Added January 20, 2018.
Beam PDE with zero initial velocity. Solve \[ u_{tt} + u_{xxxx} = 0 \] With boundary conditions \begin {align*} u(0,t) &= -12 t^2\\ f(1,t) &=1-12 t^2\\ \frac {\partial ^2 u}{\partial x^2}u(0,t) &=0 \\ \frac {\partial ^2 u}{\partial x^2}u(1,t) &=12 \end {align*}
And initial conditions \begin {align*} u(x,0) &= x^4\\ \frac {\partial u}{\partial t}u(x,0) &=0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] + D[u[x, t], {x, 4}] == 0; bc = {u[0, t] == -12*t^2, u[1, t] == 1 - 12*t^2, Derivative[2, 0][u][0, t] == 0, Derivative[2, 0][u][1, t] == 12}; ic = {u[x, 0] == x^4, Derivative[0, 1][u][x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t], 60*10]];
\[\left \{\left \{u(x,t)\to x^4-12 t^2\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)+diff(u(x,t),x$4)=0; bc := u(0,t)=-12*t^2, u(1,t)=1-12*t^2,D[1,1](u)(0,t)=0, D[1,1](u)(1,t)=12; ic := u(x,0)=x^4,D[2](u)(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t),HINT=`+`)),output='realtime'));
\[u \left ( x,t \right ) ={x}^{4}-12\,{t}^{2}\]