I was using Matlab R2014a earlier where I was able to solve this equation easily, but in Matlab R2018b it is returning the
error in sol = solve(eqn,a,[0 pi]); I couldn't figure out why. The working code in Matlab 2014a is
syms a T
v2=-2.3750
g=1;
b=0;
e2=0.5;
k=2.5;
w=-2*cos(k);
eqn = sin(3*k+a)./sin(2*k+a)==v2-w+(g.*T.^2)+(e2.*T.^2.*sin(k)^2)./(sin(2*k+a)^2+b*T.^2*sin(k).^2);
sol = solve(eqn,a,[0 pi]);
digits(5)
solutions = vpa(subs(sol),3)
Please note that "a" is to be bound to take values between 0 and pi.

3 Kommentare

Image Analyst
Image Analyst am 28 Mär. 2019
What is the exact error (ALL the red text)? Have you tried it with R2019a?
madhan ravi
madhan ravi am 28 Mär. 2019
Only you can specify the range in vpasolve() where only one parameter is not known but with solve you can't specify the bounds.
John Jarvis
John Jarvis am 28 Mär. 2019
Bearbeitet: John Jarvis am 28 Mär. 2019
@Image Analyst:
The exact errors are as follows (all the red text)
Error in solve>getEqns (line 429)
[eqns, vars] = sym.getEqnsVars(argv{:});
Error in solve (line 226)
[eqns,vars,options] = getEqns(varargin{:});
Error in SolutionsPlot (line 11)
sol = solve(eqn,a,[0 pi]);

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 Akzeptierte Antwort

Star Strider
Star Strider am 28 Mär. 2019

1 Stimme

Restrict ‘a’ using an assume call.
Try this:
syms a T
assume(a >= 0 & a <= pi)
v2=-2.3750
g=1;
b=0;
e2=0.5;
k=2.5;
w=-2*cos(k);
eqn = sin(3*k+a)./sin(2*k+a)==v2-w+(g.*T.^2)+(e2.*T.^2.*sin(k)^2)./(sin(2*k+a)^2+b*T.^2*sin(k).^2);
[sol,prms,conds] = solve(eqn,a, 'ReturnConditions',true)
digits(5)
solutions = vpa(subs(sol),3)
vpaconds = vpa(conds)

4 Kommentare

John Jarvis
John Jarvis am 28 Mär. 2019
@Star Strider Thanks. Is there a way that I can get those nice numeric solutions which I used to get with the same code in R2014a? These conditions make it hard to explicityly see the numerical solutions.
My pleasure.
The ‘conds’ result is likely as good as it gets. Note that it is a function of ‘T’, and with the inequality conditions stated in the vector, neither coeffs or any related function (such as numden or sym2poly) will produce a purely numeric result.
Adding:
Tsol(:,1) = solve(conds(1), T)
Tsol(:,2) = solve(conds(2), T)
doesn’t completely resolve the problem, since the result is now a funciton of ‘x’:
Tsol =
-2*exp(-5i/4)*(-((exp(x*2i)*exp(10i) - 1)*(4478027322974943*exp(5i/2) - 1125899906842624*exp(x*2i)*exp(15i) - 4478027322974943*exp(x*2i)*exp(25i/2) + 1125899906842624))/(4503599627370496 + 4503599627370496*exp(x*4i)*exp(20i) - 12233297969360201*exp(x*2i)*exp(10i)))^(1/2)
2*exp(-5i/4)*(-((exp(x*2i)*exp(10i) - 1)*(4478027322974943*exp(5i/2) - 1125899906842624*exp(x*2i)*exp(15i) - 4478027322974943*exp(x*2i)*exp(25i/2) + 1125899906842624))/(4503599627370496 + 4503599627370496*exp(x*4i)*exp(20i) - 12233297969360201*exp(x*2i)*exp(10i)))^(1/2)
The only other option I can provide is:
Tfcn = matlabFunction(Tsol)
that will produce an anonymous function to evaluate those.
John Jarvis
John Jarvis am 28 Mär. 2019
@Star Strider, Many thanks for your elaboration.
Star Strider
Star Strider am 28 Mär. 2019
As always, my pleasure.

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