Items tagged with equations

The method of solving underdetermined systems of equations, and universal method for calculating link mechanisms. It is based on the Draghilev’s method for solving systems of nonlinear equations. 
When calculating link mechanisms we can use geometrical relationships to produce their mathematical models without specifying the “input link”. The new method allows us to specify the “input link”, any link of mechanism.

Three-bar mechanism.  The system of equations linkages in this mechanism is as follows:

f1 := x1^2+(x2+1)^2+(x3-.5)^2-R^2;
f2 := x1-.5*x2+.5*x3;
f3 := (x1-x4)^2+(x2-x5)^2+(x3-x6)^2-19;
f4 := sin(x4)-x5;
f5 := sin(2*x4)-x6;

Coordinates green point x'i', i = 1..3, the coordinates of red point x'i', i = 4..6.
Set of x0'i', i = 1..6 searched arbitrarily, is the solution of the system of equations and is the initial point for the solution of the ODE system. The solution of ODE system is the solution of system of equations linkages for concrete assembly linkage.
Two texts of the program for one mechanism. In one case, the “input link” is the red-green, other case the “input link” is the green-blue.
After the calculation trajectories of points, we can always find the values of other variables, for example, the angles.
Animation displays the kinematics of the mechanism.

(if to use another color instead of color = "Niagara Dark Orchid", the version of Maple <17)



I have two equations (1) and (2) and i want to divide (2) with (3). A good point is that Maple understand division with equations. Nevertheless, I didn't obtain a simplified solution.

Here my code :

eq1:=-sin(alpha0(t))*cos(beta0(t)) = -sin(alpha[1](t))*cos(beta[1](t));
-sin(alpha0(t)) cos(beta0(t)) = -sin(alpha[1](t)) cos(beta[1](t))
eq2:=cos(alpha0(t))*cos(beta0(t)) = cos(alpha[1](t))*cos(beta[1](t));
cos(alpha0(t)) cos(beta0(t)) = cos(alpha[1](t)) cos(beta[1](t))

Here the result obtained :-sin(alpha0(t))/cos(alpha0(t)) = -sin(alpha[1](t))/cos(alpha[1](t))

Consequently, I would like to obtain tan(alpha0(t))=tan(alpha1(t))

Do you have ideas why I didn't obtain a simplified result ? And How can I obtain the solution with tangents ?

Thank you for your help

Hello dears! Hope everything going fine with you. I have faced problem while solving the system of equations using fsolve command please find the attacment and fixed my problem.

I am very thankful to you for this favour.

Mob #: 0086-13001903838

Dear Maple users

In order to solve a problem in Biology, I have come across four equations with three unknowns, making it overdetermined. The equation system does not have any specific solutions, but since the equations can be considered containing errors, I am looking after the best possible solution - which will minimize the coefficients in the original equation system. My question is twofold:

1. Does there exist a specific Mathematical Theory to handle non-linear equation systems of this type?

2. Which command in Maple can solve it?




Erik i can dsolve couple linear equations with power series solutions or taylor series expantion?

file attached below.



I would like to create a function from an equation.
I have define 4 functions u[1](t), ..., u[4](t).
From these 4 functions, I would like to define 4 equations. I would like to obtain this result:

My code is the following:

for i to 4
end do;

These lines are not working because the left member is not incremented (u[i](t) stays at each iteration u[i](t))

Generally, how can I transform a function to an equation ?
And, in this specific case, how can I obtain the four equations mentioned above ?

Thanks a lot for your help.



I try his code:


for i to 6 do u[i, j] := (u[i-1, j]+u[i+1, j]+lambda*(u[i, j-1]+u[i, j+1])+sigma(h, r[i+1])*(u[i+1, j]-u[i-1, j]))/mu end do;

Error, too many levels of recursion.

I want to the  results of the analytic equations, for example:

 u[1,j]= (u[0, j] + u[2, j] + lambda (u[1, j - 1] + u[1, j + 1])                                                       

      + sigma(h, r[2]) (u[2, j] - u[0, j]))/mu





Dear all
Please guide me how to convert system of expressions into system of equations, so as solve them using "solve command".

The following expressions are just coefficients extracted from certain equation.

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2], 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2]

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2], 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2]


It possible for me to write (1) in the following form

for EQ in 16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2], 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2] do EQ = 0 end do

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2 = 0


48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2] = 0


64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2] = 0


But I want to write (1) in the following form

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2 = 0, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2] = 0, 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2] = 0

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2 = 0, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2] = 0, 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2] = 0





How to find the value of alpha and beta?


u := proc (x) options operator, arrow; sin(x)+cos(x)-4*x+(alpha-4)*x end proc; v := proc (x) options operator, arrow; sin(x)-cos(x)+beta end proc; `assuming`([solve({alpha = int(u(t)+v(t), t = 0 .. Pi), beta = int(u(t)-v(t), t = 0 .. Pi)}, {alpha, beta})], [alpha <> 0, beta <> 0])

{alpha = 4+beta*Pi+(1/2)*alpha*Pi^2-4*Pi^2, beta = -beta*Pi+(1/2)*alpha*Pi^2-4*Pi^2}



thank you for helping


Hi. I want to solve a system of equations. But I got this type of error. 


