The image contains a system of differential equations with initial conditions, and the task is to solve it using the Laplace transform. Here's a breakdown of the problem:

1. System of equations: x'(t) - 2x(t) &= y(t), \\ y'(t) - y(t) &= 1. \end{aligned}$$
2. Initial conditions: $x(0) = 0, \quad y(0) = 0.$

Steps to solve using Laplace Transform:

1. Apply the Laplace Transform to both equations:

   • For $x'(t)$, use $\mathcal{L}\{x'(t)\} = sX(s) - x(0)$.
   • For $y'(t)$, use $\mathcal{L}\{y'(t)\} = sY(s) - y(0)$.

2. Transform the system:

   \mathcal{L}\{x'(t)\} - 2\mathcal{L}\{x(t)\} &= \mathcal{L}\{y(t)\}, \\ \mathcal{L}\{y'(t)\} - \mathcal{L}\{y(t)\} &= \mathcal{L}\{1\}. \end{aligned}$$ Substituting the initial conditions and known Laplace transforms: $$\begin{aligned} (sX(s)) - 2X(s) &= Y(s), \\ (sY(s)) - Y(s) &= \frac{1}{s}. \end{aligned}$$

3. Solve the second equation for $Y(s)$: $sY(s) - Y(s) = \frac{1}{s} \implies Y(s)(s - 1) = \frac{1}{s} \implies Y(s) = \frac{1}{s(s - 1)}.$

4. Solve the first equation for $X(s)$: $sX(s) - 2X(s) = Y(s) \implies X(s)(s - 2) = Y(s) = \frac{1}{s(s - 1)}.$ Therefore, $X(s) = \frac{1}{s(s - 1)(s - 2)}.$

5. Inverse Laplace Transform:

   • The function $Y(s) = \frac{1}{s(s - 1)}$ can be decomposed using partial fractions: $\frac{1}{s(s - 1)} = \frac{A}{s} + \frac{B}{s - 1},$ which gives $A = 1$ and $B = -1$, leading to: $Y(s) = \frac{1}{s} - \frac{1}{s - 1}.$ The inverse Laplace transform is: $y(t) = 1 - e^t.$
   • For $X(s)$, use partial fraction decomposition and the inverse Laplace transform.

Would you like me to continue with the detailed decomposition of $X(s)$, or do you have any specific questions? Here are some related questions:

1. Would you like to see the detailed inverse Laplace steps for $X(s)$?
2. Do you need a step-by-step explanation of the partial fraction decomposition?
3. How does the Laplace transform help in solving differential equations?
4. Do you need a refresher on solving linear systems of differential equations?
5. Would you like to explore specific applications of the Laplace transform in engineering?

Tip: The Laplace transform is particularly useful for handling systems with initial conditions, making it easier to switch to the frequency domain and solve algebraically.