Order of the solution to an IVP

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Suppose you have the following IVP

$$\dot{y}(t) = d_1 y^2 + d_2 \epsilon^{-1} y$$

with $y(0) = y_0$ and where $d_1$ and $d_2$ are two positive constants independent of $\epsilon$.

What can you conclude about the order of the solution $y$ in terms of $\epsilon$?

Any comments welcome. I no nothing of asymptotics so sorry if this is a silly question.

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The equation is separable:

$$\frac{dy}{A y^2+B y} = dt$$

where $A=d_1$ and $B=d_2/\epsilon$. By writing

$$\frac{1}{A y^2+B y} = \frac{1}{B} \left(\frac{1}{y} - \frac{A}{A y+b} \right)$$

and integrating both sides, you may derive the following solution in terms of your parameters:

$$y(t) = \frac{d_2 y_0 e^{d_2 t/\epsilon}}{d_2-\epsilon \, d_1 y_0 (e^{d_2 t/\epsilon}-1)}$$

Examining this, you can see the irregular singular behavior of this solution stemming from the behavior of the equation.