Use Laplace transform to solve:
$u_x = 2 u_t + u , u(x,0)=2e^{-3x}; $ $ u$ bounded for $x>0,t>0.$
$\frac{\bar{\partial{u}}}{\partial{x}}= 2 [s \bar{u}(x,s)-u(x,0)]+\bar{u}$
$\frac{\bar{\partial{u}}}{\partial{x}}=2[s\bar{u}-2e^{-3x}]+\bar{u}$
$\frac{\bar{\partial{u}}}{\partial{x}}=\bar{u}(2s+1)+2e^{-3x}$
$\bar{u}=\hat{u}x(2s+1)+\frac{2}{-3}e^{-3x}+B$
I dont know what comes next?
Consider $${{u}_{x}}=2{{u}_{t}}+u$$ $$u\left( x,0 \right)=2{{e}^{-3x}}$$ hence $${{U}_{x}}=2\left( sU-u\left( 0,x \right) \right)+U$$ Where U is the laplace transform of u w.r.t t. Therefore $${{U}_{x}}-\left( 2s+1 \right)U=-4{{e}^{-3x}}$$ Using an integrating factor one obtains $$\frac{\partial }{\partial x}U{{e}^{-\left( 2s+1 \right)x}}=-4{{e}^{-3x}}{{e}^{-\left( 2s+1 \right)x}}$$ Solving yeilds $$U=\frac{2}{\left( 2+s \right)}{{e}^{-3x}}+A\left( s \right){{e}^{\left( 2s+1 \right)x}}$$ Inverting we obtain $$u\left( x,t \right)=\frac{1}{2\pi i}\int\limits_{C}^{{}}{\frac{2{{e}^{-3x}}{{e}^{st}}}{\left( 2+s \right)}+A\left( s \right){{e}^{\left( 2s+1 \right)x}}{{e}^{st}}ds}$$ where the contour is the Bromwich contour. Residue calculus yields $$u\left( x,t \right)=2{{e}^{-3x-2t}}+\frac{1}{2\pi i}\int\limits_{C}^{{}}{A\left( s \right){{e}^{\left( 2s+1 \right)x}}{{e}^{st}}ds}$$ Note $$u\left( x,0 \right)=2{{e}^{-3x}}+\frac{1}{2\pi i}\int\limits_{C}^{{}}{A\left( s \right){{e}^{\left( 2s+1 \right)x}}ds}\Rightarrow \frac{1}{2\pi i}\int\limits_{C}^{{}}{A\left( s \right){{e}^{\left( 2s+1 \right)x}}ds}=0$$ And so A=0. Therefore $$u\left( x,t \right)=2{{e}^{-3x-2t}}$$.