it is having hard time with the BC and ic. But it does odetest the pde itself once you remove the casesplit. It does not seem to handle casesplit well.

restart;
eq := diff(u(x, t), t) - k*diff(u(x, t), x, x) = 0;
ic := u(x, 0) = g(x);
bc  := D[1](u)(0,t)=0,(D[1](u)(1,t)+u(1,t))=0;
sol := (simplify(pdsolve([eq, ic, bc])) assuming (0 < k, 0 < x and x < 1, 0 < t));

Just remove the last where stuff (for the eigenvalues conditions it found)

sol:=u(x,t) = Sum(4*exp(-k*lambda[n]^2*t)*lambda[n]*cos(lambda[n]*x)
*Int(g(x)*cos(lambda[n]*x),x = 0 .. 1)/(2*lambda[n]+sin(2*lambda[n])),n = 0 ..
infinity)

And now it works just for the pde itself.

pdetest(sol, eq)

     0

And for the BC, it gives  zero for the left side one. But not for the right side one

pdetest(sol,[eq,bc])

if I understand you right, one way could be to convert it to list, then use plot

A:= seq(f(i), i = 0 .. 1, 0.1);
A:=convert~([A],list);

  #A := [[0, 1], [0.1, 1.1], [0.2, 1.2], [0.3, 1.3], [0.4, 1.4], [0.5, 1.5], [0.6, 1.6], [0.7, 1.7], [0.8, 1.8], [0.9, 1.9], [1.0, 2.0]]

plot(A[1..nops(A),1],A[1..nops(A),2], style=point)

one way

restart;
A:=Matrix([[1,2,3],[4,5,6],[7,8,9]]):
add((x->x^2)~(ArrayTools:-Diagonal(A)))

    107

CAS gives generic solution for the Euler ODE. Same solution is given by Mathematica.

To get specific solutions use assumptions

restart;
ode:=x^2*diff(y(x), x, x) + 3*x*diff(y(x), x) + lambda*y(x) = 0:
dsolve(ode) assuming lambda>1;
dsolve(subs(lambda=1,ode));
dsolve(ode) assuming lambda<1;

Noticed that assuming lambda=1 does not work.

another option 

restart;
r:=theta->cos(2*theta); 
plots:-animate(plots:-polarplot, [r(theta) , theta=0..max_angle], max_angle=0..2*Pi,frames=100)

it is type=numeric  and not type=numerical 

You also have singularity. (btw, strange name you picked for your dependent variable function, n? Also need to use midpoint method for the BVP

restart;

A1 := 8*Pi^3*R^2*n(x)^4*m+(2*Pi*sin((1/2)*x)*m*omega0*p+Pi*sin((1/2)*x)*m*omega0+3*Pi^2*(diff(n(x), x, x)))*n(x)^3+(-2*sin((1/2)*x)^2*m^2*omega0^2*p^2+2*cos((1/2)*x)^2*m^2*omega0^2*p^2-2*sin((1/2)*x)^2*m^2*omega0^2*p+2*cos((1/2)*x)^2*m^2*omega0^2*p)*n(x)^2+(-4*(diff(n(x), x$2))*sin((1/2)*x)^2*m^2*omega0^2*p^2-8*sin((1/2)*x)*(diff(n(x), x))*cos((1/2)*x)*m^2*omega0^2*p^2-4*(diff(n(x), x$2))*sin((1/2)*x)^2*m^2*omega0^2*p-8*sin((1/2)*x)*(diff(n(x), x))*cos((1/2)*x)*m^2*omega0^2*p)*n(x)+8*sin((1/2)*x)^2*(diff(n(x), x))^2*m^2*omega0^2*p^2+8*sin((1/2)*x)^2*(diff(n(x), x))^2*m^2*omega0^2*p;

R := 1; m := 1; p := 10; omega0 := 1000;

bc:=n(0) = n(4*Pi), (D(n))(0) = (D(n))(4*Pi);
sol:=dsolve([A1,bc], type = numeric, range = 0 .. 4*Pi,method=bvp[middefer]):

Error, (in dsolve/numeric/bvp) singularity encountered

Maple does not seem to like your BC. Something to look into.
 

There is something very confusing here. You say "need to find involved integration constant i.e c1 to c4. "  But you are providing boundary condition? There will not be any constants of integration in the solution., 

Also, why do you call "c1" a constant of integration, when it is part of the ODE itself? You wrote 

ode1:=diff(q(y), y$2) - A*q(y) = B*(P*y + c1);

And then you called c1 constant of integration?

