THis ode looks complicated

ode := (2*x^(5/2) - 3*y(x)^(5/3))/(2*x^(5/2)*y(x)^(2/3)) + ((-2*x^(5/2) + 3*y(x)^(5/3))*diff(y(x), x))/(3*x^(3/2)*y(x)^(5/3)) = 0;

But is actually a simple first order linear ode:

RHS:=solve(ode,diff(y(x),x));
new_ode:=diff(y(x),x)=RHS;

Whose solution is 

But Maple gives this very complicated answer as shown below. When asking it to solve as linear ode, it now gives the much simpler solution.  

Maple complicated solutions are all verified OK. But the question is, why did it not give this simple solution?

Attached worksheet.  All on Maple 2024

Download why_missed_simple_solution_march_17_2024.mw

I am wondering why Maple simplifies (x^(1/3))^3 to x ,  but not (x^3)^(1/3) .
I even tried the surd function. I believe the surd function is for real number arguments, so it should simplify to x.

Download inverse1.mw

Please, Help me to fix this error..

Download Grafik_R0kurang1.mw

I cannot figure out which operand(?) is substituded here

subs(1 = 2, a*b);
                              2  2
                             a  b 

Same for

subs(1 = 3, a + b);
                           3 a + 3 b

but

subs(1 = 2, a/b);
                                2
                               a 
                               --
                               b 

subs(1 = 3, a - b);
                            3 a - b

Is this by design?

In Maple 2022

restart;

res := t^3 - 3*t^2*sqrt(t^2*(12*sqrt(2)*ln(t) + 9*ln(t)^2 + 8)^(1/3))*ln(t)/(12*sqrt(2)*ln(t) + 9*ln(t)^2 + 8)^(2/3) - 2*t^2*sqrt(t^2*(12*sqrt(2)*ln(t) + 9*ln(t)^2 + 8)^(1/3))*sqrt(2)/(12*sqrt(2)*ln(t) + 9*ln(t)^2 + 8)^(2/3);
plot(res,t=-5..1)

gives

The same exact code in Maple 2024 gives

Worksheet is below. 

Both on same PC. Windows 10.

Will report to Maplesoft, but thought to check also here is others have seen such problem before.

Btw, Maple 2022 plot is the correct one.

Download wrong_plot_V_2024_march_15_2024.mw

Maple's coulditbe  is useful. But unfortunately it does not return back to the user the conditions under which the proposition was found true. This could make it much more useful. It seems in way similar to Mathematica' Reduce but Reduce returns the conditions.

Is there a way to find the conditions which makes it true? 

I use coulditbe alot. I use it to verify that the result of odetest (I call it the residue) is zero or not. Maytimes, odetest does not return zero. And using simplify, or evalb or is to check if the residue is zero, all fail. But many times, coulditbe returns true, meaning the residue is zero. But I do not know under what conditions. In Mathematica's Reduce, it tells me the conditions. 

Here is one of hundreds of examples I have

restart;
ode:=(t^3+y(t)^2*sqrt(t^2+y(t)^2))-(t*y(t)*sqrt(t^2+y(t)^2))*diff(y(t),t)=0;
ic:=y(1)=1;
sol:=dsolve([ode,ic]);
the_residue:=odetest(sol,[ode,ic]);

You see, odetest says it could not verify the solution (the first entry above) but it did verify the solution against the initial conditions. 

Using simplify, evalb and is all also could not verify it

simplify(the_residue[1]);
evalb(the_residue[1]=0);
is(the_residue[1]=0);

Now coulditbe does:

_EnvTry:='hard':
coulditbe(the_residue[1]=0);

So the solution is correct, but I do not know under what conditions. Using Mathematica's Reduce I can find this:

So now back in Maple, I can do this

simplify(the_residue[1]) assuming t>exp(-2*sqrt(2)/3);

                      0

Actually in this example, just using assume t>0 also gives zero. But I am using Mathematica's result for illustration.

You might ask, why do I need to know for what values of the independent variable is the residue zero?

Because in some cases, the residue is zero only at single point! So it does not make sense to say the solution is verified to be correct only at one single point of the domain, right?

it needs to be some finite range at least. Here is an example of an ode whose solution is correct only at x=0

ode:=diff(y(x),x)=3*x*(y(x)-1)^(1/3);
ic:=y(3)=-7;
sol:=dsolve([ode,ic]);
the_residue:=odetest(sol,[ode,ic]);

And simplify, evalb, is all fail to verifiy this, but coulditbe says true

simplify(the_residue[1]);
evalb(the_residue[1]=0);
is(the_residue[1]=0);
_EnvTry:='hard':
coulditbe(the_residue[1]=0);

So now, we ask, is this solution then correct or not? It turns out to be zero but only at origin x=0

plot(abs(the_residue[1]),x=-1..1)

If I knew that residue is zero only at single point, then I would say this solution is not correct, right?

And that is why I need to know under what conditions coulditbe retruned true.

I tried infolevel[coulditbe]:=5 but nothing more was displayed on the screen.

Mathematica's Reduce confirms that when x=0 the residue is zero.

So my question is simply this: Can one obtain the conditions used by coulditbe to determine when result is true?

It will be useful if Maple could in future version return the value/range which makes it true.

