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Hello people in mapleprimes,

I want to solve the next system of equation for B/A and C/A.

eq1:=A+B=F+G;
eq2:=k*(A-B)=kappa*(F-G);
eq3:=F*exp(I*kappa*a)+G*exp(-I*kappa*a)=C*exp(I*k*a);
eq4:=kappa*F*exp(I*kappa*a)-kappa*G*exp(-I*kappa*a)=k*C*exp(I*k*a);


But, though it is well-known, solve({eq1,eq2,eq3,eq4},{B/A,C/A})
does not work well, as the values I want to solve it for are
expressions: B/A and C/A not variables.

Then, you might thing the next works well.
eq:=subs({B=A/t,C=A/u},{eq1,eq2,eq3,eq4}):
solve(eq,{t,u});

But, this doesn't work well, with the answer was
only the ratio of t and u expressed as the following:

t = t, u = exp(I*k*a)*(exp(-I*kappa*a)*k^2-exp(I*kappa*a)*k^2-exp(-I*kappa*a)*kappa^2+exp(I*kappa*a)*kappa^2)*t/(4*kappa*k*exp(I*kappa*a)*exp(-I*kappa*a))

Isn't there nice way to solve the above system of equation, except that
sol1:=solve({eq3,eq4},{F,G});assign(sol1);
sol2:=solve({eq1,eq2},{A,B});assign(sol2);

Best wishes
taro

hello, i went solve these equation ,with a & b take any value

b*x*ln(x)-x*ln(a)+a=0

thank you

I have a nonlinear system with 4 equations and 4 unknowns. I am using fsolve. I know that there are multiple solutions for each variable but am only getting one. I need the others. what do I do??

This is my code:

R__1 := Matrix([[1, 0] , [0, 1] ]);

R__2 := Matrix([[1/2, sqrt(3)/2] , [-sqrt(3)/2, 1/2] ]);

R__3 := Matrix([[-1/2, sqrt(3)/2] , [-sqrt(3)/2, -1/2] ]);

R__4 := Matrix([[-1, 0] , [0, -1] ]);

R__5 := Matrix([[-1/2, -sqrt(3)/2] , [sqrt(3)/2, -1/2] ]);

 

d__1 := Vector( [ 0, 5.4] );

d__2 := Vector( [ 6.4, 4.539] );

d__3 := Vector( [ 11, 4.078] );

d__4 := Vector( [ 15.5, 2.079] );

d__5 := Vector( [ 19, 1.039] );

 

a := Vector( [ a__x, a__y] );

 

A__1:=R__1.a+d__1;

A__2:=R__2.a+d__2;

A__3:=R__3.a+d__3;

A__4:=R__4.a+d__4;

A__5:=R__5.a+d__5;

 

OO:=Vector([O__x,O__y]);

 

DA1:=A__2.A__2-A__1.A__1-2*(A__2-A__1).OO;

DA2:=A__3.A__3-A__1.A__1-2*(A__3-A__1).OO;

DA3:=A__4.A__4-A__1.A__1-2*(A__4-A__1).OO;

DA4:=A__5.A__5-A__1.A__1-2*(A__5-A__1).OO;

 

fsolve({DA1,DA2,DA3,DA4},{a__x,a__y,O__x,O__y});

Thanks for any tips you may be able to offer

 

How to solve this inequality?...

November 16 2014 _zgj 80

solve(a/(x+1)>1,x)

When I type this command and run it,It returns nothing but this warning

"Warning, solutions may have been lost"

How can I get the correct solution using Maple 18

How to solve the inequality

with Maple?

My attempts were the following.

Warning, solutions may have been lost

Of course, this works

f(1);
,

but one wishes to describe the solutions in the dependence on the parameter a. Unfortunately, both

and

produce wrong outputs(An SCR has been submitted by me.).

 

 

 

 

 

 

hi..i am a problem with solving following ....please help me ....thanks alot

dsys3 := {10*f2(x)+12*(diff(f1(x), x))+14*f3(x) = 0, 2*(diff(f1(x), x, x))+4*(diff(f2(x), x))+6*(diff(f3(x), x)) = 0, 16*(diff(f3(x), x, x, x, x))+19*(diff(f3(x), x, x))+22*(diff(f1(x), x))+25*f2(x)+27*f3(x)+29*f3(x)+31+32 = 0, f1(0) = 0, f1(1) = 0, f2(0) = 0, f2(1) = 0, f3(0) = 0, f3(1) = 0, ((D@@1)(f1))(0) = 0, ((D@@1)(f1))(1) = 0, ((D@@1)(f2))(0) = 0, ((D@@1)(f2))(1) = 0, ((D@@1)(f3))(0) = 0, ((D@@1)(f3))(1) = 0}; dsol5 := dsolve(dsys3, 'maxmesh' = 500, numeric, range = 0 .. 1, abserr = .1, output = listprocedure); fy3 := eval(f3(x), dsol5); fy2 := eval(f2(x), dsol5); fy1 := eval(f1(x), dsol5)

