i have solved the coupled equations .... and want to subtract a constant

i want to subtract constant from the result of last equation

r_p_m.mw

What is the best way for Maple to create datasets with attributes?

There must be a better way than just listlists  ie [[Honda, green, 2008],[Honda, red,civic,2004],[Hoda, blue, CRV, 2WD]] etc....

It would be similar to the table of elements using get properties. 

Hi all,

I have this equation that I can not get all solutions symbolically:

restart:

eq1 := cos(lambda*ln(r1))*cos(lambda*ln(r2))+sin(lambda*ln(r1))*sin(lambda*ln(r2))-1 = 0:

solve(eq1, lambda, allsolutions) assuming r1>0, r2>0, r2>r1

when r1:=1: r2:=2: I get the solution

2*Pi*Z/ln(2)

when r1:=1.1: and r2:=2.1: # no solutions

How to get symbolique solution

Thanks

Hi.

This is probably not very much maple related question, but to some extend it is.

After failing this question on my exam I have tried to solve it, but it seems like I cant get it right.

Given a sphere z^2+r^2=4 and a cylinder r=1 I was told to set up the volume integral for the element T enclose byt the outer sphere and the inner circle.

I tried to generate a plot but my skills are rather poor in plotting, if I could get the plot right I would be able to set up the volume integral. I have also tried to figure out how to do the surface integral and chose to use a task template as it is a bit more convinient when you find the syntax hard.

I would say I am familiar with the VectorCalculus:-SurfaceInt in cartesian for when i have intersection of two surfaces given in terms of z=

but this kind of problem is new to me.

Download surface_int.mw

A population  governed by the logistic equation with a constant rate of harvesting satisfies the initial value problem . This model is typically analyzed by setting the derivative equal to zero and finding the two equilibrium solutions . A sketch of solutions  for different values of a suggests that the larger equilibrium is stable; the smaller, unstable.

When a is less that the unstable equilibrium,  becomes zero at a time , and the population becomes extinct. If  is not interpreted as pertaining to a population, its graph exists beyond , and actually has a vertical asymptote between the two branches of its graph.

In the worksheet "Logistic Model with Harvesting", two questions are investigated, namely,

1. How does the location of this vertical asymptote depend on on a and h?
2. How does the extinction time , the time at which , depend on a and h?

To answer the second question, an explicit solution , readily provided by Maple, is set equal to zero and solved for . It turns out to be difficult both to graph the surface  and to obtain a contour map of the level sets of this function. Instead, we solve for  and obtain a graph of  with  as a slider-controlled parameter.

To answer the first question, the explicit solution, which has the form , exhibits its vertical asymptote when . Solving this equation for  gives the time at which the vertical asymptote is located, a function that is as difficult to graph as . Again the remedy is to solve for, and graph, , with  as a slider-controlled parameter.

Download the worksheet: Logistic_with_Harvesting.mw

Dears

Hope everything fine with you. I want to solve the attached problem by numarically and want to plot it but failed. Please see the attachement and correct it. I am waiting your positive respone. 

System_of_ODEs.mw

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

I'm trying to approximate the solution of an IVP using Euler's method in the InitialValueProblem command and I keep getting this error (see attached worksheet). Can someone explain why? Thanks!
 

Download ivp_error.mw

Hi all. I need to solve a large number of PDEs (partial differential equations) symbolically and simultaneously, to find the linearly independant answers for all of them, I use

ans := pdsolve({seq(PDE[i]=0,i=1..d)});

The PDEs are all linear first order and it is very easy to solve them one by one by hand, but in some cases I have 100 PDEs or more, so Maple is either very slow or doesn't work. For d=120, it was evaluating for hours without a result.

