ConvIntoIneq  procedure solves the problem:

ConvIntoIneq:=proc(Rel)

local t, R, x, y;

t:=indets(Rel)[];

R:=solve(Rel);

x, y:=op(1,R), op(2,R);

if x=-infinity then if type(y,realcons) then return cat(t, "<=", y) else

cat(t, "<", op(1,y)) fi else

if type(x,realcons) then return cat(t, ">=", x) else

cat(t, ">", op(1,x)) fi; fi;

end proc:

Examples of use:

ConvIntoIneq(not x<100), ConvIntoIneq(not x>100), ConvIntoIneq(not x>=100), ConvIntoIneq(not x<=100), ConvIntoIneq(x>100), ConvIntoIneq(x<100), ConvIntoIneq(x>=100) ;

 Addition. Unfortunately, if we want to get a "nice" output, the name of variable can appear on the right side of the inequality:

map(parse, [ConvIntoIneq(not x<100), ConvIntoIneq(not x>100), ConvIntoIneq(not x>=100), ConvIntoIneq(not x<=100), ConvIntoIneq(x>100), ConvIntoIneq(x<100), ConvIntoIneq(x>=100)])[] ;

I added 2 options into your code: axes=normal and orientation=[15,50]  and used user lighting :

 Addition. The plot was saved in Paint as png - file.

Another variant. Additionally I added  color=khaki ,  style=surface , numpoints=20000  and changed  user lighting  :

plot([[0,0.36],[5,0.38],[10,0.3],[15,0.18],[20,0.96],[25,0.18],[30,0.16],[35,0.96]], style=line, thickness=5, color=red, view=[0..35,0..1], axis[1] = [gridlines = [7, thickness = 1, subticks = false, color = grey]], axis[2] = [gridlines = [6, thickness = 1, subticks = false, color = grey]], tickmarks=[[0=1,5=5,10=10,15=50,20=250,25=1000,30=10000,35=50000], spacing(0.2,0)], size=[650,350], font=[COURIER,ROMAN,14]);

 Edited.

P := (L, a, b) ->plot(L, x = a .. b, color = [seq(`if`(i::odd, red, black), i = 1 .. nops(L))]):   # L is a list of functions

Example of use:

P([sin(x), cos(x), 2, 3-x], 0, 5);

First, you can find the indices of columns that must be deleted, and then delete them.

Example:

A:=<0,1,0,3; 0,-2,0,4>;

ListTools[SearchAll](0, convert(A[1], list));

LinearAlgebra[DeleteColumn](A, [%]);

Your syntax is wrong there. Below is the text of the procedure called  P  that solves the problem for any  Time  and  solution:

P:=proc(Time, solution)

local n;

n:=nops(Time);

piecewise( seq( op([t>=Time[i] and t<Time[i+1], solution[i]]), i = 1..n-1));

end proc:

Solution of above example:

P([0,1,2,3], [t^2, 1, 3-t]);

plot(%, t=0..3, scaling=constrained);

Edited.

Example:

F:=piecewise(t>=0 and t<1, t^2, t>=1 and t<2, 1, t>=2 and t<3, 3-t);

plot(F, t=0..3, scaling=constrained);

Addition:  outside the interval t=0..3  the function  F by default considered to be equal to 0 . 

Examples of plotting  4 basic trigonometric functions:

F1 := sin(x): F2 := cos(x): F3 := tan(x): F4 := cot(x): R := x = -Pi .. 2*Pi: C := [red, blue, green, brown]: c := scaling = constrained:

v1 := plot(F1, R, color = C[1]):

v2 := plot(F2, R, color = C[2]):

v3 := plot(F3, R, -3 .. 3, color = C[3], discont):

v4 := plot(F4, R, -3 .. 3, color = C[4], discont):

plot([F1, F2, F3, F4], R, -3 .. 3, color = C, discont, c);  # All together in one plot

 plots[display](< v1, v2; v3, v4 >, c, size = [300, 300]);  # Matrix of the plots

 plots[display](< v1 | v2 | v3 | v4 >, c);  # Vector-row of the plots

 plots[display](< v1, v2, v3, v4 >, c, size = [100, 100]);  # Vector-column of the plots

If you want the tickmarks are not lost while reducing the size of the plot, reduce the font size:

combine(y(x) = (1/3)*exp(x)+_C1/(exp(x))^2);

