I try to solve a set of differential equations with boundary conditions by dsolve command but I got this error:

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odes := D2*(diff(P(x), x))/((3*D1*a+4*D2)*P(x))-(diff(S(x), x))/(q*S(x)-1) = 0, diff(S(x), `$`(x, 2))+(diff(S(x), x))*cotx+4*pi*(3*D1*a+4*D1)*P(x)/((q*S(x)-1)*D2) = 0

ics := P((1/2)*pi) = 1, S((1/2)*pi) = -1, (D(S))((1/2)*pi) = 0, (D(P))((1/2)*pi) = 0

sol := dsolve({ics, odes}, numeric);
Error, (in dsolve/numeric/process_input) input system must be an ODE system, got independent variables {x, (1/2)*pi}

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Does any body knows what the problem is??

Hello

I wantt to assigne value to a variable that doesn't exist before the radio button. How can i do that?

Could you please help me?

Hey guys. I've tried to find an answer for this, but have struggled since our learning book is in danish, so the used terms may not be technically correct, so sorry for poor phrasing.

Anyway, how would you solve this problem in maple?

Find a solution for the differential equation:

d⁴y/dt⁴ - 16y = u' + u

With the effect*  u(t) = e^3t + 3e^t

I've gotten this result (By hand calculation)

y(t) = 4/65*e^3t - 6/15*e^t

Thanks for the help. It's my first post, so let me know if I should do something different next time :)

-Alex

Bonus question:

How do you calculate the transfer function in maple:

H(s) = (s + 1)/(s⁴-16)

*Don't know if 'effect' is the correct term. 

See subject. Via Google I only found links for converting from decimal to binary. But how do I write a binary number in the first place? I don't have a decimal to convert from, I have long binaries to start with! Which I don't want to convert to decimal, since I want binaries as output too.

Not sure exactly how i could achieve this but:

how do i determine the value of k for which the graphs p(x) = x^2+2x+3 and q(x) = k+5x-7x^2 enclose an area of exactly 36?

I have to do it in maple and using i guess area under the curve.

Thanks

I need to show that the least square solution x that I obtained as x=pseudoinverse(A).b is the solution of Transpose(A).A.x=Transpose(A).b with the smallest norm. I've obtained the norm for the RHS of this expression as well as the norm of x but I'm unsure of how to conclude that this is the lowest possible norm using these values

The issue I am currently having is that, while analyticity (and physics) indicates a certain function must have roots, fsolve is having trouble finding them. In fact, I have even found roots manually in a certain region myself, simply inputting into the function various values until I found them. However, fsolve does not seem to want to find these roots, and I believe it is a numerics issue: when I changed the digits around, for extremely low values of Digits, it would find a root (even though it was incorrect). Further, this problem arose elsewhere in the domain of interest for other values of Digits (in particular, for Digits:=5, fsolve failed in a region it had not failed before).

The region of interest is the "peak" of the output of poleR(M0, 0.935, mK), which should be somewhere around M0 = 0.95 or so. However, because fsolve cannot find the roots, the plot cannot be made.

Anyone have any ideas as to why fsolve cannot find the roots? I was also experiencing issues with some of these functions having multiple roots, which itself is weird as well (note that I am working over the complex plane).

Attached is the document.

pole-dragging-mapleprimes.mw

Any ideas?

Hi everybody!

It's nice to join in this forum.

I'm trying to get the analytic solution of the Bernouilli-Euler beam equation, with the next boundary conditions:

w(x,t) = displacements.

w(0,t) = 0   -> It's a cantilever beam. At the x=0 it's clamped.

diff(w(x,t),x) = 0.   -> the gyro in the clamp is zero.

E*I*diff(w(L,t),x,x) = 0  -> the moment at the end of the beam (x=L) is zero.

E*I*diff(w(L,t),x,x,x) = 0  -> the shear at the end of the beam (x=L) is zero too.

