The differential equation I'm solving for is:

Hello fellow Maple users. 
I have been useing Maple for 5 month now and i think its a great program. But sometimes i find it a bit difficult to use. 
When i plot a function i would like to see the function on the line in the plot that represents the funtion. So if have several lines representing different functions, i would Maple to highlight the function when i hower over the function 
with the mouse .. Is this posible ? I know its posible in Inspires CAS. 

It must be posible for Maple to label the plots with the function ??? If not Maple programmers got a job...

I have a problem in excuting this differential equation in maple it takes a long time but yet no result.

> restart;

> Delta:= epsilon[2]-epsilon[1];

> epsilon[y] := epsilon[2]-(1/4)*Delta*(1-tanh(a*y))^2;

 > z:= tanh(a*y) ;

 > ODE[4]:= diff(Y(y),y,y)- ( a/2* Delta *(1-z)*(z^2-1))/(epsilon[2]- Delta*(1-z)/4)* diff(Y(y),y)-( beta^2+ mu[0]*epsilon[y]*omega^2)*Y(y) = 0;

> dsolve(ODE[4],Y(y));

does this always occur or i do have problem with my version of maple 15, 7 and 16.

Thank you, looking forward for your answers.

I have created a simple proc to return true or false depending on the value in a list of values.  now i would like to filter out the falses and return the actual value that is true from the list.

sol := dsolve([ode2, theta(0) = (1/180)*Pi*10, (D(theta))(0) = 0], numeric)

sol(0)

[t = 0., theta(t) = .174532925199433, diff(theta(t), t) = 0.]

given a numerical solution in the above form, how can I integrate a function

f(theta(t)) numerically?

I tried to isolate theta(t) first:

x:=t->rhs(op(2,sol(t)))

But I can't integrate

phi:=t->int(x(t),t=0..t)

I am having some difficulty animating the function shown in the attached file.  I am going to create an animation which will show the curve as a function of t.  My first question is that there is no way to compute K_n because the initial conditions I have are only given as arbritary functions F(z),G(z).  So I am not really sure how to proceed here.

My second question is that I also want to plot the Z dependent part of y as a function of z/b.  I have tried to incorporate this into Maple, however, all I get back is that there are 'unexpected variables present'

Thanks.

Is it possible to assign a matrix given values/expressions?

Instead of assigning each element separately like in the example below:

with(combinat):
list1 := permute([a, b, a, b, a, b], 3);
list1a := subs(b=1,subs(a=0, list1));
list1a := permute([seq(seq(k,k=0..1),k2=1..3)], 3);
list2 := permute([a, b, c, d, e, f, g, h, a, b, c, d, e, f, g, h, a, b, c, d, e, f, g, h], 3);
list3 := subs(h=18,subs(g=17,subs(f=16,subs(e=15,subs(d=14,subs(c=13,subs(b=12,subs(a=11,list2))))))));
list3 := permute([seq(seq(k,k=11..18),k2=1..3)], 3);
list5 := Matrix(nops(list1a)*nops(list3), 1);
count := 1;
for n from 1 to nops(list3) do
temp1 := subs(1=list1a[1],list3[n]);
for k from 11 to nops(list1a)+10 do
temp1 := subs(k=list1a[k-10],temp1);
od;
list5[count] := temp1;
count := count + 1;
od;
list6 := permute(list5, 2);

Error, (in combinat:-permute) 1st argument must be a list, set or a non-negative integer

I have a bit of an issue.

I'm trying to solve this:

eq1:= exp(-lambda)*exp(k)/factorial(k) > .95

lambda:=2

solve(eq1,k)

Warning, solutions may have been lost

Good afternoon sir.

I request your kind support to the above cited query.

With thanks & Regards

M.Anand

Assistant Professor in Mathematics

SR International Institute of Technology,

Hyderabad, Andhra Pradesh, INDIA.

Consider the differential equation zZ'' + Z' + a²Z = 0,  where Z = Z(z).  Using the change of variables x = \sqrt{z/b}with b a constant,  obtain the differential equation Z'' + (1/x)Z' + c²Z = 0, where Z = Z(x) and c = 2a \sqrt{b}.

I tried Maple help and it offers the dchange command, and what I have tried is shown below;

with(PDEtools):

DE:= ...

tr:= {z = x²b}

dchange(tr, DE)

This did not return anything however.  I am thinking I need to specify that b is a constant, however, I am a little unsure on how to do this. Is the above the correct way to proceed?  I don't see how I have specified anywhere that in the final PDE, I require Z=Z(x).

Thanks for any help.  This is my first post here, so apologies for the typesetting. If there is inbuilt latex, I will use it next time.

Hi;

Trying to find the maximum values of the implicit function.. I tree the following commands.

l:=1:alpha:=1:b:=100:k:=20:

(alpha+(l+alpha)*u+alpha*k*u^2)*a=u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2)));

with(plots):

implicitplot((alpha+(l+alpha)*u+alpha*k*u^2)*a=u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2))),a=0..100,u=0..20,numpoints=9000000,color=black,axes=boxed,font=[1,1,18],thickness=2,tickmarks=[3,3],view=[0..12,0..12],labels=[a,u]);

maximize((alpha+(l+alpha)*u+alpha*k*u^2)*a=u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2))), N=0.5..1, location);

Any help I'll appriciated 

hi guys i want to solve this equation with maple please help me

eq[1]:=0.223569c_1+2.35589c_2*c_1^2+0.002356c_1*c_2^2;

eq[2]:=1.277899c_1*c_3-2.350023c_2*c_3^2+7.5856c_3*c_2^2;

eq[3]:=3.225989c_1^2+-2.35589c_3*c_1^2-7.28356c_3*c_2^3;

i want solve those equations with newton method

My attempt to export array data using the Browse>Export option when looking at the data fails to work as needed.

The exported data always start in cell A1 of the Excel worksheet even a different starting cell is entered into the Matrix Browser export window.

In summary, my experience is that the Matrix Browser matrix export in Maple 16 and Maple 17 ignores the information about the intended starting destination cell in the Excel file.

Does anyone find that the Matrix Browser exports to a starting Microsoft Excel cell other than A1? 

This exporting with Matrix Browser worked find in Maple 15.  It has been broken in Maple 16 and 17, including Maple 17.02.

Thanks.

Hello every one,

I am facing problem with for loop. Please have a look.

let say I have T[0]:= 99 ; X[0]:=0; f(x):=108 + 2 x

  for i from 1 to 2 do f(x[i-1] ):

                              y[i]:=x[i-1] +1 ;

if f(y[i])<f(x[i-1]) then x[i]:=y[i] elif 0.98< e^0.99/T[0]  then x[i]:=y[i] else x[i]:=x [i-1]

end if;

end do;

after I got the answer from this loop, now I want to continue the loop for T[1]:= 0.1 T[0], 

 how  to do another loop by using the same  information but in different value of T? in other words, the only change in the looping is the value T so that the next loop I have the set of answer.

thanks.