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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]

(1)

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

(2)

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

(3)

``

 

Download Maple_Query.mw

How to find the value of alpha and beta?

restart

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}

(1)

``

thank you for helping

Download Qu_3.mw

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

>restart;

>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.

1.

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

 

2.

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

 

 

Given*the*equations

  restartreset:

with(Student[LinearAlgebra])``

(1)
Coefficient Tanle

Equation 1

Equation 2

Equation 3

Equations

`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]])

(2)

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]])

(3)

``The equations in simplified form are:

3*x+y-z = 3

(4)

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

(5)

(231/26)*z = 231/26

(6)

``

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

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

(7)

 

``

 

Download GausianFinal15Nov2015.mwGausianFinal15Nov2015.mw

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?

whether

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

or

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. 

 

Equation_questions_.mw

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:

x-2y+z=0

2y-8z=8

-4x+5y+9z=-9

The intersection point is (29,16,3)

 

Thank you !

I want to solve these two differential equations. I have the initial conditions:
x(0)=0
y(0)=0
x'(0)=5.7
t'(0)=8.1
What am I doing wrong?

Hi

Anyone could help me in solving the following system of equations to get constants C1, C2, C3 and C4. MALPE give me this "soution may have been lost".  The MAPLE sheet is also attached.

 

restart:

Eq1:=simplify(C3*exp(-(1/4)*(C2*(x^2-2*0)+sqrt(C2*(x^2-2*0)^2+4*M*(x^2-2*0)*w1*(x^2-2*0)))/w1)+C4*exp((1/4)*(-C2*(x^2-2*0)+sqrt(C2*(x^2-2*0)^2+4*M*(x^2-2*0)*w1*(x^2-2*0)))/w1)-U) = 0;

C3*exp(-(1/4)*(C2*x^2+(x^4*(4*M*w1+C2))^(1/2))/w1)+C4*exp(-(1/4)*(C2*x^2-(x^4*(4*M*w1+C2))^(1/2))/w1)-U = 0

(1)

Eq2:=simplify(exp(-(1/4)*(C2+sqrt(C2^2+4*M*w1))*(x^2-2*0)/w1)*C3*x+exp((1/4)*(-C2+sqrt(C2^2+4*M*w1))*(x^2-2*0)/w1)*C4*x+C2-V-z) = 0;

exp(-(1/4)*(C2+(C2^2+4*M*w1)^(1/2))*x^2/w1)*C3*x+exp(-(1/4)*(C2-(C2^2+4*M*w1)^(1/2))*x^2/w1)*C4*x+C2-V-z = 0

(2)

Eq3:=simplify((-2*w2*w5*ln(C3*exp(-(1/2)*sqrt(w2*w4*(w2*w4+w3*w6))*C2*(x^2-2*0)/(w2*w4*w5))-C4)*sqrt(w2*w4*(w2*w4+w3*w6))+w2*w5*(-w2*w4+sqrt(w2*w4*(w2*w4+w3*w6)))*ln(exp(-(1/2)*sqrt(w2*w4*(w2*w4+w3*w6))*C2*(x^2-2*0)/(w2*w4*w5)))+C1*w3*w6*sqrt(w2*w4*(w2*w4+w3*w6)))/(sqrt(w2*w4*(w2*w4+w3*w6))*w3*w6)-1)= 0;

(-ln(exp(-(1/2)*(w2*w4*(w2*w4+w3*w6))^(1/2)*C2*x^2/(w2*w4*w5)))*w2^2*w4*w5+C1*w3*w6*(w2*w4*(w2*w4+w3*w6))^(1/2)-2*w2*w5*ln(C3*exp(-(1/2)*(w2*w4*(w2*w4+w3*w6))^(1/2)*C2*x^2/(w2*w4*w5))-C4)*(w2*w4*(w2*w4+w3*w6))^(1/2)+ln(exp(-(1/2)*(w2*w4*(w2*w4+w3*w6))^(1/2)*C2*x^2/(w2*w4*w5)))*(w2*w4*(w2*w4+w3*w6))^(1/2)*w2*w5-(w2*w4*(w2*w4+w3*w6))^(1/2)*w3*w6)/((w2*w4*(w2*w4+w3*w6))^(1/2)*w3*w6) = 0

(3)

Eq4:= simplify((-C2*x^2*w2*w4-.50*C2*x^2*w3*w6+sqrt(w2*w4*(w2*w4+w3*w6))*C2*x^2+2.*w2*w4*w5*ln(w3^4*w6^2*(C3^2*exp(-1.0*sqrt(w2*w4*(w2*w4+w3*w6))*C2*x^2/(w2*w4*w5))-2*C3*exp(-.5*sqrt(w2^2*w4^2+w2*w3*w4*w6)*C2*x^2/(w2*w4*w5))*C4+C4^2)/(w2*w4*(w2*w4+w3*w6)*C2^2))-5.544000000*w2*w4*w5-w3^2*w6)/(w3^2*w6)) = 0;

(-C2*x^2*w2*w4-.5000000000*C2*x^2*w3*w6+(w2*w4*(w2*w4+w3*w6))^(1/2)*C2*x^2+2.*w2*w4*w5*ln(w3^4*w6^2*(C3^2*exp(-(w2*w4*(w2*w4+w3*w6))^(1/2)*C2*x^2/(w2*w4*w5))-2.*C3*exp(-.5*(w2*w4*(w2*w4+w3*w6))^(1/2)*C2*x^2/(w2*w4*w5))*C4+C4^2)/(w2*w4*(w2*w4+w3*w6)*C2^2))-5.544000000*w2*w4*w5-w3^2*w6)/(w3^2*w6) = 0

(4)

solve({Eq1, Eq2, Eq3,Eq4}, {C1, C2, C3,C4});

Warning, solutions may have been lost

 

``

``

Download solution_lost.mw

 

 

hi guys, i have a complicated 4 set equations and i want to solve it,

one way is to find a solution for an equation and check it manually in other equations but i am looking for better method

eqq.mw

 

thanks

Hello everybody.

In the attached file, you find 6 equations. All of these parameters are known except "pd" and "qd". How can I find these two unknowns from 6 equations??? It should be pointed out that, "pd" and "qd" must contain "ud", "vd","wd" and "rd".

Thanks in advance.

D.mw

hi.I want to dsolve set of nonlinear equations with one unknown parameter ...is this possible with dsolve rule.in matlab this possible with bvp4c rule..please help me for this problem.if we should another rule please attached file reform.Thanks alot12.mw

I need to solve a set of equations but changing a constant each time.

For example, x+by=0, bx-y=10 where b=10,20.

I don't want to put it in a loop because, in loop, the equations are solved repeatedly. I want Maple to solve it only once and substitute b values automatically since I want to solve big set of equations faster.

Is there any option in Maple to do so with a single command

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