@Kitonum   with a large number of lines. The arrow was made in Paint:

R := evalc(1/(1-x-I*y)):

R1 := op(1, R), coeff(R, I):

f := unapply([R1], x, y):

F := plottools[transform](f):

A := plot([seq([convert(<cos(1/8*Pi*i), -sin(1/8*Pi*i); sin(1/8*Pi*i), cos(1/8*Pi*i)> . <t, 0>, list)[], t = -.95 .. .95], i = 0 .. 15), seq([r*cos(t), r*sin(t), t = 0 .. 2*Pi], r = 0.1 .. 0.95, 0.085)], color = red, thickness = 0, view = [-1 .. 1, -1 .. 1], numpoints = 1000, view = [-3.5 .. 3.5, -3.5 .. 3.5]):

B := plots[display](F(A), view = [0 .. 7.5, -3.5 .. 3.5]):

plots[display](<A | B>, scaling = constrained, axes = box);

@Carl Love  Thank you for your helpful comment. I had never thought about the possibility of adjusting the thickness of the gridlines. Especially I like monochrome color with thin gridlines:

y := 0: z := 0:

plot3d(abs((2*(-exp(-t-x-z)+exp(t+x+z)))/(exp(-t-x-z)+tanh((1+I)*t+(1/2-1/2*I)*y+z)+exp(t+x+z))), x = -5 .. 5, t = -5 .. 5, color = khaki, thickness = 0, axes = normal, orientation = [15, 50], numpoints = 20000);

@mskalsi  I do not know how to explain this effect.

@Vic  Maple does it automatically:

Example:

solve({x+y=0, x+y=1});

Nothing was printed.

@Abdoulaye  In this case, do a procedure with formal parameter n .

@baharm31  You wrote "It is easy to do what you did when the horizontal and vertical axis are the same size."  It is not necessary.

Example:

plot(sin(x), x=-Pi/2..Pi/2, size=[157,100], tickmarks=[spacing(Pi/4),default]);  # Font by default, size=12

plot(sin(x), x=-Pi/2..Pi/2, size=[157,100], tickmarks=[spacing(Pi/4),default], font=[TIMES,ROMAN,8]);

@acer  Thank you, very useful comment.

If we want also simplify the trig term then:

A:=y(x)=1/3/2^x+_C1/(3^x)^2 + cos(x)*sin(x):

combine(A, {power, trig}); 

                           y(x) = 1/3*2^(-x)+_C1*3^(-2*x)+1/2*sin(2*x)  

@Markiyan Hirnyk  You are not right. From Collins dictionary:

partial ['pɑːʃəl] 1) relating to only a part; not general or complete a partial eclipse 2) biased a partial judge 3) (postpositive; foll by to) having a particular liking (for) 4) botany a) constituting part of a larger structure a partial umbel b) used for only part of the life cycle of a plant a partial habitat c) (of a parasite) not exclusively parasitic 5) mathematics designating or relating to an operation in which only one of a set of independent variables is considered at a time 6) Also called: partial tone music , acoustics any of the component tones of a single musical sound, including both those that belong to the harmonic series of the sound and those that do not 7) mathematics a partial derivative

See the meaning 7)

I can confirm. I already wrote about this error here.

Maple 2016.1 also fails to your task:

restart:

with(Optimization):

LPSolve(2*x+5*y, {3*x-y = 1, x-y <= 5}, assume = {integer, nonnegative});

  Error, (in Optimization:-LPSolve) no feasible integer point found; use feasibilitytolerance option to adjust tolerance

Addition: my Maples 2015 and 2016 are 32-bit

@Carl Love  

If you write as you suggest, then:

1. The original issue does not indicate that the numbers must be positive.
2. You can lose also positive solutions, see

sols := [isolve(sum(2*(m+i-1)+1,i=1..n) = 16)]:

S:=select(type, sols, set(name = posint));

nops(S);

for s in S do

[seq(eval(2*m+1+2*(n-1),s[1]), n=1..eval(n,s[2]))] ;

od;

                           S := [{m = 3, n = 2}]

                                            1

                                         [7, 9]

The solution  [1,3,5,7]  is lost.

@iman  I checked your statement.

See

Maple does not find this nontrivial solution.

Download 123456.mw

 Edited.

@Carl Love  I think the reason for the failure of OP is in using  vector  command instead of  Vector.

See:

X:=[seq(i^2, i=1..20)]:  # list 1

Y:=[seq(i^3, i=1..20)]:  # list 2

M:=vector(20, x->X[x]):

N:=vector(20, x->Y[x]):

plot(M,N); 

Good morning  Alsu! I am very pleased to see the new user from Russia in this  forum. I live right next to you in the city of Penza.

Here are some tips to improve your codes:

1. To create simple animations to be more efficient use of  plots[animate]  command, in which the frames are generated automatically:

restart;

plots[animate](plot, [[x,y, y=-20..20], linestyle=2, color=red, thickness=4], x=-10..10, frames=60);

2. I do not understand why you use plots[implicitplot] command (instead of plot command) in your further examples  transformations of simple plots defined by explicit formulas. Using a list, you can create multiple plots in one coordinate system:

restart;

plot([sin(x), sin(3*x), sin(1/3*x)], x=-3*Pi..3*Pi, y=-1.4..1.4, color=[red,blue,green], linestyle=[2,1,1], thickness=[2,3,3], scaling=constrained, legend=[typeset("  График функции  ", y=sin(x)), typeset("   График функции  ", y=sin(3*x)), typeset("  График функции  ", y=sin(1/3*x))], legendstyle=[font=[times,roman,14]]);

3. To work with the objects, depending on the parameters,  Explore command is very helpful. See help for details on this command.

@adel-00  To avoid this, it is impossible - this is the nature of the solutions to this equation. To find a solution further right of  1.5355176  you need to set new initial conditions, for example

restart:

with(plots):

ini1:=D(x)(1.54)=0, x(1.54)=1:

dsys:=diff(x(t),t,t)+(x(t)-2)*diff(x(t),t)+5*x(t)=0;

dsol1 :=dsolve({dsys,ini1}, numeric, abserr=1e-9, relerr=1e-8, maxfun=0);

plots:-odeplot(dsol1, [t,x(t)], t=1.54..4, axes=boxed, color=black, linestyle=1, tickmarks=[6, 6], axes=boxed, titlefont=[SYMBOL,12]);

We see that there is a new singularity  3.0755176, about which the plot goes to minus infinity.

@acer  Thank you for the correction. It was my carelessness when typing of the code.