The solution yt consists of two exponential terms defined by

The solution y(t) consists of two exponential terms defined by the following equation y(t) = Ae^-m_1t + Be^-m_2t We want you to write a Mat Lab script for plotting y(t) versus t for 0

Solution

a)plot the matlab values by t and Y

code:

clc;
clear all;
t=0:0.01:10;
A=8;
B=-9;
m1=3;
m2=4;
Y=(A*exp(-m1*t))-(B*exp(-m2*t))
figure
plot(t,Y,\':b\');
xlabel(\'Time\');
ylabel(\'Y(t)\');

b)

i)changing A coefficients as a random number

code:

clc;
clear all;
A=[-8 0.62 6 -20];
B=-6;
m1=-2;
m2=-3;
t=0:0.01:10;
Y1=(A(1)*exp(-m1*t))-(B*exp(-m2*t));
Y2=(A(2)*exp(-m1*t))-(B*exp(-m2*t));
Y3=(A(3)*exp(-m1*t))-(B*exp(-m2*t));
Y4=(A(4)*exp(-m1*t))-(B*exp(-m2*t));
plot(t,Y1,t,Y2,t,Y3,t,Y4);
legend(\'A=-8\',\'A=0.62\',\'A=6\',\'A=-20\');

ii)changing B coefficient values as a random number

code:

clc;
clear all;
B=[-5 0.66 6 -20];
A=8;
m1=-3;
m2=-4;
t=0:0.01:10;
Y1=(A*exp(-m1*t))-(B(1)*exp(-m2*t));
Y2=(A*exp(-m1*t))-(B(2)*exp(-m2*t));
Y3=(A*exp(-m1*t))-(B(3)*exp(-m2*t));
Y4=(A*exp(-m1*t))-(B(4)*exp(-m2*t));
plot(t,Y1,t,Y2,t,Y3,t,Y4);
legend(\'B=-5\',\'B=0.66\',\'B=6\',\'B=-20\');