>Digits := 15;
>with(plots):n:=0.7:Pr=1: mu(eta):=(diff(U(eta),eta)^(2)+diff(V(eta),eta)^(2))^((n-1)/(2)):
>Eqn1 := 2*U(eta)+(1-n)*eta*(diff(U(eta), eta))/(n+1)+diff(W(eta), eta) = 0:
>Eqn2 := U(eta)^2-(V(eta)+1)^2+(W(eta)+(1-n)*eta*U(eta)/(n+1))*(diff(U(eta), eta))-mu(eta)*(diff(U(eta), eta, eta))-(diff(U(eta), eta))*(diff(mu(eta), eta)) = 0:
>Eqn3 := 2*U(eta)*(V(eta)+1)+(W(eta)+(1-n)*eta*U(eta)/(n+1))*(diff(V(eta), eta))-mu(eta)*(diff(V(eta), eta, eta))-(diff(V(eta), eta))*(diff(mu(eta), eta)) = 0:
>Eqn4 := (W(eta)+(1-n)*eta*U(eta)/(n+1))*(diff(theta(eta), eta))-(mu(eta)*(diff(theta(eta), eta, eta))+(diff(mu(eta), eta))*(diff(theta(eta), eta)))/Pr = 0:
>bcs1 := U(0) = 0, V(0) = 0, W(0) = 0, theta(0) = 1:
>bcs2 := U(20) = 0, V(20) = -1, theta(20) = 0:
>R1 := dsolve({Eqn1, Eqn2, Eqn3, Eqn4, bcs1, bcs2}, {U(eta), V(eta), W(eta), theta(eta)}, initmesh = 20000, output = listprocedure, numeric);

Error, (in dsolve/numeric/bvp/convertsys) too few boundary conditions: expected 8, got 7

>for l from 0 by 2 to 20 do R1(l) end do;
>plot1 := odeplot(R1, [eta, theta(eta)], 0 .. 20, numpoints = 2000, color = red);


What is the problem actually because based on the paper that I refer to, there is only 7 bc. 

Can anyone help me?

Thankyou in advance.


assume there are 3 real number or floating number a, b, c

how to find the w and phi in 

a = sin(w*t1 + phi)

b = sin(w*t2 + phi)

c = sin(w*t3 + phi)

where w and phi are the same in above 3 equations, only different in t1, t2, t3



same as above , how to find w, apha and beta in below if there are a, b, c

a = alpha*cos(w*t) + beta*sin(w*t)

b = alpha*cos(w*t) + beta*sin(w*t)

c = alpha*cos(w*t) + beta*sin(w*t)

There have come unwanted lines and marks . I donot know how to remove them. Using doc.block, remove block seems to be little tough to incorporate! Please enlighten me. Modified doc. is most welcome. Thanks. Ramakrishnan V 

Gaussian Elimination Method






Coefficient Tanle

Equation 1

Equation 2

Equation 3


`m__1,1` := 3:

`m__2,1` := 2:

`m__3,1` := 1:

`m__1,1`*x__1+`m__1,2`*y+`m__1,3`*z = `m__1,4`; = 3*x__1+y-z = 3

`m__2,1`*x__1+`m__2,2`*y+`m__2,3`*z = `m__2,4`; = 2*x__1-8*y+z = -5

```m__3,1`*x__1+`m__3,2`*y+`m__3,3`*z = `m__3,4`; = x__1-2*y+9*z = 8

The equations in matrix form is given by

Matrix([[3, 1, -1, 3], [2, -8, 1, -5], [1, -2, 9, 8]])


The Gaussian Elimination gives the simplified natrix equation as given below:

Matrix([[3, 1, -1, 3], [0, -26/3, 5/3, -7], [0, 0, 231/26, 231/26]])


``The equations in simplified form are:

3*x+y-z = 3


-(26/3)*y+(5/3)*z = -7


(231/26)*z = 231/26



The aolution ia obtained by solving the above equations in reverse order

{x = 1, y = 1, z = 1}






a system of monomials, which has 5 equations for 5 variables , have more than one solutions,

first solution is my wanted solution,

how to eliminate other unwanted solutions?


1. add more equations to eliminate unwanted solutions? how to do?


2. edit existing system to eliminate unwanted solutions? how to do?

    a.  add extra terms to some equations?


what is the cause that make it having more than one solutions?

can this reason help to edit existing system?


i succeed with adding extra equation,a1+a2+a3+a4+a5-(6+s) =0  in 3 variables case, it calculate very fast within 1 second.

but when calculating 5 variables, it evaluating a very very long time, what is the problem


without extra equation a1+a2+a3+a4+a5+a6+a7-(1+2+3+4+5+s), it get result within 1 second, but after adding this extra equation, it is like dead loop,

my surface computer run with large fans noise and very hot.

Question regarding solving the equation systems. 

See maple file for equations.

Hello all

I am trying to write  a tutorial about systems of linear equations, and I want to demonstrate the idea that when you have a system of 3 euqtions with 3 unknowns, the solution is the intersection point between these planes. Plotting 3 planes in Maple 2015 is fairly easy (you plot one and just drag the others in), but I don't know how to plot the intersection point. Can you help please ?


My equations are:




The intersection point is (29,16,3)


Thank you !

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