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

Example with other powers not exp:

combine(y(x) = 1/3/2^x+_C1/(3^x)^2);

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

@Rules,  in its solution  you are using  diff  command. Unfortunately during the second differentiation is obtained an incorrect result. D command corrects the situation:

restart;

eq:=z(x,y)^2+y^3+x^4 - ln(x+y+z(x,y)):

F:=(x,y)->z(x,y)^2+y^3+x^4 - ln(x+y+z(x,y)):

D[1](F)(x,y);

D_x:=solve(%,(D[1](z))(x,y));

D[2](F)(x,y);

D_y:=solve(%,(D[2](z))(x,y));

D[1,2](F)(x,y);

D_xy:=simplify(eval(solve(%, (D[1,2](z))(x,y)),{(D[1](z))(x,y)=D_x,(D[2](z))(x,y)=D_y}));

D[2,1](F)(x,y);

D_yx:=simplify(eval(solve(%, (D[2,1](z))(x,y)),{(D[1](z))(x,y)=D_x,(D[2](z))(x,y)=D_y}));

is(D_xy=D_yx);

 Addition. Above I wrote "Unfortunately during the second differentiation is obtained an incorrect result" .  In fact, when used properly,  diff  command also gives the correct result:

restart;

Eq:=z(x,y)^2+y^3+x^4 - ln(x+y+z(x,y)):

diff(Eq,x);

D_x:=solve(%,diff(z(x,y),x));

diff(Eq,y);

D_y:=solve(%,diff(z(x,y),y));

diff(Eq,x,y);

D_xy:=simplify(eval(solve(%,diff(z(x,y),x,y)),{diff(z(x,y),x)=D_x,diff(z(x,y),y)=D_y}));

diff(Eq,y,x);

D_yx:=simplify(eval(solve(%,diff(z(x,y),y,x)),{diff(z(x,y),x)=D_x,diff(z(x,y),y)=D_y}));

is(D_xy=D_yx);

M.Hirnyk, please check this code in Maple 5 R4 .

@Rules, кстати, вы можете задавать свои вопросы по-русски на форуме по Maple на exponenta.ru

plots[inequal]  command for non-linear inequalities works only in the latest versions of Maple. For earlier versions, you can use  plots[implicitplot]  command with option  filledregions :

A := plots[implicitplot]([x^2+y^2-1, (x-1)^2+y^2-1], x = -2 .. 3, y = -2 .. 2, color = [red, blue], linestyle=3, thickness = 2, gridrefine = 3):

B := plots[implicitplot](max(x^2+y^2-1, (x-1)^2+y^2-1) < 0, x = -2 .. 3, y = -2 .. 2, filledregions, coloring = [yellow, white], gridrefine = 3):

plots[display](A, B, scaling = constrained);

Edited.

I think the solution of your equation can not be expressed explicitly. Therefore, you can solve it immediately automatically, without using explicit integration (but using implicit option):

restart;

eq:=(exp(x)+y(x))+(2+x+y(x)*exp(y(x)))*diff(y(x),x)=0;  # Replaced  dy/dx=diff(y(x),x)

dsolve({eq, y(0)=1}, y(x), implicit):

Sol:=subs(y(x)=y,%);

plots[implicitplot](Sol,x=-10..10,y=-10..10, gridrefine=3);

 Addition. I wonder what is the cause of the left branch, which is not visually goes through the point  (0, 1) ?

You forgot colon in assigning L. Also evalb command can be omitted:

f:={1,2,3,4}:

h:={1,2,4,5}:

L:=seq(i,i=1..4):

for i in L do

if  f[i]=h[i]  then

print(f[i]);

end if;

end do;

One way, you can do some calculations in for loop and in each step assign a name to your result.

An example:

for a from 0 to 3 by 0.5 do  # For each value of parameter a we find the list of roots of the equation  x^3-a*x+1=0

R[a]:=[fsolve(x^3-a*x+1)];

od;

Another way to use  seq  command (a solution of the same problem):

R:=seq([fsolve(x^3-a*x+1)], a=0..3, 0.5);

Reference to these results you have in different ways: in the first case, the indices coincide with the values of the parameter a, in the second case as the index, you just specify the number of the solution.

Edited.