I have a question on particular solution of 1-D Heat equation $$\frac{\partial^{2} T(x, t)}{\partial x^{2}}=\frac{1}{\alpha} \frac{\partial T(x, t)}{\partial t} \quad \text { in } \quad 0<x<\infty, \quad t>0$$ $$\begin{array}{ll} \text { BC1: } & T(x=0)=f(t) \\ \text { BC2: } & T(x \rightarrow \infty)=0 \\ \text { IC: } & T(x, t=0)=0 \end{array}$$
It is well known that the analytical solution of this equation is given by
$$T(x, t)=\frac{x}{\sqrt{4 \pi \alpha}} \int_{\tau=0}^{t} \frac{f(\tau)}{(t-\tau)^{3 / 2}} \exp \left[\frac{-x^{2}}{4 \alpha(t-\tau)}\right] d \tau$$ Now if $f(\tau)=Asin(\omega\tau)$, then the solution that satisfies this equation could be also written as (particular solution) $$T(x, t)=Ae^{-x / d}[\sin (\omega t-x / d)]$$ where $$d=(2 \alpha /\omega)^{\frac{1}{2}}$$since this solution also satisfies the original Heat equation. However when I compare them I noticed that they are not same (I plot them in Matlab and found that the temperature profile are not same). Since this problem is well-posed so I assume the solution is unique, but the particular solution is widely used in research, is it a wrong solution?
What's the relationship between the analytical solution and particular solution? Also the limit of analytical solution as $t\rightarrow 0$ is not $f(\tau)$ since the Heat equation is defined as $0<x<\infty$, is there a discontinuity at $t=0$?
Thank for the help.