Singular Perturbation Problem? Asymptotics? I am in the midst of solving this equation 
$\epsilon \ddot{y}+\dot{y}+1-\frac{1}{(y+1)^{2}}=0$ with the boundary condition $y(0)=1$ and $\dot{y}(0)=-1$
and $\epsilon$ is small. To start off with, I asymptotically expanded $y$ to yield
$y=y_{0}+\epsilon y_{1}$
and substituted the above and solve for $\mathcal{O}(1)$ and $\mathcal{O}(\epsilon)$ for the outer solution. The thing is, how can we make sure that the asymptotic expansion chosen for $y$ is correct? What happens if I choose to expand $y$ like 
$y=y_{0}+\epsilon ^2 y_{1}$ or even $y=y_{0}+\sqrt{\epsilon} y_{1}$ to solve the outer solution? 
 A: Assuming there is a solution with an asymptotic series $y = y_0 + \epsilon^p y_1 + \ldots$, substituting it in to the left side of the differential equation should give something whose asymptotic series is $0$ to all orders.  If you used $p > 1$, there would be a problem at order $\epsilon$ (unless $\ddot{y_0}$ happened to be $0$ - that actually does occur with the solutions $y_0=0$ and $y_0 = -2$).  If you used $0<p<1$, you might have a problem at order $\epsilon^p$.
Hmmm: actually you do get solutions with $p=1/2$, for example:
$$y(t) =  \epsilon^{1/2} e^{-2t} - \dfrac{3 }{2} \epsilon e^{-4t} + \left(\dfrac{13}{4} e^{-6t} - 4 t e^{-2t}\right) \epsilon^{3/2} + \ldots$$
EDIT: Actually, if you don't worry about initial conditions, for any $p \in (0,1)$ there will be solutions with $y(t) = y_0(t) + \epsilon^p y_1(t) + o(\epsilon^p)$ (in some neighbourhood of $t=0$).  You just specify initial conditions for $y$ to differ by something like 
$\epsilon^p$ from those for $y_0$.  But if (as you now have done) you require fixed 
initial conditions, things are different.
However, your current initial conditions will not work.  If $y_0$ is the solution for $\epsilon = 0$, $y_0(0)=1$ would imply $\dot{y_0}(0) = -3/4$, not $-1$.
Let's try initial conditions $y(0)=1$, $\dot{y}(0) = -3/4$.  If you take $y = y_0 + \epsilon^p y_1 + \ldots$ with $0 < p < 1$, the order $\epsilon^p$ term in the equation will say $\dot{y_1} + \dfrac{2 y_1}{(y_0+1)^3} = 0$, but you also
need $y_1(0)=0$, and the solution to that would be $y_1 = 0$.     
A: Your question is, in general, concerned with selecting the correct set of gauge functions to use in your asymptotic expansion. The book "Introduction to Perturbation Methods" by Mark Holmes gives a nice description of gauge functions as well as has numerous examples and problems related to the exponent matching procedure described above.