Hope this is a simple question to answer but how do you use units and show 1 of something without losing the 1.  For example, it's easy enough to show 3 hours as 3h but nobody writes 1 hour as "h".  It's just silly.  Is there a way for 1 hour to render as "1h"?  Really the question is can you get 1 of any unit to show as "1 unit" and not just "unit"? Thanks.

Hi!

I am trying to implement the algorithms given in this paper (free for download) in Maple 2015

https://www.researchgate.net/publication/374636058_A_simple_algorithm_for_computing_a_multi-dimensional_the_Sierpinski_space-filling_curve_generalization

Such algorithms, apparently very easy, provides an approximation of  a sapce-filling curve and its pseudo-inverse. I am interesting in this space--filling curve for its properties. Please, find attached the Maple file, I am not sure if the code of the paper is not fine or I am doing something wrong. 

Many thanks un advsnce for your help

Sierp_v1.mw

p1>0, p2<0 and x[0] are arbitrary constants. How can i solve this integral? 

   I want to see    

thanks in advance. 

Hi,

First of all, I'm beginner in Maple. 
I want to plot  "M" versus "kc" and "deltah" as a 3D curve (M is real and positive). Intervals of kc and deltah are  given (see attached file). These three parameters (i.e, M,kc,deltah) should satisfiy the following equation:
Eq=-(1-(deltab+deltah))-(1-(deltab+deltah))/(kc-3/2)-deltah/sigmah-deltah/(sigmah*(kh-3/2))-(1/6)*deltab*sqrt(3)*mu*(1/sqrt((M-u0b+sqrt(3)*sqrt(mu*sigmab))**2)+1/sqrt((M-u0b-sqrt(3)*sqrt(mu*sigmab))**2))/sqrt(mu*sigmab)-(1/6)*sqrt(3)*(1/sqrt((M+sqrt(3)*sqrt(sigmai))**2)-1/sqrt((M-sqrt(3)*sqrt(sigmai))**2))/sqrt(sigmai)=0.

Other parameters are as follows:
 kh=3, sigmah=4,sigmab=1,u0b=0.05,sigmai=1/300,deltab=1e-6,mu=1836.
3D.mw

How to solve and plot a ODE system in RK method.
eq1 := diff(f(x), x, x, x)-(1/2)*Sc*sin(alpha)*g(x)*(diff(g(x), x, x))+(1/2)*x*cos(alpha)*(diff(f(x), x, x))+(1/2)*sin(alpha)*f(x)*(diff(f(x), x, x)) = 0; eq2 := (diff(g(x), x, x, x))/Pm+(1/2)*x*cos(alpha)*(diff(g(x), x, x))+sin(alpha)*f(x)*(diff(g(x), x, x))-sin(alpha)*(diff(f(x), x, x))*g(x) = 0; eq3 := (diff(theta(x), x, x))/Pr+(1/2)*x*cos(alpha)*(diff(theta(x), x))+(1/2)*x*(diff(f(x), x))*(diff(theta(x), x))+sin(alpha)*(x*(diff(f(x), x))-f(x))*(diff(theta(x), x))-Nb*(diff(s(x), x))*(diff(theta(x), x))-Nt*(diff(theta(x), x))^2+(1/4)*Sc*Br*sin(alpha)^2*(diff(f(x), x))^2*(x*(diff(g(x), x))-g(x))+(diff(g(x), x))^2*(x*(diff(f(x), x))-f(x)) = 0; eq4 := diff(s(x), x, x)+S*((1/2)*cos(alpha)*x*(diff(s(x), x))+(1/2)*sin(alpha)*f(x)*(diff(s(x), x)))+Nt*(diff(theta(x), x, x))/Nb = 0

ics := f(0) = 0, (D(f))(0) = 1, g(0) = 0, (D(g))(0) = 1, theta(0) = 1, s(0) = 1; bcs := (D(f))(100) = 0, (D(g))(100) = 0, theta(100) = 0, s(100) = 0

alpha = - 30 degree, Sc = 1.0, Pm = .1, Pr = 6.2, Nb = .1, Nt = .1, Br = .5, S = 1

Hi

I would like to create/import a list from a text-file. The text-file "example.txt" has the following structur:

1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 12
11 14

On every line are two numbers, separated by a space. The list should contain only the second number, in this example

L:=[1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 14].

The original text-file has 12615 lines, the biggest number is around  3.126 * 10^139.

Thanks for your help.

Was working with some antenna design equations and noticed that in a certain plot definition, the addition of gridlines changes the shape of the plotted curve.  Is this a bug or is there something obvious here that I am overlooking?  This one really has me scratching my head.  The only difference between the two plots below is the addition of the gridline spec.

Many thanks!

gridlines_change_plot.mw

Hello everyone,

I'm currently exploring the properties of a sum in Maple, and I was wondering how to obtain the values of the following sum:

sum 1/(i*j*k*) such that 1<=i+j+k<= n 

I'm interested in the values of this sum for a given n. How can I express this efficiently in Maple?

Thanks in advance for your help!

Hi, I have an homework where I need to find the highest point and the lowest point on an ellipse form by the intersection of two equations wich are 4x-3y+8z=5 and z^2=x^2+y^2 and I have to use the LagrangeMultiplier command. I get how it works but I can't get the correct form. How should I do it ?