@nm

Thanks!

I got today the book  Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th ed. Richard Haberman + solutiom manual
Elementary Differential Equations and Boundary Value Problems, 10th Edition. William E. Boyce,200e Richard C. DiPrima do i have also

One book left to get
Looks promising the book from author Richard Haberma, because it starts not too technical and is written for education

Problem with this studymaterial( or other new ) is to understands concepts, therefore i do have a lot of studymaterial assembled.
Also a book here: : from J.kevorkian  PDE analytical solutions en hopefully i can make modest start with it too : it starts with the diffusion equation with a heated rod like your book from Haberman also starts with.  

@nm 

You seems to me a expert with PDE's. 
Do you have recommendations how to study them by myself or is this mission impossble ? 

 

@janhardo 

pdsolve can be used too to show that g(y,z) = h(z)  if  g[y] = 0 (partial derative)
Seems to me one of the most complex command in Maple if i see what the outcome can be ?

A scalar potential exists. Call it f(x,y,z). Then diff(f,x)=A. To find f, integrate A with respect to x, holding y and z constant. 

f := int(A, x) + g(y, z);
              f := y z x + exp(x) cos(y) + g(y, z)


Now diff(f,y) should be B. Obtain that derivative and compare to B.

diff(y*z*x + exp(x)*cos(y) + g(y, z), y) - B = 0;
                         d             
                        --- g(y, z) = 0
                         dy            



This says g(y,z) is independent of y, so g(y,z) = h(z), a function of z only. That makes f become

This result g(y,z) = h(z), could be done with pdsolve too
pdsolve(diff(y*z*x + exp(x)*cos(y) + g(y, z), y) = B);
                        g(y, z) = _F1(z)

@Preben Alsholm 

Thanks!

This illustrate your reasoning fine and also my geometrical idea of it, but it is more then a plane alone , because y can be any function.
That's what i ike with Maple to see directly a plot to support a idea
The plot shows that partiel directive of x =0. ( f[x]=0)   
For u = f(x,y,z) if one of the partial  directive =0 , you could plot this too then...no its beyond 3D
 

Its clear now with a function of 2 or 3 variables if one of the part directives = 0 in those functions , the remaining 2D or 3D functions are the ones with only those variables not mentioned as partial directive variable. (the same idea for integrating for the intergation constant )
This was the case with calculation made for a potential function for a vectorfield. 

Download een_part_afgeleide_is_0_voor_functie_van_twee_var.mw

@Preben Alsholm 
Thanks!
That's not that easy to follow your reasoning.
But is was thinking more on a geometrical explanation 
Suppose z =f(x,y)  and a partial derative to x and this derative = 0 
What kind of surface you get ? ..it must be a plane   parallel to the x-axis     

The part derative of y could be any number and if it is 0 then you can get a x-y plane or planes above or below  
Make this sense too ?
 

@Preben Alsholm 
Thanks!

You are right for Maple, it has already proven algorithms built in what represents a enormous amount of scientific knowledge. 
But with pdsolve it gives the result what i am looking for namely  h(z) !    

pdsolve(diff(g(y, z), y) = 0);
                        g(y, z) = _F1(z)

 The reasoning why pdsolve comes with this result h(z) , there must be some one?.i am curious what it is.

@vv 

Thanks 
Amazing procedure and if correct it shows de constants for integrating in 1 to 3 variables 
For one variable gives

F := exp(x);
f := IntWithConst(F, x);
                       f := exp(x) + _F()
But that is a number of course and for two and 3 variables it are functions as constants

The derive of a scalarfunction  out of a vectorfunction is for some points not clear 
But studying some vids makes it more clear !

See for ff : Scalar_Potential.mw

@janhardo 
Unfortanely could not follow complete the handling of the partial deratives  and integrating

In general integrating in 1,2 and 3 variables and their constants?
What says that one partial derative is zero in u=f(x,y,z) ?

Could this figure out in Maple symbolically ?..yes it can, but how
With pdsolve( two or more variables) and dsolve (one variable)

example

A := exp(x)*cos(y) + y*z

f := int(A, x) + g(y, z);
              f := y z x + exp(x) cos(y) + g(y, z)

The constant is here g(y,z) 

Now diff(f,y) should be B. Obtain that derivative and compare to B.

diff(y*z*x + exp(x)*cos(y) + g(y, z), y) - B = 0;
                         d             
                        --- g(y, z) = 0
                         dy            

This says g(y,z) is independent of y, so g(y,z) = h(z), a function of z only. That makes f become

ff := eval(f, g(y, z) = h(z));
               ff := y z x + exp(x) cos(y) + h(z)

@rlopez 

Thanks

Well, if you have reached all goals in life, then this person is a satisfied (enlighted) person i belief.
( maslow pyramid)

I noticed for your worksheet that it is maple 1d input in a 2D document and this is not how a regular student tackle his math
i think.
But i like the direct style, but for that is needed  a some more working knowledge of Maple commands     
 

@rlopez 
Thanks!

Interesting see another solving method and helpful.
I figured out how to do it in 2D mode too 
Only try to follow a bookexample and there is no curl  there mentioned
Picked up :  it is conservative field and those filelds are not rotating, so curl = 0 then    

That's a long time ago 50 years back, at that time i was a teenager of 16 years old

thomas_calculus-vb2-blz_1075_-vraag_forum_deel2.mw
   

@tomleslie 

Thanks!

With pdsolve you get the complete general function and with ScalarPotential you must add constant to the equation
Did not studied yet the outcome of a PDE to a serie of ODE 
But the idea is to this do it by first principles in 2D mode, like a student must do 
 

With the documentmode is it possible to get quick idea of a pattern, but beware.
For example i integrated  for f(x,y,z)= x+y+z with respect to x 
The pattern is that x ( whatever function x is ) is integrated and  y and  z are multiplied with only x  
x + y + z  -> int = 1/2*x^2 + y*x + z*x  
2*x + y + z-> int =x^2 + x*y + x*z
x*sin(x) + y + z -> int= y*x + z*x + sin(x) - x*cos(x)

That's brings the question can i see in documentmode to respect what variable is diferentiate/integrate , because there is no notation of it

You can do this for integration/differentation for the 3 variables x,y,z to get a idea how the final function is derived from the vectorfunction 

Or the idea is 
F is vectorfunction (given) => gradF is gradientfield  = dF/dx, dF/dy,dF/dz (round d) 
So integrate for the 3 partial deratives gives a scalar function back ( no need to look for pattern  for integrate/differentate)  

@9009134 

If you click on plot ( in document mode) , you get a context menu. There is a tickbox  with scaling constrained option too.
You can experiment with other plot options there. 

@vv 
Thanks 
For the good advice.
Understanding a math concept is the first step and then practicing by hand 
To do it on your own makes it even harder and probably for many math topics impossible without a teacher
But these day internet can be source of  studymaterial. 
Must be possible for me think as B Ed in math to study some calculus topics, if am taking the time for it of course.
 

@vv 

It is a sparringpartner Maple for understanding more  calculus, but it follows not always a textbook for integration for not adding a constant
 

@vv 

Thanks

There are two functions involved f(x) and f(y) for  z= f(x,y)  for partiel differentation
(If you consider it as two intersections curves from a surface as defining a partiele directive)
So then f[x](x,y) =0    ..y(x) is left over as constant

It s getting more complicated in another example with v(x,y,z) for a vectorfield:.has it a potential function?