Question

4. Solve the following initial-value problem: 2 2 for 3-dimensional vector X. Present the final answer in terms of єkt, ektsinrnt and ekt cos mt.

Answer 1

Solution:

According to the matrix we have three equations

the first one is easy and the solution is, r(t)32 +2e21

for the second two we apply laplace transform to both sides

$s^2Y(s)-sy(0)-y'(0)=3Y(s)-4Z(s)=>\left (s^2-3 \right )Y(s)=2s+4-4Z(s)$

$s^2Z(s)-sz(0)-z'(0)=4Y(s)+3Z(s)=>\left (s^2-3 \right )Z(s)=2+4Y(s)$

eliminating Z(s) we get,

$=>\left (s^2-3 \right )Y(s)=2s+4-4\left (\frac{2+4Y(s)}{\left (s^2-3 \right )} \right )$

$=>\left (s^2-3 \right )Y(s)=2s+4-\frac{8}{\left (s^2-3 \right )}-\frac{16Y(s)}{\left (s^2-3 \right )}$

$=>\left (s^2-3 \right )Y(s)+\frac{16Y(s)}{\left (s^2-3 \right )}=2s+4-\frac{8}{\left (s^2-3 \right )}$

2162s+ 2 3 (s2 - 3)

$=>Y(s)=\frac{\left (2s+4 \right )\left (s^2-3 \right )}{\left (s^2-3 \right )^2+16}-\frac{8}{\left (s^2-3 \right )^2+16}$

$=>Y(s)=2\left (\frac{s-2}{s^2-4s+5} \right )$

$=>Y(s)=2\left (\frac{s-2}{\left ( s-2 \right )^2+1} \right )$

taking inverse laplace transform we get,

$=>y(t)=2e^{2t}\cos (t)$

$\text{Now, }$

$=>Z(s)=\frac{2}{\left (s^2-3 \right )}+8\left (\frac{s-2}{\left (s^2-4s+5 \right )\left ( s^2-3 \right )} \right )$

$=>Z(s)=\frac{2\left (s^2-4s+5 \right )+8s-16}{\left (s^2-4s+5 \right )\left ( s^2-3 \right )}$

$=>Z(s)=\frac{2s^2-6}{\left (s^2-4s+5 \right )\left ( s^2-3 \right )}$

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

$=>Z(s)=\frac{1}{\left (s-2 \right )^2+1}$

taking inverse laplace transform we get,

z(t)et sin(t

$\text{Therefore, the three solutions are }$

$=>x(t)=3e^{2t}+2e^{-2t},y(t)=e^{2t}\cos (t),z(t)=e^{2t}\sin (t)$