Fisher’s equation. Solve for \(u(x,t)\)
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*}& \{u \left (x , t\right ) = 1\}\\& \left \{u \left (x , t\right ) = \frac {\tanh \left (-\frac {5 t}{12}+\frac {\sqrt {6}\, x}{12}+c_1 \right )^{2}}{4}-\frac {\tanh \left (-\frac {5 t}{12}+\frac {\sqrt {6}\, x}{12}+c_1 \right )}{2}+\frac {1}{4}\right \}\\& \left \{u \left (x , t\right ) = \frac {\tanh \left (\frac {5 t}{12}-\frac {\sqrt {6}\, x}{12}+c_1 \right )^{2}}{4}+\frac {\tanh \left (\frac {5 t}{12}-\frac {\sqrt {6}\, x}{12}+c_1 \right )}{2}+\frac {1}{4}\right \}\\& \left \{u \left (x , t\right ) = \frac {\tanh \left (-\frac {5 t}{12}-\frac {\sqrt {6}\, x}{12}+c_1 \right )^{2}}{4}-\frac {\tanh \left (-\frac {5 t}{12}-\frac {\sqrt {6}\, x}{12}+c_1 \right )}{2}+\frac {1}{4}\right \}\\& \left \{u \left (x , t\right ) = \frac {\tanh \left (\frac {5 t}{12}+\frac {\sqrt {6}\, x}{12}+c_1 \right )^{2}}{4}+\frac {\tanh \left (\frac {5 t}{12}+\frac {\sqrt {6}\, x}{12}+c_1 \right )}{2}+\frac {1}{4}\right \}\\& \left \{u \left (x , t\right ) = -\frac {\tanh \left (-\frac {5 t}{12}+\frac {i \sqrt {6}\, x}{12}+c_1 \right )^{2}}{4}-\frac {\tanh \left (-\frac {5 t}{12}+\frac {i \sqrt {6}\, x}{12}+c_1 \right )}{2}+\frac {3}{4}\right \}\\& \left \{u \left (x , t\right ) = -\frac {\tanh \left (\frac {5 t}{12}-\frac {i \sqrt {6}\, x}{12}+c_1 \right )^{2}}{4}+\frac {\tanh \left (\frac {5 t}{12}-\frac {i \sqrt {6}\, x}{12}+c_1 \right )}{2}+\frac {3}{4}\right \}\\& \left \{u \left (x , t\right ) = -\frac {\tanh \left (-\frac {5 t}{12}-\frac {i \sqrt {6}\, x}{12}+c_1 \right )^{2}}{4}-\frac {\tanh \left (-\frac {5 t}{12}-\frac {i \sqrt {6}\, x}{12}+c_1 \right )}{2}+\frac {3}{4}\right \}\\& \left \{u \left (x , t\right ) = -\frac {\tanh \left (\frac {5 t}{12}+\frac {i \sqrt {6}\, x}{12}+c_1 \right )^{2}}{4}+\frac {\tanh \left (\frac {5 t}{12}+\frac {i \sqrt {6}\, x}{12}+c_1 \right )}{2}+\frac {3}{4}\right \}\\ \end{align*}
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