how do i show that,
$$ u(x, t)=\frac{1}{\sqrt{4\pi t}}\int_0^\infty y^2e^{-(x-y)^2/4t} \, dy = ( x^2+2t )N\left(\frac{x}{\sqrt{2t}}\right) + \sqrt{\frac{t}{\pi}} xe^{-x^2/4t} $$
where
$$ N(z)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^z e^{-p^2/2} \, dp $$
[update]
I managed to get the answer by,
let $p=(x-y)/\sqrt{2t}$
then
$$ u(x, t)=\frac{1}{\sqrt{2\pi }}\int_{-\infty}^{\frac{x}{\sqrt{2t}}} (x-\sqrt{2t}p)^2e^{-(x-y)^2/4t} \, dy \\= \frac{1}{\sqrt{2\pi }}\int_{-\infty}^{\frac{x}{\sqrt{2t}}} (x^2- 2\sqrt{2t}px+2tp^2)e^{-(x-y)^2/4t}\\=x^2N(x/\sqrt{2t}) -2\sqrt{\frac{t}{\pi}}x\int_{-\infty}^{\frac{x}{\sqrt{2t}}}pe^{-p^2/2}dp+\frac{2t}{\sqrt{2\pi}}\int_{-\infty}^{\frac{x}{\sqrt{2t}}}p^2e^{-p^2/2}dp\\=x^2N(x/\sqrt{2t})+2\sqrt{\frac{t}{\pi}}xe^{-x^2/4t}+\left(-\sqrt{\frac{t}{\pi}}xe^{-x^2/4t}+2tN(x/\sqrt{2t})\right)\\=( x^2+2t )N\left(\frac{x}{\sqrt{2t}}\right) + \sqrt{\frac{t}{\pi}} xe^{-x^2/4t} $$
