How to find L from the given code?
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lags = 0.6;
k1 = 3;
k_1 = 1;
k2 = 2.5;
k3 = 1;
k_3 = 1;
k4 = 2;
k5 = 1;
E1 = 1;
E2 = 2;
K1 = (k_1+k2)/k1;
K2 = (k_3+k4)/k3;
k3*E2*(L^2+L*(k1*K1+k4+k1*E1)+(E1+K1)*k1*k4+k1*k2*E1)+(L+k3*K2)*(k1*k2*E1+(L+k1*K1+k1*E1)*(L-k5*E1*e^(-L*lags))) = 0;
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Accepted Answer
Torsten
on 30 Jul 2022
Edited: Torsten
on 30 Jul 2022
lags = 0.6;
k1 = 3;
k_1 = 1;
k2 = 2.5;
k3 = 1;
k_3 = 1;
k4 = 2;
k5 = 1;
E1 = 1;
E2 = 2;
K1 = (k_1+k2)/k1;
K2 = (k_3+k4)/k3;
fun = @(L) k3*E2*(L.^2+L.*(k1*K1+k4+k1*E1)+(E1+K1)*k1*k4+k1*k2*E1)+(L+k3*K2).*(k1*k2*E1+(L+k1*K1+k1*E1).*(L-k5*E1*exp(-L*lags)));
L0 = -1.5;
L1 = fsolve(fun,L0)
L0 = -2.5;
L2 = fsolve(fun,L0)
plot((-2.5:0.1:-1),fun(-2.5:0.1:-1))
More Answers (1)
Walter Roberson
on 30 Jul 2022
Edited: Torsten
on 30 Jul 2022
syms L
e = exp(sym(1))
lags = 0.6;
k1 = 3;
k_1 = 1;
k2 = 2.5;
k3 = 1;
k_3 = 1;
k4 = 2;
k5 = 1;
E1 = 1;
E2 = 2;
K1 = (k_1+k2)/k1;
K2 = (k_3+k4)/k3;
eqn = k3*E2*(L^2+L*(k1*K1+k4+k1*E1)+(E1+K1)*k1*k4+k1*k2*E1)+(L+k3*K2)*(k1*k2*E1+(L+k1*K1+k1*E1)*(L-k5*E1*e^(-L*lags)))
vpasolve(eqn)
There is a second solution near -2.3
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