Maple Questions and Posts

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Hello

I am using Maple to solve a couple of differential equations.  Here is what I did so far

k := 141/10000;
yB0 := 296/1000;
e := -148/1000;
Ff0 := 67844/1000;
Far0 := 323066/1000;
FB0 := 135688/1000;
P0 := 10;
x0 := 0;
a := 38/1000;
dsys:={diff(x(w),w)=(k*((yB0*p(w)*(1 - x(w)))/(1 + e*x(w)))^(1/3)*(Ff0/(Far0 + FB0)*p(w)*((Ff0/FB0 - 1/2*x(w))/(1 + e*x(w))))^(2/3))/FB0,
diff(p(w),w)=P0*(-a)/(2*p(w)/P0)*(1 + e*x(w)),x(0)=x0,p(0)=P0}:
dsn1:≔dsolve(dsys,numeric,[x(w),p(w)],stiff=true);

Maple returns neither an error message or a solution.   I am sure I have mistyped something or did not understand how dsolve works at all.  

Can you help me out?

Many thanks

Ed

PS. How to plot the solution?  

 

 

On CBS tv show Survivor S38E4, an interesting scenario arose. There was initially 2 teams comprising 6 persons (blue team) and 9 persons (yellow team). Then they decided to mix it up, and split it into 3 teams. Each player was randomly assigned a buff, (red, green or orange =3 colored buffs *5).= 3 teams of 5.

What was interesting was 5 out of the original 6 blue team are now in the new red team. 5 of the original yellow team are in the new green team and 4 of the original yellow team are in the new orange team. 

What is the chances of that happening?
A couple of attempts:

survivor_rand.mw

How i can generate Pr matrix when P= -x and x0=-1 , x1=-2/3 , x2=-1/3 , and x3=0 with the help loop?

Hi everybody:

I have one equation:

EQ := V^3-R*T*V^2/P-(B^2+P*B*R*T*sqrt(T)/(P-A))*V-A*B/(P*sqrt(T)) = 0;
how can I obtain these derivatives:


 

macros can be made to work like subs, you just need to know a few tricks to get it to work the same way.  macros just works in a slightly different manner and we can make it useful.

The difference is with subs, one has to keep specifying the substitution with each equation you want subbed, whereas macro will already have it defined.  As an example:

a := v^2*z^3 - 34/(5*x^2*sin(y*v^2)) + 36*v^2 - b*v^2 + 3^(v^2 - cos(v^2 + g))
                            

If we want to substitute h for v^2, then we would normally do this using subs

subs(v^2=h,a)
                          

however, we can also use macro

macro(v^2=h)
                                    

now it doesn't just automatically substitute those values so we need to coax maple a little bit.  We can do that by converting the equation to a string and parsing it.

parse(convert(a,string))
                     

so as you see we arrive at the same result.  Now there is a caveat using macro, if you've already defined a variable in a macro, subs will not work using the same variable sustitution - you first need to reset the variable in the macro back to itself. 

subs(v^2=h,a)
                      #doesn't work since the variable is defined in a macro

macro(v^2=v^2) #reset the variable in the macro

subs(v^2=h,a)
                         # now it works

we could also define a little procedure to simplify our typing, to have the macro variable work on our equation.

mvs:=proc(a) #macro variable substitution
  parse(convert(a,string));
end proc:

macro(v^2=h)
mvs(a)
            

now if we had some other existing equation before defining the macro
aa:=exp(v^2-sin(theta))+v^2*cos(theta)-1/x^sin(v^2-g)
                                   

we just have to simply apply our proc on the equation to apply the variable substitution
mvs(aa)
              


 

 

 

 

I just noticed with chagrin that one of my favorite menu commands, Edit -> Remove output -> From worksheet, is missing from Maple 2019's Standard GUI. Is there a keyboard command or toolbar item to replace it?

How can I plot Y in terms of X (Y is the vertical axis and X is the horizontal axis) in below equation with Maple?

X^3-24.478115*X^2/Y-(0.2038793409e-2+19.08455282*Y/(Y-97.539))*X-.2550630228/Y = 0

X:=0..50

Y:=0..1

How we can get matrix M with the help of loop?

Hello

Im solving a problem. When i use solve command maple dont give me whole of amswers it just give solve the obvious problem that is zero. what i should do to it give all of the answers not just zero?

Hi guys,

I'm very new to Maple and I'm trying to get  a 'diagonalisable' matrix set up.

It will help for me to attach what I currntly have for this to make sense:

What I currently have is a way of setting up a 5x5 matrix where it inserts a value at every instance where i=j, i=j+3 and i=j+6.

Currently I can only make this work for value from 1 to x and, for this to work, I basically need to make up a matrix that goes from -x to x

i.e currently I have x = 1 though 5 (5x5) and I need to set it up for x= -5 through 5 (11x11)

Every time I try and get the negative elements and 0 in there it tells me that I can't do it.

Is there any way to do this is Maple and am I going the wrong way about it?

 

Thanks very much for your time!

I'm using Maple on a compute server (via Putty from my computer which has windows system). It is an only-text mode server so it is a black screen and text and no menu bar to choose file and then save by clicking etc. You only can write commands and press Enter.

The IT section suggested to use writeto("filename.txt"), but I can't see any file on my drive, I used writeto("\drvename\filename.txt"), but still nothing.

Thanks to @acer@Joe Riel and the IT section. The problem now is solved. It was about the way that I was writing the directory path. I had to add something before the "drivename" in my case. Plus that it is a Unix path, so I should use / not \.

Please check the value of T[1] ?

Hello, i am experiencing some problems when trying to open the maple 2018 software*
I have tried unistalling and download it again.
I have tried to search for sollution but there is very ittle intel
When i open Maple 2018 it just lingeres on the start up (pic below) and just disappears after 10 seconds

Can someone please help i have a very important examination upcoming

how we can get matrix of p[2],p[0] and p[1] when p[2]=1,p[0]=1 and  p[1]=-x and x[0]=-1,x[1]=-2/3 ,x[3]=0 when the general matrices of p are given below.....and answer should be p[2]=p[0]= identity matrix and p[1]= only diagonal values are 1,2/3,1/3,0..

Diff.(u(x,t), t$alpha) = 1/2. x^2. Diff(u(x,t),x$2), 0 <x <1, t >0, alpha =1.5  or  1.75

Bcs: u(0,t) = 0, u(1,t) = 1 + sinh t,

Ics: u(x,0) = x, ut(x,0) = x^2 

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