These links can be useful when solving the problem:

Indexing Arrays, Matrices, and Vectors

?LinearAlgebra/LinearSolve 

@mag4112 there are several commands which reflect exactly the meaning of the word itself. :) You want to sort it, so look at ?sort.

@mag4112 there are several commands which reflect exactly the meaning of the word itself. :) You want to sort it, so look at ?sort.

@pagan that might be it since this would explain such view option.

@pagan that might be it since this would explain such view option.

don't you use what acer and Preben Alsholm already created for you?

With this

dsol := dsolve(dsys union isc, numeric, method = gear[polyextr],
                initstep = 0.15e-1, minstep = 10^(-11),
                abserr = 0.1e-5, relerr = 0.1e-5);
plots:-odeplot(dsol,[[t,S(t)],[t,E(t)],[t,ES(t)],[t,P(t)],[t,E(t)+ES(t)]],0..50,color=[green,blue,cyan,red,black],linestyle=[solid,dash,dash,solid,dash]);

you get the desired graphs. First, I don't understand why you would like to view the dependent variable in the interval [0..5*10^(-7)] and second: those values such as "0.15e-1" are not containing the base of the natural logarithm.

0.15e-1 means 0.015, 0.1e-5 means 0.000001 and so on, while exp(x) means the base of the natural logarithm to the power of x, see ?exp.

PS: You can also use some other stiff method such as ?dsolve/rosenbrock , which is a lot faster (at least on my computer).

don't you use what acer and Preben Alsholm already created for you?

With this

dsol := dsolve(dsys union isc, numeric, method = gear[polyextr],
                initstep = 0.15e-1, minstep = 10^(-11),
                abserr = 0.1e-5, relerr = 0.1e-5);
plots:-odeplot(dsol,[[t,S(t)],[t,E(t)],[t,ES(t)],[t,P(t)],[t,E(t)+ES(t)]],0..50,color=[green,blue,cyan,red,black],linestyle=[solid,dash,dash,solid,dash]);

you get the desired graphs. First, I don't understand why you would like to view the dependent variable in the interval [0..5*10^(-7)] and second: those values such as "0.15e-1" are not containing the base of the natural logarithm.

0.15e-1 means 0.015, 0.1e-5 means 0.000001 and so on, while exp(x) means the base of the natural logarithm to the power of x, see ?exp.

PS: You can also use some other stiff method such as ?dsolve/rosenbrock , which is a lot faster (at least on my computer).

@faisalnoufaey now I probably understand what you want, but I'm afraid it is not possible to solve it in Maple with such boundary conditions.

@Gbrandt From the second equation

x(t) has to be equal to C1*t+C2, where C1 and C2 are constants. From the initial conditions you then get C1=0 and C2=0, so x(t)=0. And therefore the system has no solution.

You uploaded a worksheet which is a little different than what you inserted before as a code image. I find it hard to understand the boundary conditions (even when combining both codes). Could you, please, explain (just briefly) the meaning of the equations and the conditions?

@Markiyan Hirnyk and it could be done by:

with(Statistics):
plot(OrderByRank(<x|y>, Rank(x)));



@Markiyan Hirnyk and it could be done by:

with(Statistics):
plot(OrderByRank(<x|y>, Rank(x)));



Can you upload a worksheet or insert your code as a text?

It would be better to upload a worksheet or insert the code as a text (not an image). However, you have only one equation for two variables x(t) and m(t)?

Haven't you asked the same question here?