Solve issue in Matlab R2018b
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John Jarvis
el 28 de Mzo. de 2019
Comentada: Star Strider
el 28 de Mzo. de 2019
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 comentarios
madhan ravi
el 28 de Mzo. de 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.
Respuesta aceptada
Star Strider
el 28 de Mzo. de 2019
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)
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