In addition, your ODE's are not coupled. The first ODE can be solved on its own, and the solution plugged into the second ODE to solve that on its own. Coupled ODE's is when both dependent variable appear in both ODE's.

But you can still solve these ODE together, it does not matter to Maple

restart;
ode1:=diff(q(y), y$2) - A*q(y) = B*(P*y + c1);
bc1 := q(-sigma) = 0, q(sigma) = 0;
ode2:=(1 + N)*diff(u(y), y$2) + N*diff(q(y), y) = P;
bc2 := u(-sigma) = 1, u(sigma) = -k;
dsolve([ode1,ode2,bc1,bc2])

And since the ODE's are not coupled they can be solved one at a time:

restart
ode1:=diff(q(y), y$2) - A*q(y) = B*(P*y + c1);
bc1 := q(-sigma) = 0, q(sigma) = 0;
ode2:=(1 + N)*diff(u(y), y$2) + N*diff(q(y), y) = P;
bc2 := u(-sigma) = 1, u(sigma) = -k;
solq:=dsolve([ode1,bc1]);

ode2:=subs(solq,ode2);
solu:=dsolve([ode2,bc2])

Which gives same solutions are above

I normally use subs and algsubs.

subs seems to have worked here

subs(5435 + 3*sqrt(515793)=t,s1)

and simplify on the above gives

You really should get into the habit of using symbols, then replace these with numbers at the end. If you do that, it will easier to see why it can't give explicit solution.  In addition, your code will be easier to modify and maintain.

Your ODE is linear first oder. But the integrating factor produces integral which Maple can't integrate.

restart;
ode1:=-diff(q(t), t) - A*q(t)*(B - C*(d*cos(f*t) + e*sin(f*t))^2) +h=0;
dsolve(ode1);

The integral of exp(.....) above is the problem. Try numerical dsolve instead.

To get to the essense of it, keep throwing away uneeded things, and you'll get to something that looks like 

int(exp((t -cos(t)- sin(t)+ sin(t) + t)), t)

which has no closed form solution. It is the integral of things like exp(cos(t)) which makes Maple produces the result it did, with `Int` in it, as there is no antiderivative for these. 

sys_ode := diff(F0(zeta), zeta, zeta)-b^2*F0(zeta)+G0(zeta)^2 = 0, diff(G0(zeta), zeta, zeta)-b^2*G0(zeta) = 0, 2*F0(zeta)+diff(H0(zeta), zeta) = 0;

ics := F0(0) = 0, G0(0) = 1, H0(0) = 0, F0(infinity) = 0, G0(infinity) = 0;

sol:= dsolve([sys_ode,ics])

simplify(sol) assuming positive

you can eliminate  the remove has calls by doing

map(x->ListTools:-SearchAll(x,L1),L2)

do not know if this will speed it up much or not.

Another apprach to the fine answers given already is to use: (can't write Latex in the form, so I put screen shot)

Using Maple

g:=x->cos(x);
limit(1/abs(g(x)),x=0)+limit(1/abs(g(x)),x=-Pi)+limit(1/abs(g(x)),x=Pi)

    3

 But the lecturer's solution was non-imaginary.

restart;
ode := x^2*diff(z(x), x, x) + (1 + gamma + beta)*x*diff(z(x), x) + gamma*beta*z(x) - cos(ln(x)):
sol:=(dsolve({ode, z(1) = 1, D(z)(1) = -1}) assuming (gamma <> beta));

simplify(evalc(sol)) assuming x>0

You get same out using

remove(`=`,op~(0,indets(x)),symbol)

{f, g, h}

It works since op(0,f(x))  gives f  while op(0,f) gives symbol

there might be a better way to do ofcourse, but this is how I see it so far. But there should be a more robust way to do this using type(....,'function')

May be something like

restart;
x := -h(U) + f( f( g( f(U+W*X)*V) + g( f(W)*g(V) ) ) ):
foo:=x-> if type(x,'function') then return(op(0,x)) fi;
convert(foo~(convert(indets(x),list)),set)

{f, g, h}

In Section 14.5 - External Calling ..., page 489

I do not use Maple on Linux. But you have one ". You need two of these, one of each side. 

Here is the example from book. Try and see if this fixes it.