It is easy to prove that your second graph is empty. Probably you rewrote something wrong. I modified the second equation a bit to make the graph look like your second graph in the picture:

Edit. Let us prove that the equation  abs(z-I) - abs(z+1)  =  2  has no solutions for complex numbers  z=x+I*y . This is most easily done using the fact that   abs(z1 - z2)  is the distance between the points  z1  and  z2 . Consider 3 points  z1=I ,  z2=-1 and  z . It follows from the triangle inequality that  abs(z-z1) - abs(z-z2) < abs(z1-z2) = sqrt(2)<2

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If you need to do this more than once, then (as a development of vv's idea) a simple procedure  Sort  can be used to reduce typing:

restart;
S:=[seq(_C||k=c[k], k=0..10)]:
sol:=dsolve(diff(y(x),x) = x+y(x),y(x));

Sort:=s->sort(subs(S,s)):

Sort(sol); 

Use  piecewise  instead of  if .. fi  :

Download if_new.mw

I do not know the reasons for this behavior, but to immediately get the desired position of the constants use the  sort  command:

restart;
ode:=diff(diff(y(x),x),x)+8*diff(y(x),x)+25*y(x) = 1:
sol:=sort(dsolve(ode));

This code in Maple 2018.2

Here is another proof without complex numbers using the well-known rotation matrix by an angle. We take the unit vector  <1,0>  and rotate it first by the angle of  2*x , we get the vector  X . Rotating the same vector  <1,0>  sequentially, first by the angle  x , and then again by x , we get a vector  Y . Obviously  X=Y . At the same time, we proved another formula  cos(2*x)=cos(x)^2-sin(x)^2 .

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You are probably interested in calculating the area of the region bounded by a red closed curve below (this is not an ellipse of course).
It's easy to do numerically:

Download int.mw

To solve your example, it is enough to know the simplest definitions related to the joint distribution of two random variables  X  and  Y . Let's introduce the notations:  f__XY  is the density function of the joint distribution of  X  and  Y ,  f__X  is the density function of  X ,  f__Y  is the density function of  Y ,  f__X_Y  is the density function of the conditional distribution of  X  given  Y ,   f__Y_X  is the density function of the conditional distribution of  Y  given  X .

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You have chosen too narrow a range of values for the contours. If we increase it, we get more interesting pictures, for example:

restart;
randomize():
a, b, c_3, c_4, A_5:='rand(-5..5)'()$5;
A := a*(A_5)^3-b; B := 3*a*(A_5)^2-3*b/(A_5); C := (c_3)*(A_5)-(c_4);
u:=3*y^2+2*x^3*B/A+3*x^2*C/A;
plots:-contourplot(u, x=-2.25..2.25, y=-2.25..2.25, axes=boxed, contours=[seq(-10..10,1)],grid=[200,200], coloring=["Blue","Red"], size=[500,500]);

Below are 2 solutions to the problem. In the first one, we randomly set all the necessary parameters. In the second one, we use  Explore  command for this:

MaplePrime did not display the result of the last command. See the attached file for this.

Download contour_new.mw

Addition - another way:

T__e_np := sqrt(r^2+1)/(r^2+sqrt(r^2+1)*r+1);
numer(T__e_np)*sqrt(r^2+1)/expand(denom(T__e_np)*sqrt(r^2+1));
simplify(%);

Download Q20201208_new.mw

Maple simply doesn't know what it should return if  k  is a symbol. Therefore, we must provide for this in the body of the procedure. Below is a possible solution to the problem:
 

Download divide5_new.mw

  BrettKnoss I have simplified  Carl's code a bit. I think this will make it clearer for you. To study the convergence at the ends of the convergence interval, we used the n-th term test: If , then the series diverges  (from wiki)  . Since the results are not 0, the series diverges at both ends.

restart;
f:= (x-1)/(x+2);
s:= convert(f, FormalPowerSeries);
c:= unapply(op([2,1],s), k);
L:= limit(abs(c(k+1)/c(k)), k= infinity);
solve(L < 1, x);
``;
# Investigation on the ends
RealDomain:-limit(eval(c(k),x=-2),k=infinity); 
RealDomain:-limit(simplify(eval(c(k),x=2)),k=infinity);

Download ic.mw

Procedure  P  returns the list of  N  pairs of such numbers  [p, q] :

Edit.

Download P1.mw

In fact, we have a nonlinear inequality with the parameter c . Maple usually cannot solve such inequality in a closed form. But it is easy to write a procedure  (named  F  below) that, for each value of the parameter c , finds the corresponding range for  r . Now you can use this procedure by specifying values for c:

restart;
F:=c->solve({-r+c+1 + sqrt(r^2 - 2*r*c -r +2*c )<1, 0<r, r<1}, r):

Examples of use:

F(0.1), F(0.5), F(0.9);

             {r <= .2000000000, .1900000000 < r}, {.7500000000 < r, r < 1.}, {.9900000000 < r, r < 1.}

Using the  rand  command, we give your parameters some random values from the appropriate ranges:

If you want your initial parameters to take not only integers, but also fractional values in float format, then write  a_1:=rand(1...5.)() and so on. I have not set values for  f  and  g  parameters, so Maple builds not one, but a whole family of curves. Of course, you yourself can set these parameters in the same way.

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