Added December 27, 2018.
Taken from https://en.wikipedia.org/wiki/List_of_nonlinear_partial_differential_equations
Fisher’s equation. Solve for \(u(x,t)\) \[ u_t = u(1-u)+u_{xx} \]
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == u[x, t]*(1 - u[x, t]) + D[u[x, t], {x, 2}]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, t], {x, t}], 60*10]];
\begin {align*} & \left \{u(x,t)\to \frac {1}{4} \left (1+\tanh \left (\frac {1}{12} \left (5 t-\sqrt {6} x-12 c_3\right )\right )\right ){}^2\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {1}{12} \left (5 t-i \sqrt {6} x-12 c_3\right )\right )\right ) \left (1+\tanh \left (\frac {1}{12} \left (5 t-i \sqrt {6} x-12 c_3\right )\right )\right )\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {1}{12} \left (5 t+i \sqrt {6} x-12 c_3\right )\right )\right ) \left (1+\tanh \left (\frac {1}{12} \left (5 t+i \sqrt {6} x-12 c_3\right )\right )\right )\right \}\\& \left \{u(x,t)\to \frac {1}{4} \left (1+\tanh \left (\frac {1}{12} \left (5 t+\sqrt {6} x-12 c_3\right )\right )\right ){}^2\right \}\\& \left \{u(x,t)\to \frac {1}{4} \left (1+\tanh \left (\frac {5 t}{12}-\frac {x}{2 \sqrt {6}}+c_3\right )\right ){}^2\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {5 t}{12}-\frac {i x}{2 \sqrt {6}}+c_3\right )\right ) \left (1+\tanh \left (\frac {5 t}{12}-\frac {i x}{2 \sqrt {6}}+c_3\right )\right )\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {5 t}{12}+\frac {i x}{2 \sqrt {6}}+c_3\right )\right ) \left (1+\tanh \left (\frac {5 t}{12}+\frac {i x}{2 \sqrt {6}}+c_3\right )\right )\right \}\\& \left \{u(x,t)\to \frac {1}{4} \left (1+\tanh \left (\frac {5 t}{12}+\frac {x}{2 \sqrt {6}}+c_3\right )\right ){}^2\right \}\\ \end {align*}
Maple ✓
restart; pde := diff(u(x,t),t)= u(x,t)*(1-u(x,t))+ diff(u(x,t),x$2); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',PDEtools:-TWSolutions(pde,u(x,t))),output='realtime'));
\begin {align*} & \left \{ u \left ( x,t \right ) =1 \right \} \\& \left \{ u \left ( x,t \right ) ={\frac {1}{4} \left ( \tanh \left ( -{\frac {5\,t}{12}}+{\frac {\sqrt {6}x}{12}}+{\it \_C1} \right ) \right ) ^{2}}-{\frac {1}{2}\tanh \left ( -{\frac {5\,t}{12}}+{\frac {\sqrt {6}x}{12}}+{\it \_C1} \right ) }+{\frac {1}{4}} \right \} \\& \left \{ u \left ( x,t \right ) ={\frac {1}{4} \left ( \tanh \left ( {\frac {5\,t}{12}}-{\frac {\sqrt {6}x}{12}}+{\it \_C1} \right ) \right ) ^{2}}+{\frac {1}{2}\tanh \left ( {\frac {5\,t}{12}}-{\frac {\sqrt {6}x}{12}}+{\it \_C1} \right ) }+{\frac {1}{4}} \right \} \\& \left \{ u \left ( x,t \right ) ={\frac {1}{4} \left ( \tanh \left ( -{\frac {5\,t}{12}}-{\frac {\sqrt {6}x}{12}}+{\it \_C1} \right ) \right ) ^{2}}-{\frac {1}{2}\tanh \left ( -{\frac {5\,t}{12}}-{\frac {\sqrt {6}x}{12}}+{\it \_C1} \right ) }+{\frac {1}{4}} \right \} \\& \left \{ u \left ( x,t \right ) ={\frac {1}{4} \left ( \tanh \left ( {\frac {5\,t}{12}}+{\frac {\sqrt {6}x}{12}}+{\it \_C1} \right ) \right ) ^{2}}+{\frac {1}{2}\tanh \left ( {\frac {5\,t}{12}}+{\frac {\sqrt {6}x}{12}}+{\it \_C1} \right ) }+{\frac {1}{4}} \right \} \\& \left \{ u \left ( x,t \right ) =-{\frac {1}{4} \left ( \tanh \left ( -{\frac {5\,t}{12}}+{\frac {i}{12}}\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}}-{\frac {1}{2}\tanh \left ( -{\frac {5\,t}{12}}+{\frac {i}{12}}\sqrt {6}x+{\it \_C1} \right ) }+{\frac {3}{4}} \right \} \\& \left \{ u \left ( x,t \right ) =-{\frac {1}{4} \left ( \tanh \left ( {\frac {5\,t}{12}}-{\frac {i}{12}}\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}}+{\frac {1}{2}\tanh \left ( {\frac {5\,t}{12}}-{\frac {i}{12}}\sqrt {6}x+{\it \_C1} \right ) }+{\frac {3}{4}} \right \} \\& \left \{ u \left ( x,t \right ) =-{\frac {1}{4} \left ( \tanh \left ( -{\frac {5\,t}{12}}-{\frac {i}{12}}\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}}-{\frac {1}{2}\tanh \left ( -{\frac {5\,t}{12}}-{\frac {i}{12}}\sqrt {6}x+{\it \_C1} \right ) }+{\frac {3}{4}} \right \} \\& \left \{ u \left ( x,t \right ) =-{\frac {1}{4} \left ( \tanh \left ( {\frac {5\,t}{12}}+{\frac {i}{12}}\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}}+{\frac {1}{2}\tanh \left ( {\frac {5\,t}{12}}+{\frac {i}{12}}\sqrt {6}x+{\it \_C1} \right ) }+{\frac {3}{4}} \right \} \\ \end {align*}
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