Determine inner and outer expansion for boundary layer

Question

I have a brief question in regards to defining outer and inner expansion in a boundary layer problem. We have given:

$\epsilon \frac{d^{2}f}{dx^2} + (2+x)\frac{df}{dx} + f = 1, \quad 0 < x < 1$

$0 < \epsilon \ll 1, \quad f(0) = 2, \quad f(1) = 0$

I am able to define the outer expansion, but I have a question regarding the inner expansion. Here I am supposed to let $s = x/\epsilon$. This will then give me:

$\frac{1}{\epsilon}\frac{d^{2}f}{ds^2} + (2 + \epsilon s)\frac{1}{\epsilon}\frac{df}{ds} + f = 1$

Multiplying every term with $\epsilon$ yields:

$\frac{d^{2}f}{ds^2} + 2\frac{df}{ds} + \epsilon s \frac{df}{ds} + \epsilon f = \epsilon$

If we then define $f = f_0 + \epsilon f_1 + . . .$

Should the $\epsilon$ on the right side of this equality be included in the equation for $f_0$ or $f_1$? In other words should I let:

$O(1): \frac{d^{2}f_0}{ds^2} + 2\frac{df_0}{ds} = 0$

or

$O(1): \frac{d^{2}f_0}{ds^2} + 2\frac{df_0}{ds} = \epsilon$

And likewise, should I let:

$O(\epsilon): \frac{d^{2}f_1}{ds^2} + 2\frac{df_1}{ds} = -s\frac{df_0}{ds} - f_0$

or:

$O(\epsilon): \frac{d^{2}f_1}{ds^2} + 2\frac{df_1}{ds} = -s\frac{df_0}{ds} - f_0 + 1$

If anyone can explain this to me I will be very grateful!