Given $g:[0,\infty] \rightarrow \mathbb{R}$ with $g(0)=0,$ derive the formula $$u(x,t)=\frac{x}{\sqrt{4 \pi}} \int_{0}^{t} \frac{1}{(t-s)^{3/2}} e^{\frac{-x^2}{4 (t-s)}} g(s)~ds,~~x > 0$$ for a solution of the IVP/BVP $$ \begin{cases} u_t - u_{xx}=0~~&\text{in}~~\mathbb{R}_{+} \times (0,\infty) \\ u=0~~&\text{on}~~\mathbb{R}_{+} \times \{ t=0 \}, \\ u=g~~&\text{on}~~\{ x=0 \} \times [0,\infty). \end{cases} $$ (Hint. Let $v(x,t):=u(x,t)-g(t)$ and extend $v$ to $\{ x< 0 \}$ by odd reflection.)
My approach: As suggested in the Hint, define $$ v(x,t)= \begin{cases} u(x,t)- g(t)~~& x >0, \\ -u(-x,t)- g(t)~~& x \leq 0. \end{cases} $$ So we have $$ v_t(x,t)= \begin{cases} u_t(x,t)- g'(t)~~& x >0, \\ -u_t(-x,t)- g'(t)~~& x \leq 0. \end{cases} $$ and $$ v_{xx}(x,t)= \begin{cases} u_{xx}(x,t)- g(t)~~& x >0, \\ -u_{xx}(-x,t)- g(t)~~& x \leq 0. \end{cases} $$ Thus $v$ satisfies the IVP/BVP $$ \begin{cases} & v_t(x,t)-v_{xx}(x,t)= \begin{cases} -g'(t)~~& x >0, \\ g'(t)~~& x \leq 0. \end{cases} \\ &v(x,0)=0, \\ &v(0,t)=0. \end{cases} $$ Using the solution formula for $v,$ we get \begin{align*} v(x,t)&=\int_{0}^{t} \frac{1}{\sqrt{4 \pi (t-s)}} \bigg\{ \int_{-\infty}^{\infty} e^{\frac{-(y-x)^2}{4 (t-s)}} g'(x)~dy \bigg\}~ds \\ &=\int_{0}^{t} \frac{1}{\sqrt{4 \pi (t-s)}} \bigg\{ \int_{-\infty}^{0} e^{\frac{-(y-x)^2}{4 (t-s)}} g'(x)~dy - \int_{0}^{\infty} e^{\frac{-(y-x)^2}{4 (t-s)}} g'(s)~dy \bigg\}~ds. \end{align*}
I'm stuck here. Any help in continuing from here onwards to obtain the desired result is much appreciated.