I'm not able to introduce the second and the third derivatives as boundary conditions in the pdsolve equation. I post the hole code:

restart;
ode := I*E*(diff(w(x, t), x, x, x, x))+m*(diff(w(x, t), t, t)) = 0;

s := pdsolve(ode, w(x, t));

ode1 := op([2, 1, 1], s);

ode2 := op([2, 1, 2], s);

f1 := op(4, rhs(ode1));

f2 := op(2, rhs(ode2));

sol1 := dsolve(ode1, f1);

sol2 := dsolve(ode2, f2);

sol := rhs(sol1)*rhs(sol2);

conds := [w(0, t) = 0, (D[1](w))(0, t) = 0, eval(I*E*(D[1, 1](w))(x, t), x = L) = 0, eval(I*E*(D[1, 1, 1](w))(x, t), x = L) = 0];

pde := [ode, conds];

pdsolve(pde, w(u, t));

And I get this error:

"Error, (in PDEtools/pdsolve) invalid input: `pdsolve/sys` expects its 1st argument, SYS, to be of type {list({`<>`, `=`, algebraic}), set({`<>`, `=`, algebraic})}, but received [I*E*(diff(diff(diff(diff(w(x, t), x), x), x), x))+m*(diff(diff(w(x, t), t), t)) = 0, [w(0, t) = 0, (D[1](w))(0, t) = 0, I*E*(D[1, 1](w))(L, t) = 0, I*E*(D[1, 1, 1](w))(L, t) = 0]]"

It's seems I'm introducing the Boundary conditions of the second and third derivatives in a wrong way, but I can't discover how to do it.

Thanks very much in advance to everybody!!

Ger89

P.D. - I have use this "tutorial" to write the code ( http://homepages.math.uic.edu/~jan/mcs494f02/Lec34/pde.html ). Thanks very much again. 

Hooke's joint and type of Hooke's joint.  


https://vk.com/doc242471809_363263648
https://vk.com/doc242471809_363264824

Type_of_Hooke.mw

Hi

What can i do to prevemt maple from resizing my drawings and inserting empty space beneath drawings?

It happens when i export to pdf or print the document out.

Is there any other way i can export to a readable format without this happening?

I've got a vector x=[0.36,1.3279,1.6882] (1*3) obtained as x:=pseudoInverse(A)*b where A=[<2,4,1>|<1,-1,1>|<3,3,2>] where <2,4,1>,<1,-1,1>,<3,3,2> are the columns of A and b is [8,5,4](1*3). Now, when I find the rowspace of A using RowSpace(A) I get the row vectors <1,0,1> and <0,1,1>, neither of which are equivalent to x. How do I arrive at my result that x is in the rowspace?

where A^T is the transpose of A and it's given that (A^T).A is not invertible.

I am stuck as to how to arrive at the solution for x in this case. I initially thought I could multiply both sides by the inverse of A^T reducing it to Ax=b but that was obviously wrong since A^T is itself not invertible(it is singular).

Hi,

I'm pretty new into MAPLE andI'm trying to get into it.

I have these four equations:

eq1:=1.6*10^(-7)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)-1.6*10^(-14)*R^2*cos(t)^2+4.2*10^(-14)*R^2-1.3+2.1*10^(-9)*R*cos(t)=0;

eq2 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.9*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0;

eq3 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.2*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0;

eq4 :=2.1*10^(-9)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)+1.6*10^(-14)*R^2*cos(t)^2+2.6*10^(-14)*R^2-1.3+1.6*10^(-7)*R*cos(t)=0;

I am a middle school teacher trying to program an app where a student will type in a linear equation in any form. They are to solve this for the variable Y and graph the line. I have most of this up and running, but I would like to include a step by step help system, where after entering the equation, the students can click on a button and it will display the 1st step to solve the equation. I am using the built-in function “LinearSolveSteps” in the Students [Basics] package. Every repeated click of the button would reveal the next step. I have this working to a point, but I don't seem to be able to add the next step into the math container without deleting the contents that are already in there. I have tried rebuilding the string by using the "+" but the + shows up in the equation and it is not on a new line. Any ideas would be appreciated. Thank you.

It seems a frequent issue that exported 3d plots are not shown as wished. I experience the same problem. Although I exported in the .eps format into a .tex latex-file the resulting .pdf-file shows a somewhat pixelated image of my 3d plot as if it was created in "Paint". Is there a solution for this in Maple13?