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Added April 5, 2019.
Solve for \(u(x,t)\) in \[ u_t = u_{xx} - u_x \]
For \(t>0,x>0\). With boundary conditions \(u(0,t)=0\) and intitial conditions \(u(x,0)=f(x)\)
Mathematica ✗
ClearAll[x, t, f]; pde = D[u[x, t], t] == D[u[x, t], {x, 2}] - D[u[x, t], x]; ic = u[x, 0] == f[x]; bc = u[0, t] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}, Assumptions -> {x > 0, t > 0}], 60*10]];
\[ \text {Failed} \]
Maple ✓
unassign('x,t,f'); pde:=diff(u(x,t),t)=diff(u(x,t),x$2)- diff(u(x,t),x); ic:=u(x,0)=f(x); bc:=u(0,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc], u(x, t))assuming t>0,x>0),output='realtime'));
\[ u \left ( x,t \right ) ={{\rm e}^{x/2}}{\it invlaplace} \left ( {\frac {1}{\sqrt {{{\rm e}^{x\sqrt {1+4\,s}}}}\sqrt {1+4\,s}}\int ^{0}\!{\frac {{{\rm e}^{{\it \_a}}}{\it \_C2}+f \left ( {\it \_a} \right ) -{\it \_C2}}{\sqrt {{{\rm e}^{{\it \_a}\,\sqrt {1+4\,s}}}}\sqrt {{{\rm e}^{{\it \_a}}}}}}{d{\it \_a}}},s,t \right ) -{{\rm e}^{x/2}}{\it invlaplace} \left ( {\frac {1}{\sqrt {{{\rm e}^{x\sqrt {1+4\,s}}}}\sqrt {1+4\,s}}\int ^{0}\!{\frac {\sqrt {{{\rm e}^{{\it \_a}\,\sqrt {1+4\,s}}}} \left ( {{\rm e}^{{\it \_a}}}{\it \_C2}+f \left ( {\it \_a} \right ) -{\it \_C2} \right ) }{\sqrt {{{\rm e}^{{\it \_a}}}}}}{d{\it \_a}}},s,t \right ) -{{\rm e}^{x/2}}{\it invlaplace} \left ( {\frac {{{\rm e}^{1/2\,x\sqrt {1+4\,s}}}\int \! \left ( {{\rm e}^{x}}{\it \_C2}+f \left ( x \right ) -{\it \_C2} \right ) {{\rm e}^{-1/2\, \left ( 1+\sqrt {1+4\,s} \right ) x}}\,{\rm d}x}{\sqrt {1+4\,s}}},s,t \right ) +{{\rm e}^{x/2}}{\it invlaplace} \left ( {\frac {{{\rm e}^{-1/2\,x\sqrt {1+4\,s}}}\int \! \left ( {{\rm e}^{x}}{\it \_C2}+f \left ( x \right ) -{\it \_C2} \right ) {{\rm e}^{1/2\, \left ( -1+\sqrt {1+4\,s} \right ) x}}\,{\rm d}x}{\sqrt {1+4\,s}}},s,t \right ) -{{\rm e}^{x}}{\it \_C2}+{\it \_C2} \]