with(Statistics):
Phi:= unapply(CDF(Normal(0,1),x), x);

Phi := x -> 1/2+1/2*erf(1/2*x*2^(1/2))

Here's a procedure to make that list.

> maketicks:= proc(a::numeric, b::numeric)
    local L, j;
    L:= sort([seq(seq(i*10^j, j = ceil(log[10](a/i)) .. floor(log[10](b/i))), i = [1.0,2.0,4.0,6.0,8.0])]);
    map(v -> (v=sprintf("%3.0e",v)), L);
  end proc;

             

For example:

> plot(10^(-rho),rho=3..7,
  y=10^(-7)..10^(-3), axesfont=[HELVETICA,15],  
  ytickmarks=maketicks(10^(-10), 10^(-3)),
  axis=[mode=log, gridlines=true]);

Here's a procedure to make that list.

> maketicks:= proc(a::numeric, b::numeric)
    local L, j;
    L:= sort([seq(seq(i*10^j, j = ceil(log[10](a/i)) .. floor(log[10](b/i))), i = [1.0,2.0,4.0,6.0,8.0])]);
    map(v -> (v=sprintf("%3.0e",v)), L);
  end proc;

             

For example:

> plot(10^(-rho),rho=3..7,
  y=10^(-7)..10^(-3), axesfont=[HELVETICA,15],  
  ytickmarks=maketicks(10^(-10), 10^(-3)),
  axis=[mode=log, gridlines=true]);

At least in the English-speaking countries, North America and South America are generally considered as separate continents.  I realize that this view is not popular in many other countries.  The division of the world's land masses into continents is rather arbitrary, and there's no point in arguing about it.

group[mulperms]([[1,2]],[[1,3]]);

                             [[1, 2, 3]]

group[mulperms]([[1,2]],[[1,3]]);

                             [[1, 2, 3]]

The point is not a lack of initial conditions, it is that the equations are not differential equations.  So advice about solving differential equations is not going to be useful.

The point is not a lack of initial conditions, it is that the equations are not differential equations.  So advice about solving differential equations is not going to be useful.

I think you mean something like this.

> for i from 1 to 10 do
    A[i]:= ...
    B[i]:= ...
    h[i]:= A[i]*cos(c*x) + B[i]*sin(d*x);
  end do:
  plot([h[i] $ i=1..10], x = a .. b);

I think you mean something like this.

> for i from 1 to 10 do
    A[i]:= ...
    B[i]:= ...
    h[i]:= A[i]*cos(c*x) + B[i]*sin(d*x);
  end do:
  plot([h[i] $ i=1..10], x = a .. b);

I don't understand the relation between your three systems.  You've told us one system,

dsys1 := diff(u(t), t) = -(3/2-sqrt(2))*u(t)+5*v(t), diff(v(t), t) = -5*u(t)-(3/2+sqrt(2))*v(t)-20*cos(10*t)*w(t), diff(w(t), t) = -.5-3*w(t)+20*cos(10*t)*v(t)

AFAIK this can't be solved with method=laplace because of the non-constant coefficients.

I don't understand the relation between your three systems.  You've told us one system,

dsys1 := diff(u(t), t) = -(3/2-sqrt(2))*u(t)+5*v(t), diff(v(t), t) = -5*u(t)-(3/2+sqrt(2))*v(t)-20*cos(10*t)*w(t), diff(w(t), t) = -.5-3*w(t)+20*cos(10*t)*v(t)

AFAIK this can't be solved with method=laplace because of the non-constant coefficients.

Yes, it is.

Yes, it is.

Faa di Bruno's formula.