Try this:

> factorielle:=proc(n)
       if(n=1) then 1; else factorielle(n-1)*n; end if;
   end proc:

>factorielle(10);

Good Luck,

Sebastián.

I don´t recommend use Dsolve, becouse if you use Dsolve, you will not know what is doing Maple.

Try this:

If you want more Accuracy, you can do more iterations:

Good Luck,

Sebastián.

You can use the Eval Command to solve your problem in the proc.

>restart:
 with(plots):
 yexa:=proc(X)
 local A,a,w,ode,ics,sol:
 A:=I:
 a:=50:
 w:=100:
 ode:=diff(y(x),x)=A*y(x)+a*sin(w*x):
 ics:=y(0)=1:
 sol:=t->eval(subs(dsolve({ode,ics},y(x)),y(x)),x=t):
 sol(X);
 end proc:
 
> evalf(yexa(2))

Good Luck.

Sebastián.

>ode:=diff(y(x),x,x)=2*y(x)+1:
>ics:=y(0)=1,D(y)(0)=0:
> f:=x->subs(dsolve({ode,ics},y(x)),y(x)):
> plot(f(x),x=1..5);
 

Good Luck,

Sebastián.

Very good idea. I will try it.

Thank you.

if you calculate the limit of the function when X tends to zero by left:

>limit(f(x),x=0,left);

8

And if you calculate the limit of the function when X tends to zero by right:

>limit(f(x),x=0,right);

∞

Hence, the limit of the function when X tends to zero doesn´t exist. For that, when you calculate, with maple, the limit of the function, your result is "undefined".

Moreover, f(0) is equal to 0, because you didn´t define, in the piecewise, a value to f(0).

Good Luck,

Sebastián.

You can use two commands to determine the function derived (D or diff):

The WorkSheet is a picture that I uploaded with the FileManager.

Good Luck

Sebastián.

If you don´t specify a value for f(a) when you define the function, Maple assumes that the value of f(a) is 0.

Good Luck,

Sebastián.

a)

>f:=x->1/(3*x+1):

>f´:=x->limit((f(x+h)-f(x))/h,h=0):
f´(x);

b)

>f´(1);

Good Luck.

Sebastián.

Thank you very much.

That's perfect.

Thank you very much

Using Rectangular Coordinates:

>evalf(int(int(int(1,z=-sqrt(4-x^2-y^2)..sqrt(4-x^2-y^2)),x=2-sqrt(1-y^2)..sqrt(4-y^2)),y=-sqrt(15)/4..sqrt(15)/4));

Using Cilyndrical Coordinates:

>evalf(int(int(int(r,z=-sqrt(4-r^2)..sqrt(4-r^2)),r=2*cos(theta)-(4*cos(theta)^2-3)^(1/2)..2),theta=-arctan(sqrt(15)/7)..arctan(sqrt(15)/7)));

The easiest way is transform the nonlinear equation to a linear equation around the critics points.

Our nonlinear equation is y´´+(y´)^5+y=0 (1). We can transform (1) to a system of differential equations:

y´=u
u´=-y-u^5

The critics points are the points that meet with the following equations:

f(y,u)=y´=u=0
g(y,u)=u´=-y-u^5

Hence, the only critic point is (0,0).

The Diferential Equations System around (0,0) is:

| y´ |      |  (df/dy)(0,0)       (df/du)(0,0)  |     | y  |
|     | =   |                                                  |  * |     |
| u´ |      | (dg/dy)(0,0)      (dg/du)(0,0) |     | u |

| y´ |      |  0    1  |     | y  |
|     | =   |            |  * |     |
| u´ |      |  -1   0 |     | u |

k1*sin(x) - k2*sin(2*x)=cos(x-Pi/2)*(k1-2*k2*cos(x))