Maybe you should switch gears and try a shooting method. For example, replace the boundary condition y(0)=0 with y(0.5)=c. Then guess around until you get a solution that satifies y(0)=0. It looks like c is about 1.25. Here is "poor man's way" to do this: >ODE:= y(f)*(diff(y(f), f, f))+50*f -50*f^2= 0; >S:= c->dsolve({ODE, (D(y))(.5) = 0, y(0.5) = c}, numeric, y(f), range = 0 .. 0.5); > S(1.25)(0);#replace 1.25 by other values for c > plots[odeplot](S(1.25),[f,y(f)],f=0..0.5);

cos*t should be cos(t), and your third line should be r2:=rhs(s2);

Dave, It would be helpful if you included some specific commands that you are trying, but I bet your problem will be cleared up if you use "disk" instead of "circle".

Maybe you are looking for this: is(diff(f(k),k),negative) assuming k>0;

Here is how I would modify your set up of your spherical Bessel n[l](x) > n:=proc(l) proc(x) local f; f:=x->cos(x)/x; if l=0 then -f(x); else -(-x)^l*(1/x)^l*(D@@l)(f)(x) fi; end; end; >n(0)(x); >n(1)(x); So I would replace the array idea n[l] with a proc n(l), that returns a proc. I would also use D instead of diff.

This kind of stuff `#mover(mi(θ),mo("∧"))` is MathML, and so you need to Google MathML to find documentation. It's not Maple syntax. Don't you wish you could just use LaTeX? $\widehat\theta$

You must have a vector-valued position function. Try something like this: plot([t^2-3, 1/2*t^2+2*t+1, t=-1..1]); Note the placement of [ and ].

Make sure you make it clear that x is a function of t. Lots of times when people enter differential equations into Maple they enter something like diff(x,t) = x^2 because this is the informal notation we use in textbooks and on paper. But Maple needs to know x is a function of t, so the correct way to enter the diffeq is diff(x(t),t)=x(t)^2. For your animation of equation 4 adapt David Clayworth's set up: e4 := diff(x(t), t) = x(t)^2+a^2; s4 := dsolve([e4, x(0)=1]); r4 := rhs(s4); animate(plot,[r4, t=0..0.75],a=0..1)

I think the answer to your question is explained by the following two or three lines. $ans=maple("{($a)... and inequality signs.

Nasos, The command implicitplot will work, but this might be better: >solve (V=1.186*10^3*exp(-9.478*10^(-3)*V) +749.579*0.455*exp(-t/0.245)+0.0375*exp(1/(4*t+1.2)), V); You get a solution for V involving LambertW, and then you can just say plot(%,t=0..10);

There is a "closed form" solution that appears as a finite sum. You can see it at this link: Medhurst JSTOR Mathematics of Computation, Vol. 19, No. 89 (Apr., 1965)

I could not figure out how to coax Maple into giving this result. If somebody could show the way, we would all stand to learn something.

Modify your proc as follows: MakeData := proc () local i; global ggg; ggg:=Array(1..5,1..2); for i to 5 do ggg[i, 1] := i; ggg[i, 2] := 2*i; end do; ggg; end proc; MakeData( ); So in the proc you set up ggg as a blank array, fill it, then return the filled array. With the proc MakeData as you wrote it, you can get a little action by MakeData( ); print(ggg); but you get a table instead of a list, and it might look peculiar. If you really insist on returning a list instead of an Array, modify your proc as follows: >MakeData := proc ( ) local i; global ggg; ggg:=[ ]; for i to 5 do ggg:=[op(ggg),[i,2*i]]; end do; ggg; end proc; >MakeData( ); Personally I would get rid of the do loop, and write the proc as >MakeData := proc ( ); [seq([i,2*i],i=1..5)]; end proc; MakeData( );

I see you are using document mode. I never use document mode. It always causes unexpected problems like this. Start a new "thing" in worksheet mode and you will not have the parsing problem.

I did just simply copy and paste from Maple 10 (classic and postmodern) to Mapleprimes, and I do not see what you mean by "not parsing." I used local i: global j: Colon vs. semi-colon seems to make no difference. Maybe "restart" will set things right. restart; proc1:=proc() local i: global j: i:=1; j:=2; end proc;

I think your third example does work properly. Look at this: >proc1:=proc( ) local i: global j: i:=1; j:=2; end proc; > proc1(); 2 > i; i >j; 2 So i was local to proc1 and j is global. You have to call the proc once to get j assigned to 2.