Solve $y'' + \epsilon y'^2 + 1 = 0$ with initial conditions $y(0) = 0$ and $y'(0) = 1$

Question

Let $\epsilon \ll 1$.

I guess I'm trying to use perturbation method but I've been getting really weird numbers when I'm determining the initial conditions. Can someone perhaps help me with this?

So what I have is the following:

1. Set $y(t)=y_0(t)+\epsilon y_1(t)+\epsilon^2 y_2(t)+...$
2. Plug in $y(t)$ into $y'$ and compare $\epsilon^{0}, \epsilon^{1}, \epsilon^{2}$: $$(y_0''+\epsilon y_1''+\epsilon^{2}y_2''+...)+\epsilon(y_0'+\epsilon y_1'+\epsilon^{2}y_2'+...)^2+1=0$$

Next we collect all terms multiplied by $\epsilon$. What I am doing here is basically finding all the coefficients of the $\epsilon$ and putting them together in a group: \begin{alignat}3 \epsilon^{0}&:& y_0''+1&=0& ~~ \text{where} ~~ y_0(0)&=0& ~~ \text{and} ~~ y_0'(0)&=1 \\ \epsilon^{1}&:& y_1''+y_0'^2&=0& ~~ \text{where} ~~ y_1(0)&=0& ~~ \text{and} ~~ y_1'(0)&=1 \\ \epsilon^{2}&:&~ y_2''+2y_0'y_1'+y_1'&=0& ~~ \text{where} ~~ y_2(0)&=0& ~~ \text{and} ~~ y_2'(0)&=1 \end{alignat} I think I would treat them like regular ODEs but the numbers I'm getting are a bit weird. Can anyone help me out?