For everyone's convenience, here I transcribe Mathematica's solution to Burgers' equation into the Maple notation and verify that indeed it is correct.

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@Mariusz Iwaniuk Here Maple is really confused because u(x,t)=1 does not satisfy the initial condition.  As to the HINT=f(x)/f(t), that's a good try, but in view of the exact solution, we should be looking for HINT=f(x/t).  Unfortunately that does not result in any solution.

As Pólya is reported to have said: For any problem that you cannot solve, there is an easier problem that you cannot solve.  Solve that one first!

OK, so here is an easier problem obtained by setting μ=0 and reversing the 0 and 1 in the initial condition:

restart;
pde := diff(u(x,t),t) + u(x,t)*(diff(u(x,t),x)) = 0;
ic := u(x,0) = piecewise(x <= 0, 0, 1);

This is the classic Riemann problem.  Its exact solution is u(x,t)=F(x/t), where
F := s -> piecewise(s<=0, 0, s<1, s, 1);

I am unable to coax Maple's pdsolve() to arrive at that solution.  Perhaps someone else can. I would have expected HINT=f(x/t) to do the job but it doesn't.

@devraj You need to supply an initial condition as well.  I will take c(y,0) = cos(Pi*y).  Change as needed.

Now you may solve the problem numerically as follows:

restart;
pde := Omega*(diff(c(y, t), t))-Omega*Lambda*sin(t) = diff(c(y, t), y, y);
bc := (D[1](c))(0, t) = 0, (D[1](c))(1, t) = 0;
ic := c(y,0) = cos(Pi*y);
params := Omega=100, Lambda=0.001, pe=10;
sol := pdsolve(eval(pde, {params}), {bc, ic}, numeric);
sol:-plot3d(y=0..1, t=0..30, style=patchcontour);

This problem is very different from that you originally asked, so it's doubtful whether this solution is of any value.

@devraj Let's look at your new set of equations (1) and (2).

Differentiating equation (1) with respect to x leaves us with du/dx=0.  That says the term du/dx in equation (2) is zero and should be dropped.

Somehow I don't think that this is what you are looking for.  Ask someone who knows more about this to help you.

@devraj You should have a closer look at your equations because as they are, they make no sense at all.

Let's take, for example, the equations that you have listed as (1.) and (2.) in your post.  Subtract them.  You get:
u = pe*diff(u(x,y), x) .

On the other hand, you also say:

u=6*Lambda*(-y^2+y)*cos(2*Pi*x)+1

These two expressions for u directly contradict each other.

I suggest that you forget about Maple for now.  Figure out what your equations should be, or better yet, talk with someone who can guide you with those.  Once you have a logically consistent problem, come back here and then we will see how Maple can help.

@devraj You forgot to answer my question "What is x?".  I am referring to the cos(2*Pi*x) that appears in your PDE.

@vv Ah, now that's better :-)

@vv That expression looks nice but unfortunately it's not very practical as can be seen when trying
plot(answer, x=Pi/2-0.2..Pi/2+0.2);

@nm  In this problem the initial and boundary conditions are artificially designed/fine-tuned to yield a simple analytical solution.  The slightest change, for instance replacing the 12 by 13, will destroy the balance and the simple solution will be lost.  In the general case the solution would be expressed as an infinite sum—a Fourier series.

As to your question on how I thought of the "+" form of the solution, I looked at the boundary conditions and saw that the two ends move together, one end moving as −12 t², and the other lagging by 1 relative to the other.  That led me to think of a rigid translational motion.  Since the initial condition is given as x⁴, the natural candidate would be x⁴ − 12 t².  I tried that in my head, and saw that it works.

That gave me the idea of the HINT=`+` in Maple, and that yielded the same result.

This does not make the HINT=`+` option a useful choice for such problems.  That choice works in this case by "accident" because of the artificially fine-tuned conditions.

@assma Did you try saving it as EPS?

@assma I don't have Maple 18 to test, but it is likely that it should be able to handle the saving of a 3D figure as EPS.

Note, however, that the EPS format cannot handle transparency.  If your figure has transparent parts, then you will have to save it in the JPG format which is not the best thing since it is a lossy format and results in somewhat blurry images.

@josephrajck The equation and the boundary conditions are correct, but Initial conditions are missing.  Additionally, you can't get a numerical solution if you leave the coefficient "a" and the length "l" inspecified.  Here I will take both of these to be 1.  You may change then as needed.

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@vv Uh, OK, I see it now and I agree.  Also thanks to Preben for his explanations.

@Preben Alsholm Aren't all variables at the top level global by default?
restart;
type(a, `global`);          #true
type(whatever, `global`);   #true
type(_Z, `global`);         #true

I don't understand what it is so special about singling out _Z as being global.