MaplePrimes Questions

my"+cy'+ky=F(t)      y(0)=y_0    and y'(0) =y'_0    (1)

Where F(t) is an external force, m, c and k are positive constants, y is a function of t and cy' is the damping term. Use equation (1) for your topic and F(t) = 0.

Task 2 : Assume that m = 1 unit in equation (1). For each of the following cases (all with m = 1), perform 5 steps of the modified Euler method applied to the coupled first order system you obtained in part (i), taking the step size h = 0.02 and working to 5 decimal places at each stage :

c = 0,  k = 4  , y(0)=1   and y'(0) =0;

c = 2,  k = 0,   y(0)=1   and y'(0) =0;

c = 2,  k = 1,   y(0)=1   and y'(0) =0;

c = 4, k = 6.25,  y(0)=1   and y'(0) =0.

Summarise your step-by-step results for each example in a table like the blank template below.

c = (Value),  k = (Value), h = 0.02,




































This is a question based on rewriting the equation as a pair of coupled(simultaneous) first order ODEs of the general form.

If anyone could please help to proceed with this question, as I am not quiet sure how to approach this differential equation. I will really appreciate your help please.

Thank you

I am solving some system of linear equations using fsolve(), I will like to obtain my answers without the attached label, i.e. instead of saying {x=2, y = 3} I need my result to be {2,3} because I need reuse the result. Please, help. Thanks.

I seem to be getting false solutions with Maple 2019.2 from solve and cannot figure out why.

Example :

f := x -> x*sqrt(4*x^2 + 1) + arcsinh(2*x);
sol := solve(f(x) = 5., x);
sol := 1.247747277, -1.839602523, 0.2519279800 - 1.361107684 I,
  0.2519279800 + 1.361107684 I
f(sol[1]); =    4.999999999
f(sol[2]); =   -9.027641618

Thus the second reported solution is not a solution after all.

Any ideas are most welcome.


when calculating the exponential integral in maple. Help me please!

Let Q be a quadrilateral which has a inscribed circle and a circumscribed circle. Show that the centers of these 2 circles and the point of intersection of the diagonals
of the quadrilateral are aligned.

restart; unprotect(D, O);

with(plots); with(LinearAlgebra);
Soit Q un quadrilatère qui possède un cercle inscrit et un cercle circonscrit. Montrer que les centres de ces 2 cercles et le point d'intersection des diagonales
du quadrilatère sont alignés.
_EnvHorizontalName := 'x';

_EnvVerticalName := 'y';

Vdot := proc (U, V) add(U[i]*V[i], i = 1 .. 2) end proc;
dist := proc (M, N) sqrt(Vdot(expand(M-N), expand(M-N))) end proc;
EQ := proc (M, N) local eq; eq := (y-M[2])/(x-M[1]) = (N[2]-M[2])/(N[1]-M[1]) end proc;
Pour un quadrilatère A'B'C'D' circonscrit autour d'un cercle avec points de contact A, B, C, D, les droites A'C', B'D', AC, BD sont concourantes
varphi1 := (1/3)*Pi; varphi2 := varphi1+(1/4)*Pi+.1; varphi3 := varphi2+3*Pi*(1/5); varphi4 := 2*Pi-varphi1-varphi2-varphi3; R := 5;
                    varphi4 := -0.514159263
O := [0, 0];
for i to 4 do M || i := [R*cos(varphi || i), R*sin(varphi || i)]; OM || i := plot([O, M || i], color = blue, linestyle = 3); cfOM || i := -1/(diff(solve(EQ(O, M || i), y), x)); eq || i := y = cfOM || i*x+R*sin(varphi || i)-cfOM || i*R*cos(varphi || i); T || i := plot(cfOM || i*x+R*sin(varphi || i)-cfOM || i*R*cos(varphi || i), x = -10 .. 10, color = green) end do;

for i to 4 do j := `mod`(i+1, 4); if j = 0 then j := 4 end if; sol || i := evalf(op(solve([eq || i, eq || j], [x, y]))); P || i := [subs(sol || i, x), subs(sol || i, y)] end do;
for i to 4 do Q || i := geometry:-point(Q || i, P || i[1], P || i[2]) end do;
geometry:-AreConcyclic(Q1, Q2, Q3, Q4, 'cond');
AC := plot([M1, M3], color = blue);
BD := plot([M2, M4], color = blue);
diago1 := plot([P1, P3], color = coral); diago2 := plot([P2, P4], color = coral);
quadri := plot([seq(M || i, i = 1 .. 4), M1], color = black);
cer := plot([R*cos(t), R*sin(t), t = 0 .. 2*Pi], color = blue);
Points1 := pointplot([seq(M || i[], i = 1 .. 4)], symbol = solidcircle, color = [blue], symbolsize = 15);
Points2 := pointplot([seq(P || i[], i = 1 .. 4)], symbol = solidcircle, color = [green], symbolsize = 15);
Tt1 := plots:-textplot([[M1[], "A"], [M2[], "B"], [M3[], "C"], [M4[], "D"]], font = [times, 10], align = {below, right});
Tt2 := plots:-textplot([[P1[], "A'"], [P2[], "B'"], [P3[], "C'"], [P4[], "D'"], [O[], "O"]], font = [times, 15], align = {above, right});
display([cer, quadri, seq(OM || i, i = 1 .. 4), seq(T || i, i = 1 .. 4), AC, BD, diago1, diago2, Points1, Points2, Tt1, Tt2], axes = none, view = [-9 .. 7, -12 .. 6], scaling = constrained, size = [500, 500]);

I was trying this in Maple to check Mathematica's result, and got this error. But googling and the link Maple gives lead to nothing.

Any one knows why Maple gives this error message here?


A := <
          <0, 0, exp(I*k1) + m1, exp(I*k2) + m2>|
          <0, 0, exp(I*phi)*(exp(-I*k2) + m2), exp(-I*k1) + m1>|
          <exp(-I*k1) + m1, exp(-I*phi)*(m2 + exp(I*k2)), 0, 0>|
          <exp(-I*k2) + m2, exp(I*k1) + m1, 0, 0>

Error, (in LinearAlgebra:-Eigenvectors) multiplicity mismatch


fyi, Mathemtica's result is

ClearAll[k1, m1, m2, k2];
Phi = Pi;
H = {{0, 0, Exp[I k1] + m1, Exp[I k2] + m2}, {0, 0, 
    Exp[I Phi] (Exp[-I k2] + m2), Exp[-I k1] + m1}, {Exp[-I k1] + m1, 
    Exp[-I Phi] (m2 + Exp[I k2]), 0, 0}, {Exp[-I k2] + m2, 
    Exp[I k1] + m1, 0, 0}};


I tried tracing in Maple, but so far no useful result:


simplify/size: [1/(2+m1^2+m2^2+2*m1*((1/2)*exp(I*k1)+(1/2)*exp(-I*k1))+2*m2*((1/2)*exp(I*k2)+(1/2)*exp(-I*k2)))^(1/2) exp(-I*k1) exp(-I*k2) exp(I*k1) exp(I*k2) _t[1] _t[2] m1 m2 k1 k2]
simplify/size: [exp(-I*k1) exp(-I*k2) exp(I*k1) exp(I*k2) m1 m2 _t[1] _t[2] k1 k2]
simplify/size: [1/(2+m1^2+m2^2+2*m1*cos(k1)+2*m2*cos(k2))^(1/2) exp(-I*k1) exp(-I*k2) cos(k1) cos(k2) _t[1] _t[2] m1 m2 k1 k2]
simplify/do: applying  commonpow  function to expression
simplify/do: applying  power  function to expression
simplify/do: applying simplify/size function to expression

Error, (in LinearAlgebra:-Eigenvectors) multiplicity mismatch


Maple 2019.2.1 on windows


Could anyone help me with the following task?

The simplest case is, for example, 

    If y=y(x). Denote y'(x)=Dx(y). Then we make a change of veriables :

    x=φ(z,w), y=ψ(z,w), where z independent and w=w(z),

    Now the ODE: y'(x)=1 turns to:

    Dz(y)/Dz(x)=1,  i.e. ψz+w'(z)ψwz+w'(z)φw.

