The general solution of x²y'' + 3xy' + λy = 0 is (from Example 4 here)

Why does dsolve(x^2*diff(y(x), x, x) + 3*x*diff(y(x), x) + lambda*y(x) = 0) give only the third case instead of the whole solution?

How to find all solutions [x_i,y_i] to the following system of equations  

x³y=1

y-sin(12x)=0

which satisfies condition 1,2<x<2  

I want to show the animation of the polar equation r=cos(2theta) be plotted from theta=0 to 2pi

Hello,The system of equations is as follows：

I'd like to find all integer solutions in [1,20], when I use isolve ,the results are not good.  Since at least one variable is equal to 0 in evey solution.

 In maple I did not  want to use less efficient for-loop like C programing as following.

Does Maple have more good ways to solve that?

Hello

I had to save previous results of a calculation to files as the number of elements is too big for my computer to handle in one go.   Here it is an idea of what I am doing to read the files and perform the calculations.

mainproc:=proc(arg1,arg2,arg3,arg4) ... end proc:

Grid:-Set(mainproc):
Grid:-Set(arg2,arg3,arg4):   #  They don't change ever.

for i from 1 to number_of_files do 
   read(...):  # it reads arg1 from a file
   Grid:-Set(arg1):
   ans:=Grid:-Seq(mainproc(arg1[i],arg2,arg3,arg4),i=1..numelems(arg1))):
   Grid:-Wait():
   save ans, ....:
   unassign('arg1'):
   unassign('ans'):
   gc():  # An attempt
end do:

The actual code works but, for every step in the loop, the memory used by Maple increases by a certain amount that seems to be mostly related to arg1 (as if arg1 is piling up from iteration to iteration). 

I read some of the earlier posts on a similar subject dated 5 to 10 years old.  I wonder if there is something new that can be done to minimize the usage of memory.   

Many thanks

Ed

PS. I am aware of tasksize, numcpus and Threads.   

I have a simple matrix, 11 rows x 3 columns, with header row.

I multiply column 1 and 2 to get

Now, I just want to add up these elements, but 'sum' doesn't work 'Sum' doesn't work either. Is there a simpel way to do this?
 

Download Untitled_(3).mw

I must do some formula manipulation
 

Download vraag_herleiding_conic_sections_formule.mw

Hello everybody,

i am trying to use PDEchangecoords to transform a system of differential equations from Cartesian coordinates to toroidal coordinates. However, when i use a user defined coordinate transform from toroidal to Cartesian, i don't get the initial equations. Please find attached a minimal working environment.

i would highly appreciate your hints and suggestions!

Thank you

Best regards,

F

question.mw
 

Download question.mw

Dear Community,

I would like to have an inverse interpolation with a 2D lookup table in MapleSim. The usual 2D lookup interpolation has u1 column values, and u2 row values, and a corresponding   y   table values, i.e. y = f(u1,u2). Now I would like to go the reverse way. Suppose I know y, and u2, and I would like to get u1. How can I implement this in MapleSim? An additional information is, that the 3D surface represented by the f(u1, u2) function is a smooth, slightly curved, monotonically sloping surface, so to any  y  value a unique pair of u1 and u2 values belong.

With my specific data, which are attached in the TestVLP.xlsx file:

• u2 values are: 1^st row, cells 2 .. 14
• u1 values are  1^st column, rows 2 .. 11
• tabulated values are all the rest, i.e. from cell(2,2) .. cell(11,14)

Now for my case:

• u2 = 42000.0
• y   = 94.5614

How much is u1? I know the correct answer from elsewhere, it should be something very close to u1 ~ 85. This  y  value is in the rectangle bordered by columns 7 .. 8  and rows 5 .. 6 in the Excel file.

So my question is how do I correctly implement this in MapleSim? As visible from the attached TestInverse2DInterpolation.msim file I tried to do it with an Inverse Block Constraints component and a horizontally flipped 2D Lookup Table (VLP) and Constants for y and u2, but upon running MapleSim I get an error message that no solution is found, although the solution is relatively easy and straightforward. May I ask you to have a look at it, what can be the problem here, what do I do wrong? I’m using MapleSim 2019.2.

