Por supuesto usar separación de variables:
Que $f(t,x)=T(t)X(x)$,
Entonces $T'(t)X(x)-T(t)X''(x)=0$
$T'(t)X(x)=T(t)X''(x)$
$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore f(t,x)=C_1x+C_2+\int_sC_3(s)e^{-ts^2}\sin xs~ds+\int_sC_4(s)e^{-ts^2}\cos xs~ds$ or $f(t,x)=C_1x+C_2+\sum_sC_3(s)e^{-ts^2}\sin xs+\sum_sC_4(s)e^{-ts^2}\cos xs$
Esta es ya la solución general en $[0,\infty)\times(-\infty,\infty)$, no sólo en $[0,T)\times(-\infty,\infty)$.
Existan más $f(0,x)=0$, es mejor elegir $f(t,x)=C_1x+C_2+\int_0^\infty C_3(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds$
Entonces el substituto $f(0,x)=0$:
$C_1x+C_2+\int_0^\infty C_3(s)\sin xs~ds+\int_0^\infty C_4(s)\cos xs~ds=0$
$C_1x+C_2+\mathcal{F}_{s,s\to x}\{C_3(s)\}+\mathcal{F}_{c,s\to x}\{C_4(s)\}=0$
$\mathcal{F}_{s,s\to x}\{C_3(s)\}=-C_1x-C_2-\mathcal{F}_{c,s\to x}\{C_4(s)\}$
$C_3(s)=C_1\delta'(s)-\dfrac{2C_2}{\pi s}-\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}$
$\therefore f(t,x)=C_1x+C_2+C_1\int_0^\infty\delta'(s)e^{-ts^2}\sin xs~ds-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds=C_1x+C_2-C_1\Biggl.\dfrac{d}{ds}(e^{-ts^2}\sin xs)\Biggr|_{s=0}-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds=C_1x+C_2-C_1x-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds=C_2-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds$