MaplePrimes Questions

I'm trying to solve the following linear system

eq1 := -t*x + y*z = j;
eq2 := t*x + y*z = -m;
eq3 := t*z - x*y = -b;
eq4 := t*z + x*y = a;

I have made many unsuccessful attempts.

Would anyone have the solution to the problem?



The following maple command returns an error in Maple 14 (internal error in Typesetting ... "invalid subscript selector") but not in Maple 2017.   

f:=(l::list)-> eval([y,y*z-x,-15*x*y-x*z-x],[x,y,z]=~l):

What did I miss?


Many thanks









Hello. I want to solve a differential equation on a thickening grid. That is, to be solved first at N=10, then N=20, then N=40 and so on.
Tried the design
for N from 10 to 160 by 2*N do
but for Maple it is difficult. Do you have any ideas how to implement this type of cycle?

I do not understand why Maple sometimes shows singular solution to Clairaut ODE and sometimes not.

Clairaut ODE has the form y(x) = x y'(x) + G(x, y')

In the following ODE when I ask Maple to dsolve it as is, it does give singular solution. Next, when solving explicity for y(x) first, which will generate 2 ODE's, each is Clairaut ODE, then ask Maple to dsolve each, now Maple no longer gives the singular solution. But when I solve each one of these ODE's, I see that there is the singular solution there. It must be there, since this is Clairaut ODE and it has singular solution.

When I do PDEtools:-casesplit on each of the two ODE's generated by solving for y(x) first, I see the singular solution there.

The question is, why Maple dsolve does not show the singular solution in the second case? And how to make it show it? Or did I do something wrong?



ode:=x^2*diff(y(x),x)^2-(1+2*x*y(x))*diff(y(x),x)+1+y(x)^2 = 0;

x^2*(diff(y(x), x))^2-(1+2*x*y(x))*(diff(y(x), x))+1+y(x)^2 = 0


[[_1st_order, _with_linear_symmetries], _rational, _Clairaut]

Vector([dsolve(ode,y(x))]); #now it shows singular solution (first one below)

Vector(3, {(1) = y(x) = (1/4)*(4*x^2-1)/x, (2) = y(x) = _C1*x-sqrt(_C1-1), (3) = y(x) = _C1*x+sqrt(_C1-1)})


`casesplit/ans`([(diff(y(x), x))^2 = (2*y(x)*(diff(y(x), x))*x+diff(y(x), x)-y(x)^2-1)/x^2], [2*(diff(y(x), x))*x^2-2*x*y(x)-1 <> 0]), `casesplit/ans`([y(x) = (1/4)*(4*x^2-1)/x], [])

ode:=convert(ode,D): #solve for y(x) first, this will generate 2 ODE's

Vector(2, {(1) = y(x) = (diff(y(x), x))*x+sqrt(diff(y(x), x)-1), (2) = y(x) = (diff(y(x), x))*x-sqrt(diff(y(x), x)-1)})


[[_1st_order, _with_linear_symmetries], _rational, _Clairaut]

[[_1st_order, _with_linear_symmetries], _rational, _Clairaut]

dsolve(odes[1],y(x)); #where is singular solution?

y(x) = _C1*x+(_C1-1)^(1/2)

dsolve(odes[2],y(x)); #where is singular solution?

y(x) = _C1*x-(_C1-1)^(1/2)


`casesplit/ans`([diff(y(x), x)-1 = (-(diff(y(x), x)-1)^(1/2)+y(x)-x)/x, diff(y(x), x) = (-(diff(y(x), x)-1)^(1/2)+y(x))/x], [1+2*x*(diff(y(x), x)-1)^(1/2) <> 0]), `casesplit/ans`([y(x) = (1/4)*(4*x^2-1)/x], [])


`casesplit/ans`([diff(y(x), x)-1 = ((diff(y(x), x)-1)^(1/2)+y(x)-x)/x, diff(y(x), x) = ((diff(y(x), x)-1)^(1/2)+y(x))/x], [2*x*(diff(y(x), x)-1)^(1/2)-1 <> 0]), `casesplit/ans`([y(x) = (1/4)*(4*x^2-1)/x], [])




I would like to draw an ellipse by orthonal affinity of a circle. Thank you.

Dear Maple users

I wanted to export two long lists X and Y of data into two columns in Excel by using the Export command from the ExcelTools package. Data was however exported into rows! I converted the lists into vectors and I succeeded in doing it the way I wanted:

X1 := convert(X, Vector);
Y1 := convert(Y, Vector);
Export(X1, "data1.xlsx", 1, "A1");
Export(Y1, "data1.xlsx", 1, "B1");

But is it really necessary to convert to vectors in order to accomplish this task? 





This is a problem that intrigues me.

I have been thinking about the distribution of n points, p[1], p[2], ..., p[n] in a scatterplot and I am interested in finding the location of a focal point, A, such that the distance from A to each neighboring point is a minimum. I have worked on a routine and this is given in the attached worksheet. In this example, n=20 and the position of A is determined to be [-3.25, 0.99].

Now - here's my question. Suppose I am interested in n-large and, instead of locating one focal point, A, I wish to obtain several,i.e. A, B, C, etc. such that the distances from each of these to their respective neighboring points is also a minimum.

Does any interested party know if this is possible to do and if so, can anyone suggest an approach or routine? If so, I'd be delighted to understand how to solve for this.

Thank you for reading!



The commands


give the output

Matrix(2, 2, {(1, 1) = 1, (1, 2) = 1, (2, 1) = 4, (2, 2) = 5}, datatype = anything, storage = rectangular, order = Fortran_order, shape = [])


Why won't it display properly?


