How can i solve this equation ?
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Mallouli Marwa
am 18 Jan. 2017
Kommentiert: Jorge Mario Guerra González
am 20 Jan. 2017
Hi
How ca i solve the attached equation to find lamda ?
When all the other terms are known.
0 Kommentare
Akzeptierte Antwort
Jorge Mario Guerra González
am 20 Jan. 2017
If you want to get multiple answers use vpasolve function. Set the number of answers you want yo get and the code will pick some then in a random way.
syms lambda
Mt=1; % The value of your variables
m=1;
L=1;
lt=1;
eqn= 1+(cos(lambda)*cosh(lambda))+((lambda*Mt/(m*L))*(cos(lambda)*sinh(lambda)-sin(lambda)*cosh(lambda)))-((lt*lambda^3)/m*L^3)*(cosh(lambda)*sin(lambda)+sinh(lambda)*cos(lambda))+ ((Mt*lt*lambda^4)/m^2*L^4)*(1-cosh(lambda)*cos(lambda)) ==0;
for n = 1:50
vpasolve(eqn,lambda,'random',true)
end
this example gives you 50 answers. The answer before gives the first it finds.
3 Kommentare
Jorge Mario Guerra González
am 20 Jan. 2017
syms lambda
Mt=1; % The value of your variables
m=1;
L=1;
lt=1;
eqn= 1+(cos(lambda)*cosh(lambda))+((lambda*Mt/(m*L))*(cos(lambda)*sinh(lambda)-sin(lambda)*cosh(lambda)))-((lt*lambda^3)/m*L^3)*(cosh(lambda)*sin(lambda)+sinh(lambda)*cos(lambda))+ ((Mt*lt*lambda^4)/m^2*L^4)*(1-cosh(lambda)*cos(lambda)) ==0;
k=zeros(1,50);
for n = 1:50
k(i)=vpasolve(eqn,lambda,'random',true)
end
that should create the vector
Weitere Antworten (1)
Jorge Mario Guerra González
am 19 Jan. 2017
Just use the function solve, which uses numerical techniques to find variables.
try this.
syms lambda
Mt=1; % The value of your variables
m=1;
L=1;
lt=1;
%check if I wrote this correctly
eqn= 1+(cos(lambda)*cosh(lambda))+((lambda*Mt/(m*L))*(cos(lambda)*sinh(lambda)-sin(lambda)*cosh(lambda)))-((lt*lambda^3)/m*L^3)*(cosh(lambda)*sin(lambda)+sinh(lambda)*cos(lambda))+ ((Mt*lt*lambda^4)/m^2*L^4)*(1-cosh(lambda)*cos(lambda)) ==0;
value=solve(eqn,lambda) %solution for lamda
2 Kommentare
John D'Errico
am 19 Jan. 2017
You cannot find an infinite number of solutions. If you tried to write them all down, it would take infinitely long.
Siehe auch
Kategorien
Mehr zu Calculus finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!