    So we have the relation: Dz=(φz+w'(z)φw)Dx.

    The task is how to define the operators (Dx, Dz) to transform more complecated ODE of (x,y) (e.g. (Dx+F(x,y))5(y)=0,   

     F arbitrary) to its counterpart of (z,w)?

Thank you in advance!

so I can get inert operations but what about parenthesis and built ins?




I can't remember the specific code but what I was getting was something like


1 + 3 + 4 + 5 + 6 /4

rather than (1 + 3 + 4 + 5 + 6)/4

and the parenthesis was not showing up and I couldn't use things like add or mul in an inert like form(I think I had to write a loop).


What I had to do was multiply by 1 inertly and then it added the parenthesis.


If I could add inert parenthesis it would have solved the problem and also get built in functions to be inert but only on their ineternal operations.





Help me to solve this problem with pdsolve

i have N+1 equation like following

i want to solve them. how can i do it?

i think i should first convert them to sequense and then use fsolve code ,but my code will get error!! could you help me?




I'm trying to plot:

y = x(x^4 - 10*x^2 + 39)/30

I keep getting a what looks like a straight line. it should be a curved line.

I got it to work on :

y = x(x^4 - 10*x^2 + 39)/30


I'm wanting to use nonlinear least squares fittting to minimise the error between two curves.

One curve, f, I know the parameters for and the other, f2 , I do not, which is what I am wanting to find through this methood.



Any help would be muchly appricated, thank you in advance! 

I am trying to recreate an example,  See attached worksheet below.  I can't figure out what I am doing wrong here.  I have tried using the command StepProperties on other discrete transfer functions, without any problems.  Having a function that plots the Step Response of a system, Continuous or Discrete, should be built in.


Trying to recreate step response of discrete time system as in this example below:



`Standard Worksheet Interface, Maple 2019.2, Windows 7, November 26 2019 Build ID 1435526`




"module() ... end module"



HFloat(0.0), [undefined, undefined], [undefined, undefined], [undefined, undefined], [undefined, undefined], [undefined, undefined], [undefined, undefined]








I explore the "PracticeSheet" command of the "Basic" package, and I want to build two sheets on the Discriminant of polynomial ( send degree) and tangent to curves ( Derivative application). Do you have any ideas?

Many thanks!


I am Seonhwa Kim, a mathematical researcher in Korea. Recently, I have extensively used Maple to compute character varieties of 3-manifolds. Several months ago, I obtained some strange results in Maple which implies a contradiction in theory.  I have been struggling with these issues since it is usually about enormous polynomial systems.  Eventually, I could figure out that the issues are caused by a defect in Maple and were able to construct a minimal working example to produce wrong computations in Maple.  I am writing this post to report them.


This is mainly about the PolynomialIdeal package.  Along with the documentation in Maple, If an ideal J is radical, PrimeDecomposition and PrimaryDecomposition should have the same result.  But, as we see the following, the result of PrimeDecomposition and PrimaryDecomposition are different although J is a radical ideal.

The problem seems to be that the PrimaryDecomposition command in Maple sometimes produces incorrect results.

We can compute the primary decomposition of J by hand.  It should be <x> and  <y, x-1>.

I double-checked this by the other software;Macauley2, Singular, and Magma, for example, you can see it as follows.




Secondly, not only for PrimaryDecomposition but also PrimeDecomposition may produce an incorrect result.

Here is a minimal working example.

Maple tells us a compatible result of prime and primary decomposition of a radical ideal J.

But the first component of J,  < b-1, c-a+1 >, contains the third component < a, b-1, c+1 >.

It contradicts with the definition of Primary decomposition. So the correct answer should be  < b - 1, c - a +1 >, <a,b,c>.


I also checked that  Macaluey2, Singular and Magma. They all say that my hand computation is correct. as follows.


I have used Maple 2017 by the license of my institute (Korea Institute for Advanced Study).

When I noticed these defects, I thought it would be fixed in the newest Maple version.

So, I have tried my examples by Maple2019  free trial, but It also has the same problem. 

I guess this problem is not reported or recognized yet. 


I hope this problem will be fixed as soon as possible.

Thank you for attention.




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