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The original question is from here, I am trying to work out a full solution following the hint in the answer and the comments.

Given the ODE $$y' - (y-1)^2 -\epsilon\frac{y^2}{x^2} = 0, \quad y(1) = 1,$$ we would like to find a uniform approximation for $0\leq x \leq 1$.

To approximate the solution outside of the boundary layer $$y_{out} = f_0 + \epsilon f_1 + O(\epsilon^2).$$ The equation for $f_0$ is $f_0' - (f_0 -1)^2 = 0$, we get $$f_0 \equiv 1 \quad \text{ or } \quad f_0(x) = 1-\frac{1}{x-A}.$$ Now I assumed there is no boundary layer at $x=1$, with the condition $f_0(1) = 1$ we see that $f_0\equiv 1$. And we will find that at $\epsilon$ level, we have $f_1' - {1}/{x^2} = 0$, and therefore $f_1(x) = 1-{1}/{x}$ with the condition $f_1(1) = 0$. So $$y_{out} = 1 + \epsilon \left( 1-\frac{1}{x}\right) + O(\epsilon^2).$$

Next we see that there is a problem when $x$ is close to zero. so assume there is a boundary layer at $x=0$. We substitute $z=x/\epsilon^p$ and the ODE becomes $$\epsilon^{-p}y'(z) - (y(z) -1)^2 - \epsilon^{-2p+1}\frac{y^2}{z^2} = 0.$$ Now assume $y_{in} = g_0 + O(\epsilon)$, and with dominate balance, we see that $p=1$, thus $g_0' - g_0^2/z^2 = 0$ and we get $$g_0 = \frac{1}{1/z+B}.$$ Now we will do the matching at $O(\epsilon)$ $$\lim_{z\rightarrow \infty} \frac{1}{1/z+B} = \lim_{x\rightarrow 0} 1$$ and we get $B = 1 = y_{match}$.

So $$y_{uniform} = y_{out} + y_{in} - y_{match} = \frac{1}{\frac{\epsilon}{x}+1} + O(\epsilon).$$

Is my solution correct?

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    $\begingroup$ There's nothing stopping you from satisfying the boundary condition with your inner expansion, so $$y=\frac{1}{\frac{\epsilon}{x}+1-\epsilon}$$ is a better solution, the outer expansion is unnecessary. Even for $x$ near 1 it's a better approximation than the outer solution $y_{out}=1$. $\endgroup$ – David Aug 8 '18 at 20:10
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    $\begingroup$ Additionally, notice that your outer solution $y_{out}=1$ does not give a dominant balance, since $y'=0$ and $(y-1)^2=0$, the dominant term is actually $\epsilon/x^2=O(\epsilon)$. This is why the outer solution doesn't help. I think if the boundary condition was not 1, then things might be different because the outer solution would be non-trivial. $\endgroup$ – David Aug 9 '18 at 20:22

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