Why is the Wronskian of these two functions equal to $\frac{2}{\sqrt{\pi}}$

Question

I can't seem to get the right answer

$ f = e^{\frac{y^2}{2}}$ and $g =e^{\frac{y^2}{2}}erf(y)$ where $erf(y) = \frac{2}{\sqrt\pi}\int_{0}^{y}e^{-\alpha^2}d\alpha$

I get $W = \frac{2}{\sqrt\pi} - ye^{y^2}erf(y)$

Am I making some kind of stupid mistake that I can't get the answer.

Thanks

Answer 1

I get $$ W(f,g)(y)=f(y)g'(y)-g(y)f'(y)= e^{\frac{y^2}{2}} \left(e^{\frac{y^2}{2}} y \text{ erf}(y)+\frac{2 e^{-\frac{y^2}{2}}}{\sqrt{\pi }}\right)-e^{y^2} y \text{ erf}(y) ={2\over \sqrt{\pi}}. $$

Answer 2

I think you may have calculated $g'$ incorrectly. By the Fundamental Theorem of Calculus,

$$g'(y) = yg(y) + e^{y^2/2}\frac{2}{\sqrt{\pi}}e^{-y^2} = yg(y) + \frac{2}{\sqrt{\pi}}e^{-y^2/2} $$

Try recalculating now the Wronksian.