@acer Sorry, I mixed them,thank you for point that

@acer thank you, I think that's the point.

hybrid symbolic and numeric algorithm is a big chance for maple.

would you like to show at least one example with every enhancement?

for example, Rendering of 2-D plots and then follow a plot, that will be more intuitive.

and Math engine: Laplace transforms, follow a comparison between the old and new.

that's better, I think.

@Preben Alsholm 

Dear friends, the original is here

http://www.wolfram.com/mathematica/new-in-9/advanced-hybrid-and-differential-algebraic-equations/friction-models.html

I want  resolve the last model, the first three were done in

http://www.mapleprimes.com/questions/145201-Ode-With-Events.

any idea is appreciated.

thank you  in advance.

@Carl Love 

The problem is range=0..8.

My computer can't give the solution beyond t=1.001885

@Preben Alsholm 

oh, I got it.

thank you very much for your quick reply and big help.

@Preben Alsholm 

thank you very much, my friend.

you really give me a big help.

By the way, I found a bug in piecewise command.

restart;
F1 := piecewise(0 <= t, 0, 2 <= t, t^3, 2.1 <= t, exp(t));
plot(F1,t=0..3,thickness=2,numpoints=500);
F2:=piecewise(0 <= t and t<2, 0, 2 <= t and t<2.1, t^3, 2.1 <= t, exp(t));
plot(F2,t=0..3,thickness=2,numpoints=500);

the two figs are different.

It seems like the unilateral constraint in F1 does have some problems,

the bilateral constraint in F2 does have the right answer.

i7, win7, Maple 12

@Preben Alsholm 

Thank you, that's exact I want.

the physical process is that a mass point is pushed in positive direction with force F1, when x(tf)=30, another force F2 is taking effect. It means F2 triggered by x(tf)=30.

I just split the whole process into two equations.

1.Is it possible to use dsolve with one equation ?

F3:=piecewise(t>=0,0,t>=tf,y(t)^3,t>=tf+0.1,exp(y(t)));#y(tf)=30;

eq:=diff(y(t),t$2)=subs(x=y,F1)-F3; 

ic:=y(0)=1.2,D(y)(0)=0;

dsolve({eq,ic},numeric)??

2. And, if there is not only one point meet x(tf)=30 and i want choose the second tf, how to resolve it?

I'm very appreciated for your methods.

thank you in advance.

@Mac Dude 

Sorry, I missed something.

1. eq1 and eq2 is connected by tf where x(tf)=30;

In fact, eq1 and eq2 can be combined in one;

F3:=piecewise(t>=0,0,t>=tf,y(t)^3,t>=tf+0.1,exp(y(t)));

eq:=diff(y(t),t$2)=subs(x=y,F1)-F3; 

ic:=y(0)=1.2,D(y)(0)=0;

but when using dsolve solve this equation, I got some problem, so I split it into two equations.

2.MM(tf)=x(tf),NN(tf)=D(x)(tf). 

if you have some ideas to share, I'm very appreciated.

@Carl Love 

thanks for your attention.

I don't know much about the initial guess,

yes,they're all real,

use the command fsolve({f1,f2,f3},{x=0,y=0,z=0}) in maple can't give an answer.

how to resolve it with maple

although this is an old topic which was published about four years ago, I think Maple has a big step on it.

It runs faster than it has been.

Standing on today's view, maple lacks some longer planning for its development.

Now mma 10 contains fem package, which was suggested by Art Kalb in 2008

http://www.mapleprimes.com/posts/39372-Finite-Element-Analysis#

Combining with system modeler and large scale computation performance, mma will have a broader market than ever before.

I wish maple go faster and have a long planning, otherwise it will like today's mathcad.

Thank you for your help, my friends.
I searched the Maple help.
At the page below, I found the property MutuallyExclusive,it's a keyword.
So, would someone like to show an example that use this property in Optimization package or DirectSearch?
I appreciated for that, thank you in advance.
http://www.maplesoft.com/support/help/Maple/view.aspx?path=property&term=MutuallyExclusive

@Markiyan Hirnyk 

The results I got from fortran is below.
The numbers in the brackets under "residual" is the residual by insert the root, the two numbers is realpart and imagpart seperately.
The numbers in the brackets under "root" is the point near the exact root(I got it by mesh point and muller method), and the same to the residual, first number means realpart, second number means imagpart(because I made a transformation so it's minus).
there are two roots obviously.

thank you very much for your help, Markiyan Hirnyk.

still want to know how to solve this problem by Maple.
Any advice is apprieciated.

@Markiyan Hirnyk 

The results I got from fortran is below.
The numbers in the brackets under "residual" is the residual by insert the root, the two numbers is realpart and imagpart seperately.
The numbers in the brackets under "root" is the point near the exact root(I got it by mesh point and muller method), and the same to the residual, first number means realpart, second number means imagpart(because I made a transformation so it's minus).
there are two roots obviously.

thank you very much for your help, Markiyan Hirnyk.

still want to know how to solve this problem by Maple.
Any advice is apprieciated.

@Markiyan Hirnyk 

I got two point coordinates with fortran,
d=400e-10,betar=2.13972055888224,betai=2.90030000251869e-3
d=400e-10,betar=2.43413173652695,betai=1.89019000247816e-2

thank you for your comments,Markiyan Hirnyk.

still need method with maple.