He was so lucky!

The "mistake" leads to the equation

$2xe^{x^{2}}g(x) + e^{x^{2}}g'(x) = 2xe^{x^{2}}g'(x)$ .

Now since $e^{x^{2}} \gt 0$ for all real $x$ we can cancel that factor from all the terms. Letting $y = g(x)$ we then end up with the differential equation

$2xy = (2x - 1)\frac{dy}{dx} \Longrightarrow \frac{dy}{y} = \frac{2x}{2x - 1} dx \Longrightarrow \ln(y) = x - \frac{1}{2} + \ln(\sqrt{2x - 1}) + C$ .

Now with $y = e$ when $x = 1$ we have that

$\\ln(e) = 1 - \frac{1}{2} + \ln(1) + C \Longrightarrow C = \frac{1}{2}$ ,

and so $\ln(y) - \ln(\sqrt{2x - 1}) = x \Longrightarrow y = \sqrt{2x - 1}*e^{x}$ .

Thus $g(5) = \sqrt{2*5 - 1}*e^{5} = 3e^{5}$ , making $\alpha + \beta = 3 + 5 = \boxed{8}$ .