How can i solve this equation ?
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Mallouli Marwa
el 18 de En. de 2017
Comentada: Jorge Mario Guerra González
el 20 de En. de 2017
Hi
How ca i solve the attached equation to find lamda ?
When all the other terms are known.
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Respuesta aceptada
Jorge Mario Guerra González
el 20 de En. de 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 comentarios
Jorge Mario Guerra González
el 20 de En. de 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
Más respuestas (1)
Jorge Mario Guerra González
el 19 de En. de 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 comentarios
John D'Errico
el 19 de En. de 2017
You cannot find an infinite number of solutions. If you tried to write them all down, it would take infinitely long.
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