To help decide the type of ODE, I need a function which moves all derivatives to lhs and everything else to the rhs of the ODE. This way I can more easily analyze the ODE.

The ode will be only first order. Any term which contains  (y')^n, is to be moved to the lhs. Rest to rhs. Couple of examples will help illustrate the problem. This is done in code only, without looking at the screen. The dependent variable is always y and the independent variable is x.  So I need to turn the ODE to the form 

                    G(y',y'^2,y'^3,.....,y'^n)  = F(x,y)

I can find all derivatives in the ODE using


But not sure how this will help me do what I want. isolate does not help, since it only takes one term at a time. Collect also did not help for same reason.  If the ODE contains only ONE derivative, then it is easy to do. But the question is about how to do it for ODE which contains more than y' term of different powers as the examples below show.

What methods to do this in Maple? I looked at DEtools but so far did not see anything.

Example 1



3*(diff(y(x), x))^2+(diff(y(x), x))^3+sin(x)+y(x) = x*y(x)+x*(diff(y(x), x))


{(diff(y(x), x))^2, (diff(y(x), x))^3, diff(y(x), x)}

ode_wanted:= 3*diff(y(x),x)^2+diff(y(x),x)^3-x*diff(y(x),x)=-sin(x)-y(x)+x*y(x)

3*(diff(y(x), x))^2+(diff(y(x), x))^3-x*(diff(y(x), x)) = -sin(x)-y(x)+x*y(x)

Example 2



3*(diff(y(x), x))^2+diff(y(x), x) = x*(diff(y(x), x))+5


{(diff(y(x), x))^2, diff(y(x), x)}

ode_wanted:= 3*diff(y(x),x)^2+diff(y(x),x)-x*diff(y(x),x)=5

3*(diff(y(x), x))^2+diff(y(x), x)-x*(diff(y(x), x)) = 5





i want to solve the systems of diff equation what's the problem

Dear Maple users

Some students have come to us to report, that something doesn't seem to work properly in Maple 2019.1 in Document Mode. And they seem to be right: writing an passive math formula by using Shift+F5 (the formula is gray, not blue), then using F5 to get out of that Math field and back into Text Mode. Using the Enter key to go to the next line: It doesn't work! The cursor stays in the same line. This behavior is new in Maple 2019. It worked properly in Maple 2018 and earlier. I assume it is not the intention? 

I know it can easily be dealt with by making a new Paragraph by using the shortcut Ctrl+Shift+J. I call the assumed bug 'severe' though, because it will severely delay the workflow for many students. They are used to deliver a document mixed with formulas (active or passive) and text. 

NB! I have tested it on several computers (Mac and Windows), and it doesn't work on any of them.


Erik V.

Hi I have the following ln function that I want to differentiate wrt variable c:


I2 := -sqrt(-c^2+1)*ln(abs((.9*sqrt(-c^2+1)-c*sqrt(1-.9^2))/(-.9*sqrt(-c^2+1)-c*sqrt(1-.9^2))))


When I differentiate I obtain an expression that involves 

...abs(1, (.9*sqrt(-c^2+1)-.4358898944*c)/(-.9*sqrt(-c^2+1)-.4358898944*c))...

Why does it give a comma at the abs expression? how to get rid of that.

i have a question about sets. how can i keep set members in order of addition not the defualt maple ordering.
in maple help, i saw the command setsort=0..8 but i do not know how to use it.
consider exmaple below:




for j in [3,5,6,1] do
L:=`union`(L ,{j});







{3, 5}




{3, 5, 6}




{1, 3, 5, 6}



how can i keep L as order of addition? L={3,5,6,1}. thanks in advance


Hi I am using Maple 18 and am trying to make my looping work backward.


Digits := 20;
n, K := 5, 4;
for k from 0 to n do a[k] := (int((T(2, t)-T(0, t))*T(k, t)/sqrt(-t^2+1), t = -1 .. 1))/Pi end do

First I use the above command to generate a0, a1, a2, a3, a4 and a5. Now i want to do a back substitute using the following recurrence relation to determine the values of b5, b4, b3, b2, b1 and b0.

b[n] := 0; (to find b5)

b[n-1] := a[n]; (to find b4)

(to find b3, b2, b1, b0)

Can someone assist me with the backward looping please. I does not seem to work.



I calculate volume of tetrahedron SABC by this way. Now, I want to find one option  of coordinates of all vertices of this tetrahedron. How can I get the coordinates of the points S, A, B, C so that coordinates  center of circumcircle of the triangle ABC is (0,0,0)?


SA := 3: 
SB := 5: 
SC := 7:
AB := 3:
BC := 4: 
AC := 5:
V := (1/12)*sqrt(SA^2*BC^2*(AB^2+AC^2-BC^2-SA^2+SB^2+SC^2)+SB^2*AC^2*(AB^2-AC^2+BC^2+SA^2-SB^2+SC^2)+SC^2*AB^2*(-AB^2+AC^2+BC^2+SA^2+SB^2-SC^2)-SA^2*AB^2*SB^2-SB^2*BC^2*SC^2-SA^2*AC^2*SC^2-AB^2*BC^2*AC^2);

From here
I find coordinates of the point A, B, C. Now I tried

point(A, -3/2, -2, 0):
point(B, 3/2, -2, 0):
point(C, -3/2, 2, 0):
point(S, x, y, z):
solve([distance(S, A) = 3, distance(S, B) = 5, distance(S, C) = 7], [x, y, z], explicit);
point(A, -3/2, -2, 0):
point(B, 3/2, -2, 0):
point(C, -3/2, 2, 0):
point(S, x, y, z):
solve([distance(S, A) = 5, distance(S, B) = 3, distance(S, C) = 7], [x, y, z], explicit);

Use the above code, I obtain the result.

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