Exercise 13

Consider the linear system $$\begin{aligned} u' + u - v & = 0 \\ v' - u + v & = 2 \\ u(0) = 1, & \quad v(0) = 2. \end{aligned}$$ If we let ${\mathcal L}(u(s)) = U(s)$ and ${\mathcal L}(v(s)) = V(s)$ and take the Laplace transform of the two differential equations, we obtain $$\begin{aligned} [sU(s) - 1] + U(s) - V(s) & = 0 \\ [sV(s) - 2] - U(s) + V(s) & = \frac{2}{s}. \end{aligned}$$

1. Verify that if we solve the system $$\begin{aligned} [sU(s) - 1] + U(s) - V(s) & = 0 \\ [sV(s) - 2] - U(s) + V(s) & = \frac{2}{s} \end{aligned}$$ for $U(s)$ and $V(s)$, we obtain $$U(s) = \frac{s+1}{s^2} \quad \text{and} \quad V(s) = \frac{2s + 1}{s^2}.$$

2. Find $u(t)$ and $v(t)$.

3. Solve the linear system $$\begin{aligned} u' + 4u - 6v & = 0 \\ v' + 3u - 5v & = 0 \\ u(0) = 3, & \quad v(0) = 2 \end{aligned}$$ using Laplace transforms.

4. Solve the linear system $$\begin{aligned} w' - y & = 0 \\ w + y' + z & = 1 \\ w - y + z' & = 2 \sin t \\ w(0) = 1, & \quad y(0) = 1, \quad z(0) = 1 \end{aligned}$$ using Laplace transforms.