# Asymptotic solution

I am looking for asymptotic solutions to the equation $$\alpha^{-1}x+\sqrt{\pi}\frac{\sqrt{x}}{2}\text{erf}\left(\frac{\sqrt{x}}{2}\right)=\beta^{-1}e^{-x/4},\qquad \alpha\ll1,\beta\gg1.$$ When $$\alpha$$ is large and the first term is negligible, this is easy to do, but I don't know how to proceed with the opposite case.

What I've tried for now is the following: For $$\alpha,\beta^{-1}=0$$, which is the limiting case, I get $$x=0$$, hence I have to introduce a scaling $$x=\epsilon\hat x$$, where $$\epsilon=\epsilon(\alpha,\beta)\ll1$$. Introducing this into the equation above allows me to simplify terms and reduce the equation (if I'm not wrong) to $$\epsilon(1\color{red}{+}\alpha/2)\hat x=\alpha\beta^{-1},$$ therefore I can balance the equation by choosing $$\epsilon=\alpha\beta^{-1}$$ and finally $$\hat x\approx\color{red}{2/(2+\alpha)}\qquad\Rightarrow\qquad x\approx\alpha\beta^{-1}.$$

Is this correct? Any hints or help on this?

$$\color{red}{\text{Edit: The leading order term had a mistake, I have corrected it.}}$$
Let $$\tilde \beta = 1/\beta$$. Multiplying by $$\alpha$$ and getting rid of the square roots, we can rewrite the equation as $$x - \alpha \tilde \beta e^{-x/4} + \alpha x \int_0^{1/2} e^{-x t^2} d t = 0.$$ Now we can look for $$x$$ in the form $$\sum c_{i,j} \alpha^i \tilde \beta {}^j$$ by substituting the sum into the equation and taking the bivariate Taylor expansion around $$\alpha = 0, \,\tilde \beta = 0$$.
Taking $$x = \alpha \tilde \beta + \sum_{i = 0}^3 c_{i, 3 - i} \alpha^i \tilde \beta {}^{3 - i}$$ gives $$c_{3, 0} \alpha^3 + \left( c_{2, 1} + \frac 1 2 \right) \alpha^2 \tilde \beta + c_{1, 2} \alpha \tilde \beta {}^2 + c_{0, 3} \tilde \beta {}^3 = 0,$$ therefore we get one third-order term $$-\alpha^2 \tilde \beta/2$$.
On the next step we get two fourth-order terms, which gives the approximation $$x \approx \frac \alpha \beta -\frac {\alpha^2} {2 \beta} + \frac {\alpha^3} {4 \beta} - \frac {\alpha^2} {4 \beta^2}.$$