Asymptotic Expansion, Regular Perturbation Regular perturbation. Find the first two terms in an asymptotic expansion of the small parameter $ϵ$ of the solution of
  $$
xy'+y=ϵy^{1/2},\quad x&gt;0,\quad y(1)=1.
$$
  Explain why the expansion is not valid as $x\to\infty$. What form of rescaling would be necessary to examine behaviour for large $x$? 


I've been learning about how to construct asymptotic solutions to regularly perturbed DEs, but I'm unsure of how to treat the $$y^{1/2}$$ on the right hand side. 

So I divide through by $x$, and set an expansion for $y$ in terms of $y_i$, but how do I do the square root of a series?

Thank you for any help given!
The lowest order solution is obviously $y=x^{-1}$ leading to the first order equation
$$
(xy_1)'=x^{-1/2}, \; y_1(1)=0
\implies xy_1=2(x^{1/2}-1)
\implies y_1=2(x^{-1/2}-x^{-1})
$$
In the next approximation one gets 
$$
(x(y_0+ϵy_1+ϵ^2y_2))'=ϵy_0^{1/2}(1+\tfrac12ϵy_0^{-1}y_1)
\\
(xy_2)'=1-x^{-1/2}, y_2(1)=0
\\
y_2=1-2x^{-1/2}+x^{-1}
$$
Thus the first three terms of the asymptotic expansion are
$$
y(x)=x^{-1}+2ϵ(x^{-1/2}-x^{-1})+ϵ^2(1-2x^{-1/2}+x^{-1})
$$



For the exact solution you get
$$
(\sqrt{xy})'=\frac12\frac{(xy)'}{\sqrt{xy}}=\frac12ϵx^{-1/2}
\implies
\sqrt{xy}=ϵx^{1/2}+1-ϵ
\\
\implies y=x^{-1}+2ϵ(x^{-1/2}-x^{-1})+ϵ^2(x^{-1}-2x^{-1/2}+1)
$$

So the above three term expansion is already the exact solution.
With regard to the later part of the question, the closed form of the solution by @LutzL doesn't seem to have issue when ${x \to ∞}$. Nevertheless, I think that what it may be getting at is that this expansion is asymptotic, so $$\epsilon (\frac{1}{√x}-\frac{1}{x}) = o(\frac{1}{x})$$ as ${\epsilon \to 0} $. However, for $x=O(\frac{1}{\epsilon^2})$ this is not the case, so the expansion is invalid as ${x \to ∞}$. Thus mutliscale perturbation theory is required, which is enough to say for the question.

I'm not sure if this is right, but this is my answer for when I hand this in 
(DEs II has got lousy).