For example I have d=120 PDEs, and 200 variables. It works for d=30 of them (takes 13 minutes on my 16GB RAM  windows 7 computer). So if I do this:

ans1 := pdsolve({seq(PDE[i]=0,i=1..30)});

ans2 := pdsolve({seq(PDE[i]=0,i=31..60)});

ans3 := pdsolve({seq(PDE[i]=0,i=61..90)});

ans4 := pdsolve({seq(PDE[i]=0,i=91..120)});

Then how can I have only one vector of the linearly independent answers of all of them?

And in general, is Maple supposed to do this kind of calculations at all?

If yes, do you have any ideas on how to improve this procedure? 

If not, do you know in which software or programming package I can solve a large number of PDEs symbolically?

Your help is much appreciated. 

I have noticed the second item in Description in

http://www.maplesoft.com/support/help/Maple/view.aspx?path=convert/binary

The item is: 

“The binary number is returned as a base 10 number consisting of the digits 1 and 0 only.”

Does the existence of this item mean the following behavior will not be corrected ? ("10^(-6)" in the following should be "2^(-6)" .)

******************************************************

Hi, I am new in Maple. If I have an electric network as in the figure, I want to get the Transfer Function V2(s)/Vi(s) from this equation system:

Vi=R1.I1+V1

Vi=R2.I2+V2

I1-I2=C1.dV1/dt

I2=C2.dV2/dt

Which are the commands that I may write to get this?? Before hand, Thanks by your answers!

i don't know much about maple, i need to solve the following odes system... I study a little on the help page of maple about numeric[midrich] that takes bvp and deal singularity as well but dint know how to used in the following system

odes.mw

I use pdsolve to solve this system of equation but the graph I have is different from the author's graph. I think I'm missing out on something. Can anyone help me out with the solution using any Maple command and module.

PDE := {diff(phi(x, t), t) = (diff(phi(x, t), x, x))/S__c-K__r*phi(x, t)+S__r*(diff(theta(x, t), x, x)), diff(u(x, t), t) = diff(u(x, t), x, x)-M^2*(u(x, t)-m*w(x, t))/(m^2+1)-u(x, t)/`ϰ`-2*Omega^2*w(x, t)+Gr*theta(x, t)+Gm*phi(x, t), diff(w(x, t), t) = diff(w(x, t), x, x)+M^2*(m*u(x, t)-w(x, t))/(m^2+1)-w(x, t)/`ϰ`+2*Omega^2*u(x, t), diff(theta(x, t), t) = lambda*(diff(theta(x, t), x, x))/P__r}

With Inittial and boundary condition : {phi(0, t) = 1, phi(9, t) = 0, phi(x, 0) = 0, u(0, t) = t, u(9, t) = 0, u(x, 0) = 0, w(0, t) = 0, w(9, t) = 0, w(x, 0) = 0, theta(0, t) = 1, theta(9, t) = 0, theta(x, 0) = 0}

With the following parameter declared as:

I will appreciate the graph of the solution with time t:0.3, 0.5, 0.7 and 1.0.

Thanks.

When trying to perform the following:

p1 := proc (x, y) if x^2+y^2 <= 1 then x*y-y^2 else 0 end if end proc;
plot3d(p1, x = -1 .. 1, y = -1 .. 1);
Error, (in plot3d) expected ranges but received x = -1 .. 1 and y = -1 .. 1
 

I get this strange error message. To the best of my knowledge x and y ARE provided as ranges. What am I missing/not understanding?

If I omit the ranges in plot3d Maple returns a correct plot, but the default range (-10 .. 10) does not display sufficient details

This is what I'm trying to solve:

I literally just want it to plug in those values for n and spit out a value. This is easy enough for me to do by hand, but I obviously wanted to speed up the process by doing it with Maple. What's wrong here? I clicked the pink links and that didn't help.

I'm going back over QM homeworks to study for an exam. We're allowed to use Maple on our exam so I want to be sure I'm prepared for that part. I have the solution to a problem I'm working on, but I can't get Maple to simplify it how I want it.

Is that simplification something that Maple should be able to do or is it something I would need to know how to do because Maple isn't capable? (I assume it is.)

Thanks!