos(x)^3=0,y_1(0)=0,y_1'(0)=0$, which is standard albeit a bit tedious.
Then you continue up to higher powers; it becomes cumbersome eventually because it is difficult to expand the cube for a large sum, but it can be done.
This is called regular perturbation theory. When the problem with $\varepsilon=0$ is not well-posed, the techniques used are called singular perturbation theory.