Let $U(t,x)=T(t)X(x)$ ,
Then $T''(t)X(x)+aT'(t)X(x)=b^2T(t)X''(x)$
$(T''(t)+aT'(t))X(x)=b^2T(t)X''(x)$
$\dfrac{T''(t)+aT'(t)}{T(t)}=\dfrac{b^2X''(x)}{X(x)}=\dfrac{4s^2-a^2}{4}$
$\begin{cases}T''(t)+aT'(t)+\dfrac{a^2-4s^2}{4}T(t)=0\\X''(x)+\dfrac{a^2-4s^2}{4b^2}X(x)=0\end{cases}$
$\begin{cases}T(t)=\begin{cases}c_1(s)e^{-\frac{at}{2}}\sinh ts+c_2(s)e^{-\frac{at}{2}}\cosh ts&\text{when}~s\neq0\\c_1te^{-\frac{at}{2}}+c_2e^{-\frac{at}{2}}&\text{when}~s=0\end{cases}\\X(x)=\begin{cases}c_3(s)e^{\frac{x\sqrt{4s^2-a^2}}{2|b|}}+c_4(s)e^{-\frac{x\sqrt{4s^2-a^2}}{2|b|}}&\text{when}~s\neq\pm\dfrac{a}{2}\\c_3x+c_4&\text{when}~s=\pm\dfrac{a}{2}\end{cases}\end{cases}$
$\therefore U(t,x)=\int_\frac{a}{2}^\infty C_1(s)e^{-\frac{a|b|t+x\sqrt{4s^2-a^2}}{2|b|}}\sinh ts~ds+\int_\frac{a}{2}^\infty C_2(s)e^{-\frac{a|b|t+x\sqrt{4s^2-a^2}}{2|b|}}\cosh ts~ds+\int_\frac{a}{2}^\infty C_3(s)e^{-\frac{a|b|t-x\sqrt{4s^2-a^2}}{2|b|}}\sinh ts~ds+\int_\frac{a}{2}^\infty C_4(s)e^{-\frac{a|b|t-x\sqrt{4s^2-a^2}}{2|b|}}\cosh ts~ds$
$U(t,x)\to 0$ as $x\to\infty$ :
$C_3(s)=0$ , $C_4(s)=0$
$\therefore U(t,x)=\int_\frac{a}{2}^\infty C_1(s)e^{-\frac{a|b|t+x\sqrt{4s^2-a^2}}{2|b|}}\sinh ts~ds+\int_\frac{a}{2}^\infty C_2(s)e^{-\frac{a|b|t+x\sqrt{4s^2-a^2}}{2|b|}}\cosh ts~ds$