@Oliveira Sure, you get a Fourier series representation of the solution.  That works on a rectangle but your original question was about solving over a triangle.

@bstuan Let f(x) be your integrand and let m be the minimum value of (x+2)/(2x^3+1) on the interval (1,2). Since f(x) > m/ln(x), the integral of f(x) is more than the integral of m/ln(x) on (1,2). But the integral of m/ln(x) on (1,2) is infinity, and therefore so is the integral of f(x).

Hint: Examine the convergence of the integral of 1/ln(x) over the interval (1,2).  That should be good enough to reach a conclusion about your original problem.

You need to specify a boundary condition along the bottom edge as well.

Maple is good in dealing with symbolic expressions involving scalars.  It can also handle vectors and matrices of explicitly specified sizes and contents.  But it does not have a facility to deal with purely symbolic vectors and matrices which you need in this instance.

I hope that such facilities will be added some time in the future, but I wouldn't expect it soon.

There may be a problem if the running command or proc refers to the previous calculations through their labels.

@Carl Love Oh, okay, and I have no idea regarding the validity of what Maple is showing as solutions.

You refer to an ODE but I see no ODEs in what you have shown.

Why don't you just upload a complete, self-contained worksheet like others do in this forum?  Look for a big fat green arrow in the dialog box where you reply to this request. Click on it to upload your worksheet.

All that and more can be done in Maple, but you need to supply the values of the coefficients, ca, cb, Ka, Kb, Kw and Va .  The optimal approach may vary, depending on what those coefficients are.

@WA573 In your worksheet, the expression for Phi is defined with a plut/minus sign.  I chose the plus sign.  Then, to obtain the level curves labeled h=−1,0.1 in the image that you have posted, the right-hand side will have to be h+2.

If you choose the minus sign, you will obtain the curves in the image when the right-hand side is h.

I have attached a modified worksheet with the minus sign for Phi, and the right-hand side set to h.

Download: mw2.mw

@Preben Alsholm Yes, I see the issue; I had checked the solution by hand and I had assumed that Maple will do the same thing, but as you have noted, (−1)^(1/3) is not −1 in Maple and one needs to watch out for that.

Your suggestion of rewriting the original ODE as ode := diff(v(t),t) = -2*(v(t)^2)^(1/3) addresses and resolves that issue perfectly.  As to the alternative suggestion, yes, that's also perfect, and in looking at that, I see that in offering my solution I have misread the requested initial condition v(0)=5 as v(0)=0.

How familiar are you with Maple?  Do you know how to plot any function at all?

@Earl That's a very nice demo.  As to your comment re "standard plot commands", I suppose you wish to replace your use of odeplot() with something else.  If you want, you may replace it with spacecurve(), or tubeplot().  Here is one possibility:

DispGeo := tubeplot(eval(convert(x(u, v(u)), list), dsol1), u = Pi/4 .. (3*Pi)/4, radius = 0.02, color = red, style = surface);

@Ahmed111 As Carl has pointed out, the trouble in Maple 18 may be due to "with(Physics)".  But that package is not needed in Linearize.mw.   Remove and try it again.

@Carl Love I misunderstood the question.  Then perhaps this is what is wanted:

Download: Lineaize.mw