Question

NOTE: I need the correct answer with every single details

The two coupled differential equations: *1 + 5x1 - 2x2 = 2e-t 32 - 2x1 + 2x2 = 0 Are subjected to initial conditions: x1(0) = 0 , x2(0) = 0 ,*1(0) = 0 ,*2(0) = 0 a) Find the laplace transform of the system and solve for X1(s) and X2(s). (2 points). b) Use MATLAB to find the inverse laplace transform. (2 points). c) Plot the solution from part (6) using MATLAB Your plot should have, xlabel, ylabel title and legend. (2 points)

Answer 1

The given differential equations are

$\ddot{x}_{1}+5x_{1}-2x_{2}=2e^{-t}\rightarrow (1)$

22 - 2.11 + 2.12 = 0 → (2)

Taking Laplace transform of equation (1), we get

$s^{2}X_{1}(s)-sx_{1}(0)-\dot{x}_{1}(0)+5X_{1}(s)-2X_{2}(s)=\frac{2}{s+1}$

Since $x_{1}(0)=\dot{x}_{1}(0)=0$ , we get

$\left ( s^{2}+5 \right )X_{1}(s)-2X_{2}(s)=\frac{2}{s+1}\rightarrow (3)$

Taking Laplace transform of equation (2), we get

$s^{2}X_{2}(s)-sx_{2}(0)-\dot{x}_{2}(0)-2X_{1}(s)+2X_{2}=0$

Since $x_{2}(0)=\dot{x}_{2}(0)=0$ , we get

$X_{1}(s)=\frac{\left ( s^{2}+2 \right )}{2}X_{2}(s)\rightarrow (4)$

Substituting for X1(s) from equation (4) in equation (3), we get

$\left [ \left (s^{2}+5 \right )\left ( s^{2}+2 \right )-4 \right ]X_{2}(s)=\frac{4}{s+1}$

$X_{2}(s)=\frac{4}{ \left (s^{2}+5 \right )\left ( s^{2}+2 \right )\left ( s+1 \right )-4 \left ( s+1 \right )}\rightarrow (5)$

Substituting for X2(s) from equation (5) in equation (4), we get

$X_{1}(s)=\frac{2\left ( s^{2}+2 \right )}{ \left (s^{2}+5 \right )\left ( s^{2}+2 \right )\left ( s+1 \right )-4 \left ( s+1 \right )}\rightarrow (6)$

Taking the inverse laplace transform, we get

$x_{1}(t)=\frac{3}{7}e^{-t}-\frac{8}{35}cos(\sqrt{6}t)+\frac{sin(t)-cos(t)}{5}+\frac{4\sqrt{6}}{105}sin(\sqrt{6}t)$$x_{2}(t)=\frac{2}{7}e^{-t}+\frac{4}{35}cos(\sqrt{6}t)+\frac{2sin(t)-2cos(t)}{5}-\frac{2\sqrt{6}}{105}sin(\sqrt{6}t)$

The plots of $x_{1}(t)$ and $x_{2}(t)$ are shown below

Time response time (s) (1)*x*(1) 'x

MATLAB CODE

% For inverse laplace transform

syms s
D = (((s^2)+5)*((s^2)+2)*(s+1))-(4*(s+1));
X1 = (2*((s^2)+2))/D;
X2 = 4/D;
x1 = ilaplace(X1)

x2 = ilaplace(X2)

% For plotting

clear

t=0:0.1:10;

x1=zeros(length(t),1);
x2=zeros(length(t),1);
for i = 1:length(t)
x1(i)=(3*exp(-t(i)))/7 - (8*cos(6^(1/2)*t(i)))/35 - cos(t(i))/5 + sin(t(i))/5 + (4*6^(1/2)*sin(6^(1/2)*t(i)))/105;
x2(i)=(2*exp(-t(i)))/7 + (4*cos(6^(1/2)*t(i)))/35 - 2*cos(t(i))/5 + 2*sin(t(i))/5 - (2*6^(1/2)*sin(6^(1/2)*t(i)))/105;
end
figure (1)
plot(t,x1,'-r')
hold on
plot(t,x2,'-g')
grid on
xlabel ('time (s)');
ylabel ('x_{1}(t),x_{2}(t)')
legend ('x_{1}(t)','x_{2}(t)')
title ('Time response')