ERROR.mw

Is it possible in Maple 15 to solve an equation with a parameter for a given set of parameters? How can this be passed to the solve function, should I use some kind of list?  After obtaining the solution how can I assign the solutions to variables such as x1 for the first value of the parameter, x2 for the second value of the parameters and so on. Furthermore, is this possible with the fsolve command?

 

Thanks

I am numerically solving a nonlinear system of nine equations. How long can I expect it to take?

I have run it for 30 minutes and it has not solved yet.

Here is the system of equations:

0=Lambda-mu.*S-beta.*(H+C+C1+C2).*(S./N)-tau.*(T+C).*(S./N);

0=tau.*(T+C).*(S./N)-beta.*(H+C+C1+C2).*(T./N)-(mu+mu_T).*T;

0=beta.*(H+C+C1+C2).*(S./N)-tau.*(T+C).*(H./N)-(mu+mu_A).*H;

0=beta.*(H+C+C1+C2).*(T./N)+tau.*(T+C).*(H./N)-(mu+psi.*mu_A+mu_T+lambda_T).*C;

0=lambda_T.*C-(mu+mu_A+rho_1+eta_1).*C1;

0=rho_1.*C1-(mu+mu_A+rho_2+eta_2).*C2;

0=eta_1.*C1-(mu+rho_1+gamma).*CT1;

0=eta_2.*C2-(mu+rho_2+gamma.*(rho_1)./(rho_1+rho_2)).*CT2+(rho_1).*CT1;

0 = N - S - T - H - C - C1 - C2 - CT1 - CT2;

and I have numeric values for Lambda, beta, tau, mu, mu_T, mu_A, rho_1, rho_2, psi, gamma. The only parameters left are eta_1, eta_2.

Thank you.

Hi,

 

So, I have the following problem. In my research, I am trying to prove that something is impossible. I do that by encoding it as a set of equations, and asking Maple to solve them, and if it finds that the system has no solutions, I got what I wanted. However, I really wanted to see Maple output this fact explictly, more or less like Mathematica (where I either get an error message, or I get "False" if I use the Reduce command). The reason for this is that I will embed this procedure in a loop that varies some of my starting assumptions, and I want Maple to output that the system has no solutions at each iteration. This is useful since each iteration takes more than half an hour to run, and there are dozens of them, so I want the program to provide me some intermediate feedback, rather than waiting for everything to finish to get an answer.

 

Is the question clear?

 

Thanks a lot for the help!

I'm trying to solve a series of equations and then graph them. I'm trying to solve for the variables involved:

values := solve({eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9,

eq10, eq11, eq12, eq13, eq14, eq15, eq16}, {a, b, b, c, d, e, f,

g, h, i, j, k, l, m, n, o, p});

 

. . . but it gives me this:

Warning, solving for expressions other than names or functions is not recommended. 

values := 

Archimedes supposedlly, was asked to determine whether a crown made for the king consisted of pure gold. According to 

legend, he solved this problem by weighing the crown first in air and then in water. Suppose the scale read 7.84 N when the 

crown was in the air and 6.84 N when it was in water.

 

What should Archimedes have told the king ?

 

hi.i am a problem with following dsolve.please help me....thanks alot

dsys3 := {10*f2(x)+12*(diff(f1(x), x))+14*f3(x) = 0, 2*(diff(f1(x), x, x))+4*(diff(f2(x), x))+6*(diff(f3(x), x)) = 0, 16*(diff(f3(x), x, x, x, x))+19*(diff(f3(x), x, x))+22*(diff(f1(x), x))+25*f2(x)+27*f3(x)+29*f3(x)+31+32 = 0, f1(0) = 0, f1(1) = 0, f2(0) = 0, f2(1) = 0, f3(0) = 0, f3(1) = 0, ((D@@1)(f1))(0) = 0, ((D@@1)(f1))(1) = 0, ((D@@1)(f2))(0) = 0, ((D@@1)(f2))(1) = 0, ((D@@1)(f3))(0) = 0, ((D@@1)(f3))(1) = 0}; dsol5 := dsolve(dsys3, 'maxmesh' = 500, numeric, range = 0 .. 1, abserr = .1, output = listprocedure); fy3 := eval(f3(x), dsol5); fy2 := eval(f2(x), dsol5); fy1 := eval(f1(x), dsol5)ERROR.mw

