Vectors must be the same length.

lamda=0.2;
mu=1-lamda;
P1=mu^2/(mu+lamda)^2;
P2=2*lamda*mu/(mu+lamda)^2;
P3=lamda^2/(mu+lamda)^2;
display(P1);
display(P2);
display(P3);
t = 1:5:100 ;
P11 = zeros(size(t)) ; 
P22 = zeros(size(t)) ; 
P33 = zeros(size(t)) ; 
for i = 1:length(t)    
P11(t)=(lamda^2/(lamda+mu)^2)*exp(-2*(lamda+mu)*t)+((2*mu*lamda)/(lamda+mu)^2)*exp(-(mu+lamda)*t)+mu^2/(lamda+mu)^2;
P22(t)=((2*mu*lamda)/(lamda+mu)^2)+(((2*lamda*(lamda-mu))/(lamda+mu)^2)*exp(-(mu+lamda)*t(i)))-2*((lamda^2/lamda+mu)^2)*exp(-2*(mu+lamda)*t(i));
P33(t)=((lamda^2/(lamda+mu)^2)*exp(-2*(mu+lamda)*t(i)))-(2*lamda^2/(lamda+mu)^2)*exp(-(mu+lamda)*t(i))+lamda^2/(lamda+mu)^2;

end
A=P11;
B=P33;
C=P22;
plot(A,t,B,t,C,t)
legend({'A','B','C'})

how can i solve this problem? Vectors must be the same length.??

[alex brown]

Do you need something like this figure?

lamda=0.2;
mu=1-lamda;
P1=mu^2/(mu+lamda)^2;
P2=2*lamda*mu/(mu+lamda)^2;
P3=lamda^2/(mu+lamda)^2;
display(P1);
display(P2);
display(P3);
t = 1:5:100 ;
P11 = zeros(size(t)) ; 
P22 = zeros(size(t)) ; 
P33 = zeros(size(t)) ; 
for i = 1:length(t)    
P11(t)=(lamda^2/(lamda+mu)^2)*exp(-2*(lamda+mu)*t)+((2*mu*lamda)/(lamda+mu)^2)*exp(-(mu+lamda)*t)+mu^2/(lamda+mu)^2;
P22(t)=((2*mu*lamda)/(lamda+mu)^2)+(((2*lamda*(lamda-mu))/(lamda+mu)^2)*exp(-(mu+lamda)*t(i)))-2*((lamda^2/lamda+mu)^2)*exp(-2*(mu+lamda)*t(i));
P33(t)=((lamda^2/(lamda+mu)^2)*exp(-2*(mu+lamda)*t(i)))-(2*lamda^2/(lamda+mu)^2)*exp(-(mu+lamda)*t(i))+lamda^2/(lamda+mu)^2;
end
A=P11;
B=P33;
C=P22;
plot(A)
hold on
plot(B)
hold on
plot(C)
legend({'A','B','C'})

Answer 1

[KSSV]

YOu need to check the loop part.....

lamda=0.2;
mu=1-lamda;
P1=mu^2/(mu+lamda)^2;
P2=2*lamda*mu/(mu+lamda)^2;
P3=lamda^2/(mu+lamda)^2;
display(P1);
display(P2);
display(P3);
t = 1:5:100 ;
P11 = zeros(size(t)) ;
P22 = zeros(size(t)) ;
P33 = zeros(size(t)) ;
for i = 1:length(t)
    P11(i,:)=(lamda^2/(lamda+mu)^2)*exp(-2*(lamda+mu)*t)+((2*mu*lamda)/(lamda+mu)^2)*exp(-(mu+lamda)*t)+mu^2/(lamda+mu)^2;
    P22(i,:)=((2*mu*lamda)/(lamda+mu)^2)+(((2*lamda*(lamda-mu))/(lamda+mu)^2)*exp(-(mu+lamda)*t(i)))-2*((lamda^2/lamda+mu)^2)*exp(-2*(mu+lamda)*t(i));
    P33(i,:)=((lamda^2/(lamda+mu)^2)*exp(-2*(mu+lamda)*t(i)))-(2*lamda^2/(lamda+mu)^2)*exp(-(mu+lamda)*t(i))+lamda^2/(lamda+mu)^2;
end
A=P11;
B=P33;
C=P22;
plot(A,t,B,t,C,t)
legend({'A','B','C'})

[Jonathan Ortiz]

whats the use of 1 in 1: length(t)?

[KSSV]

It is a loop indexing, where loop starts from 1 till length(t), this will ensure the arrays formed will be of dimension of t.