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Taken from Mathematica DSolve help pages.
Solve for \(u\left ( x,y\right ) \) \begin {align*} u_{xx}+ u_{yy} + 5 u(x,y) & = 0 \end {align*}
Boundary conditions \begin {align*} u(x,0) &= \text {UnitTriangle[x-2]} \\ u(x,2) &= 0 \\ u(0,y) &= 0 \\ u(4,y) &=0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = {Laplacian[u[x, y], {x, y}] + 5*u[x, y] == 0}; bc = {u[x, 0] == Piecewise[{{-1 + x, x > 1 && x < 2}, {3 - x, x > 2 && x < 3}}], u[x, 2] == 0, u[0, y] == 0, u[4, y] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]]; sol = sol /. K[1] -> n
\[\left \{\left \{u(x,y)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {64 \left (\cos \left (\frac {n \pi }{8}\right )+\cos \left (\frac {3 n \pi }{8}\right )\right ) \text {csch}\left (\frac {1}{2} \sqrt {n^2 \pi ^2-80}\right ) \sin ^3\left (\frac {n \pi }{8}\right ) \sin \left (\frac {n \pi x}{4}\right ) \sinh \left (\frac {1}{4} \sqrt {n^2 \pi ^2-80} (2-y)\right )}{n^2 \pi ^2}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,y),x$2)+diff(u(x,y),y$2)+5*u(x,y)=0; bc := u(x,0)=piecewise( x>1 and x<2, -1+x,x>2 and x<3 ,3-x), u(x,2)=0, u(0,y)=0, u(4,y)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime'));
\[u \left ( x,y \right ) =\sum _{n=1}^{\infty }32\,{\frac {\sin \left ( 1/4\,n\pi \,x \right ) \left ( 1/2\, \left ( \sin \left ( 1/2\,n\pi \right ) -1/2\,\sin \left ( 1/4\,n\pi \right ) -1/2\,\sin \left ( 3/4\,n\pi \right ) \right ) \sin \left ( 1/2\,\sqrt {-{\pi }^{2}{n}^{2}+80} \right ) \cos \left ( 1/4\,\sqrt {-{\pi }^{2}{n}^{2}+80}y \right ) +\sin \left ( 1/4\,n\pi \right ) \cos \left ( 1/4\,n\pi \right ) \cos \left ( 1/2\,\sqrt {-{\pi }^{2}{n}^{2}+80} \right ) \sin \left ( 1/4\,\sqrt {-{\pi }^{2}{n}^{2}+80}y \right ) \left ( \cos \left ( 1/4\,n\pi \right ) -1 \right ) \right ) }{\sin \left ( 1/2\,\sqrt {-{\pi }^{2}{n}^{2}+80} \right ) {n}^{2}{\pi }^{2}}}\]
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Added December 27, 2018.
Solve for \(u\left ( x,y\right ) \) \begin {align*} u_{xx}+u_{yy} + 5 u(x,y) & = 0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = {Laplacian[u[x, y], {x, y}] + 5*u[x, y] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];
\[\left \{\left \{u(x,y)\to e^{\sqrt {c_5} (-x)} \left (c_1 e^{2 \sqrt {c_5} x}+c_2\right ) \left (c_4 \cos \left (\sqrt {5+c_5} y\right )+c_3 \sin \left (\sqrt {5+c_5} y\right )\right )\right \}\right \}\] why? It solved earlier with BC?
Maple ✓
restart; pde := diff(u(x,y),x$2)+diff(u(x,y),y$2)+5*u(x,y)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y),'build')),output='realtime'));
\[u \left ( x,y \right ) ={ \left ( \left ( {{\rm e}^{\sqrt {{\it \_c}_{{1}}}x}} \right ) ^{2}{\it \_C1}+{\it \_C2} \right ) \left ( {\it \_C3}\,\sin \left ( \sqrt {{\it \_c}_{{1}}+5}y \right ) +{\it \_C4}\,\cos \left ( \sqrt {{\it \_c}_{{1}}+5}y \right ) \right ) \left ( {{\rm e}^{\sqrt {{\it \_c}_{{1}}}x}} \right ) ^{-1}}\]
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Added December 20, 2018.
Example 24, taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
Solve for \(u\left ( x,y\right ) \) \begin {align*} \frac {\partial ^{2}u}{\partial x^{2}}+\frac {\partial ^{2}u}{\partial y^2} - k u(x,y) & = 0 \end {align*}
With \(k>0\). It is called reduced Helmholtz, because of the minus sign above. Otherwise, standard Helmholtz has a positive sign.
Boundary conditions \begin {align*} u(x,0) &= 0 \\ u(x,\pi ) &= 0 \\ u(0,y) &= 1 \\ u(\pi ,y) &=0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = Laplacian[u[x, y], {x, y}] - k*u[x, y] == 0; bc = {u[x, 0] == 0, u[x, Pi] == 0, u[0, y] == 1, u[Pi, y] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> k > 0], 60*10]];
\[\left \{\left \{u(x,y)\to \underset {K[1]=1}{\overset {\infty }{\sum }}-\frac {2 \left (-1+(-1)^{K[1]}\right ) \text {csch}\left (\pi \sqrt {K[1]^2+k}\right ) \sin (y K[1]) \sinh \left ((\pi -x) \sqrt {K[1]^2+k}\right )}{\pi K[1]}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x, y), x$2)+diff(u(x, y), y$2)-k*u(x, y) = 0; bc_left_edge:=u(0, y) = 1; bc_lower_edge:=u(x, 0) = 0; bc_top_edge:=u(x,Pi)=0; bc_right_edge:=u(Pi,y)=0; bc:=bc_left_edge,bc_lower_edge,bc_top_edge,bc_right_edge; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc ], u(x, y)) assuming k>0),output='realtime'));
\[u \left ( x,y \right ) =\sum _{n=1}^{\infty }-2\,{\frac {\sin \left ( ny \right ) \left ( -1+ \left ( -1 \right ) ^{n} \right ) \left ( {{\rm e}^{- \left ( -2\,\pi +x \right ) \sqrt {{n}^{2}+k}}}-{{\rm e}^{\sqrt {{n}^{2}+k}x}} \right ) }{ \left ( {{\rm e}^{2\,\sqrt {{n}^{2}+k}\pi }}-1 \right ) \pi \,n}}\]