Hi there,

I'm doing my first steps with Maple and am a little confused with a problem:
I want to solve something like

and expect the result to be z=-m*a. However I get empty brackets as result.
I tried to dig a little and I find that maple has problems isolating x1+x2+x3+x4


Error, (in isolate) x1*a+x2*a+x3*a+x4*a does not contain x1+x2+x3+x4


 

When I isolate it myself Maple even tells me they wouldn't be the same


false

I even reduced the problem to x1*a+x2*a and Maple still has problems with that. As this is a really simple problem I strongly assume the issue lies with the user. Can anyone help me there?

Thanks

 

 

With your help I have a solution to a system of three equations:

(parameters are calculated on the basis of the data (for different values) - one example below)
A1=0.00002072968491, A2=0, A3=0.001946449287, A4=0.01946449287

B1=, B2=0, B3=0.0004773383613, B4=0.00004773383613

C1=, C2=0, C3=, C4=0.00009087604510

 

eqa1: = A1 * (diff (Tg (x), x, x)) + A2 * (diff (Tg (x), x)) + (A3 + A4) * tan (x) + A3 * Tg (x) + A4 * Tw (x) = 0;

eqa2: = B1 * (diff (Tw (x), x, x)) + B2 * (diff (Tw (x), x)) + (B3 + B4) * Tw (x) + B3 * Tg (x) + B4 * tan (x) = 0;

eqa3: = C1 * (diff (Tz (x), x, x)) + (C3 + C4) * Tg (x) + C3 * tan (x) + C4 * Tw (x) = 0;

 

indets ({eqa1, eqa2, eqa3}) minus {x};

res: = Dsolve (eval ({eqa1, eqa2, eqa3}) union {boundary conditions ??}, numeric);

 

for k from 0 to 20 evalf (res (k), 4); from;

c1:= 0.524:

c2:=0.05:

m: = 0;

for m from 0 to 20 and

T (m): = c1 * rhs (op (6, res (m))) + c2 * rhs (op (2, res (m))) + (1-c1-c2) * rhs (op (4, res (m))); print (m, T (m)); end to:

 

How and what type boundary conditions (I was thinking about the simplest or third type) to be able to determine the values on the y-axis on the graph. For example, the values started at -10, and ended at 10 (at a point (x, -10), (x, 10) in the coordinate system for a predetermined x, for example, from 0 to 20 which start at the point (0, -10 ) and stop at the point (20,10)). My main purpose is to collect these three solutions  to one equation T (x) = az * Tz (x) + and * Tw (x) + ag * Tg (x), and the ends of the graph, they should be in the above-mentioned points (0, -10 ) - start and (20,10) - stop.

 

Now thank you very much for the advice.

Ewa.

Hi there,

 

I'd like to solve 7th order implicit simultaneous equation such as below, so I tried to do it by solve command.

However the calculation wasn't over although three hours passed.

 

eq1 := f1(a,b,c,d,e,f,g) = 0;

eq2 := f2(a,b,c,d,e,f,g) = 0;

.

.

.

eq7 := f7(a,b,c,d,e,f,g) = 0;

 

Just for your information, the eq1 and eq6 are written as follows specifically.

eq1 := -a-b-c-d-e-f-g+0.501857 = 0

eq6 := a*b*c*d*e*f+a*b*c*d*e*g+a*b*c*d*f*g+a*b*c*e*f*g+a*b*d*e*f*g+a*c*d*e*f*g+b*c*d*e*f*g+a*b*c*d*e+a*b*c*d*f+a*b*c*d*g+a*b*c*e*f+a*b*c*e*g+a*b*c*f*g+a*b*d*e*f+a*b*d*e*g+a*b*d*f*g+a*b*e*f*g+a*c*d*e*f+a*c*d*e*g+a*c*d*f*g+a*c*e*f*g+a*d*e*f*g+b*c*d*e*f+b*c*d*e*g+b*c*d*f*g+b*c*e*f*g+b*d*e*f*g+c*d*e*f*g-0.5281141885e-3+1.01894577*10^(-12)*I = 0

 

And the program code I used is:

solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7],[a,b,c,d,e,f,g]);

 

Here is the specification of my computer.

OS: Windows 7 Enterprise 64bit

CPU: Intel Core i7-3520M 2.90 GHz

Memory: 4.00 GB

 

How can I handle this problem? Is the specification not enough to solve the equation? Do I need to leave my computer more and more time?

Any help would be appriciated.

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