3.26 How to solve heat PDE in 1D in Maple 2017?

to solve \(\frac {\partial u(x,t)}{\partial t}=k \frac {\partial ^2 u(x,t)}{\partial x^2}\) with homogeneous dirichlet boundary conditions \(u(0,t)=0,u(L,t)=0\) the commands are

restart; 
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2); 
bc:=u(0,t)=0,u(L,t)=0; 
sol:=pdsolve([pde,bc]) assuming 0<L:
 

Which gives

\[ u \left ( x,t \right ) =\sum _{{\it \_Z1}=1}^{\infty }{\it \_C1} \left ( {\it \_Z1} \right ) \sin \left ( {\frac {\pi \,{\it \_Z1}\,x}{L}} \right ) {{\rm e}^{-{\frac {k{\pi }^{2}{{\it \_Z1}}^{2}t}{{L}^{2}}}}} \]

Which can be made more readable as follows

sol:=algsubs(_Z1=n,sol): 
sol:=algsubs(Pi*n/L=lambda(n),sol);
 

\[ u \left ( x,t \right ) =\sum _{n=1}^{\infty }{\it \_C1} \left ( n \right ) \sin \left ( x\lambda \left ( n \right ) \right ) {{\rm e}^{-kt \left ( \lambda \left ( n \right ) \right ) ^{2}}} \]

For homogeneous Neumann B.C., at \(x=0\), let \(\frac {\partial u}{\partial x}=0\) and at \(x=L\) let \(u(L,t)=0\), the solution it gives looks different than my hand solution

restart; 
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2); 
bc:=D[1](u)(0,t)=0,u(L,t)=0; 
pdsolve([pde,bc]) assuming 0<L;
 

It gives

\[ u \left ( x,t \right ) ={\it \_C3}\,{\it \_C2}\, \left ( {{\rm e}^{1/4\,{\frac {2\,i\pi \,xL-k{\pi }^{2}t}{{L}^{2}}}}}+{{\rm e}^{-1/4\,{\frac {\pi \, \left ( 2\,ixL+k\pi \,t \right ) }{{L}^{2}}}}} \right ) \]

I need to look more into the above and see if this comes out to be the same as my hand solution.

Another example, with initial conditions now given

restart; 
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2); 
bc:=D[1](u)(0,t)=0,u(L,t)=0; 
ic:=u(x,0)=f(x); 
sol:=pdsolve([pde,bc,ic],u(x,t)) assuming 0<L; 
sol1:=algsubs(_Z2=n,sol);
 

The result is

\[ u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac {1}{L}{{\rm e}^{-1/4\,{\frac {k{\pi }^{2}t \left ( 1+2\,n \right ) ^{2}}{{L}^{2}}}}}\cos \left ( 1/2\,{\frac {\pi \,x \left ( 1+2\,n \right ) }{L}} \right ) \int _{0}^{L}f \left ( x \right ) \cos \left ( 1/2\,{\frac {\pi \,x \left ( 1+2\,n \right ) }{L}} \right ) \,{\rm d}x} \right ) \]

Another example

restart; 
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2); 
bc:=D[1](u)(0,t)=0,u(L,t)=0; 
ic:=u(x,0)=3*sin(Pi*x/L)-sin(3*Pi*x/L); 
sol:=pdsolve([pde,bc,ic],u(x,t)) assuming 0<L; 
sol1:=algsubs(_Z2=n,sol);
 

\[ u \left ( x,t \right ) =\sum _{n=1}^{\infty }768\,{\frac {1}{\pi \, \left ( 16\,{n}^{4}+32\,{n}^{3}-136\,{n}^{2}-152\,n+105 \right ) }{{\rm e}^{-1/4\,{\frac {k{\pi }^{2}t \left ( 1+2\,n \right ) ^{2}}{{L}^{2}}}}}\cos \left ( 1/2\,{\frac {\pi \,x \left ( 1+2\,n \right ) }{L}} \right ) } \]

Another example

restart; 
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2); 
bc:=u(0,t)=0,u(L,t)=0; 
ic:=u(x,0)=3*sin(Pi*x/L)-sin(3*Pi*x/L); 
sol:=pdsolve([pde,bc,ic],u(x,t)) assuming 0<L;
 

\[ u \left ( x,t \right ) =\sin \left ( {\frac {\pi \,x}{L}} \right ) {{\rm e}^{-9\,{\frac {{\pi }^{2}kt}{{L}^{2}}}}} \left ( -2\,\cos \left ( 2\,{\frac {\pi \,x}{L}} \right ) +3\,{{\rm e}^{8\,{\frac {{\pi }^{2}kt}{{L}^{2}}}}}-1 \right ) \]

The above answer seems wrong. There is not even a summation in it. It is different from my hand solution. Look more into it.