Added December 20, 2018.
Second order PDE in Polar coordinates. Solve for \(u(r,\theta )\) \[ u_{rr} + u_{\theta \theta } = 0 \] With boundary conditions \begin {align*} u(2,\theta )&=3 \sin (2 \theta )+1 \end {align*}
Mathematica ✗
ClearAll["Global`*"]; pde = D[u[r, theta], {r, 2}] + D[u[r, theta], {theta, 2}] == 0; bc = u[2, theta] == 3*Sin[2*theta] + 1; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[r, theta], {r, theta}], 60*10]];
Failed
Maple ✓
restart; pde := diff(u(r, theta), r$2)+diff(u(r, theta), theta$2) = 0; bc := u(2, theta) = 3*sin(2*theta)+1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc], u(r,theta),method = Fourier)),output='realtime'));
\[u \left ( r,\theta \right ) =3/2\,i{{\rm e}^{2\,r-4-2\,i\theta }}-3/2\,i{{\rm e}^{-2\,r+4+2\,i\theta }}+1\]
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