Your kind help is appreciated in advance

best regards

Andras

TestInverse2DInterpolation.msim

TestVLP.xlsx

Hello Everyone!

I have one more challenge for you.

How can I find for a Free-Pinned-Pinned-Free (3-span) beam (Picture A below) using the Krylov–Duncan Method (Literature links and references below):

- the matrix of the system?

- the transcendental equation in order to determine the natural frequencies?

- the first three mode shapes?

I tried to do it as you can see from my MAPLE (file below), but I got stuck when I use the command "determinant" and it did not find the transcendental equation.

Krylov_function_free_pinned_pinned_free_beam.mw

Picture_A

References:

Krylov–Duncan Method
https://link.springer.com/content/pdf/10.1007%2F978-1-4419-1047-9_14.pdf
 

Krylov–Duncan Functions - page 96

You can find that book using the https://libgen.is/
https://www.amazon.com/Formulas-Structural-Dynamics-Tables-Solutions/dp/0071367128

Hello.

I would like to solve numerically highly nonlinear and cumbersome the second order differential equation. 

Applying the numerical procedure I got an error "Error, (in dsolve) found wrong extra argument(s): range = 0 .. 4*Pi, type = numerical". The similar problem has been described earlier here however I can't realize my problem. 

Below is my code

restart;

A1 := 8*Pi^3*R^2*n(x)^4*m+(2*Pi*sin((1/2)*x)*m*omega0*p+Pi*sin((1/2)*x)*m*omega0+3*Pi^2*(diff(n(x), x, x)))*n(x)^3+(-2*sin((1/2)*x)^2*m^2*omega0^2*p^2+2*cos((1/2)*x)^2*m^2*omega0^2*p^2-2*sin((1/2)*x)^2*m^2*omega0^2*p+2*cos((1/2)*x)^2*m^2*omega0^2*p)*n(x)^2+(-4*(diff(n(x), x, x))*sin((1/2)*x)^2*m^2*omega0^2*p^2-8*sin((1/2)*x)*(diff(n(x), x))*cos((1/2)*x)*m^2*omega0^2*p^2-4*(diff(n(x), x, x))*sin((1/2)*x)^2*m^2*omega0^2*p-8*sin((1/2)*x)*(diff(n(x), x))*cos((1/2)*x)*m^2*omega0^2*p)*n(x)+8*sin((1/2)*x)^2*(diff(n(x), x))^2*m^2*omega0^2*p^2+8*sin((1/2)*x)^2*(diff(n(x), x))^2*m^2*omega0^2*p;

R := 1; m := 1; p := 10; omega0 := 1000;

A2 := A1;

with(plots):
A3 := dsolve({A2, n(0) = n(4*Pi), (D(n))(0) = (D(n))(4*Pi)}, type = numerical, range = 0 .. 4*Pi):

odeplot(A3);

I appreciate for any help and suggestion.

Help required to plot the real part only in  multiple plots. i have written some codes  please help me to rectify the errors.Thanks in advance

restart:
with(plots):
n:=0.75:Eh:=100:mn:=1:t0:=0.2:
a1:=(mn/t0)^((n-1)/(n))*(tb)^(1/n):a3:=Eh/tb:a4:=(Eh)^2:
U1:=(a1/Eh)*(-1+(a3*r/Eh))^(1/n)*(n*a4/(1+n))*(1/a3-(r/Eh))-a1*(-1+(1/tb))^(1/n)*(n/(n+1)*(tb-1)):
plot([Re(seq(eval(U1,tb=j),j in[0.8,0.9,1.0]))],r=0..1,legend = [tb =0.8, tb=0.9,tb =1.0],  labels = ["z ", "U"], labeldirections = ["horizontal", "vertical"],  linestyle = [solid,dash,dot],color = [black, red,green]);

I want to shade the area where the two polar curves overlap. The first curve is r=2 and the other curve is r=2(1-cos(theta)). How do I do this?

I want to find the surface area of this parametric curve revolving around the x-axis. I was able to plot the 2D rendering but I want to show the plot so that I can see the surface area. I suspect that would be the 3D rendering.

x=cos(t), y=2+sin(t), 0<=t<=2pi, x-axis