The problem in the image asks to solve a differential equation using the Laplace transform method. The equation is:

$x''(t) - x'(t) = e^t$ with initial conditions: $x(0) = 0, \quad x'(0) = 1$

Solution Using Laplace Transform:

1. Take the Laplace transform of both sides:

   • The Laplace transform of $x''(t)$ is $s^2 X(s) - s x(0) - x'(0)$.
   • The Laplace transform of $x'(t)$ is $s X(s) - x(0)$.
   • The Laplace transform of $e^t$ is $\frac{1}{s-1}$.

   So, applying the Laplace transform to the given equation: $\mathcal{L}\{x''(t) - x'(t)\} = \mathcal{L}\{e^t\}$ This gives: $(s^2 X(s) - s x(0) - x'(0)) - (s X(s) - x(0)) = \frac{1}{s-1}$

2. Substitute the initial conditions: Since $x(0) = 0$ and $x'(0) = 1$, we substitute these values: $s^2 X(s) - 1 - s X(s) = \frac{1}{s-1}$ Simplify: $X(s)(s^2 - s) = \frac{1}{s-1} + 1$ $X(s)(s^2 - s) = \frac{s}{s-1}$

3. Solve for $X(s)$: $X(s) = \frac{s}{(s-1)(s^2 - s)}$

4. Simplify the expression: Factor the denominator: $X(s) = \frac{s}{(s-1)s(s-1)} = \frac{1}{s-1}$

5. Inverse Laplace transform: The inverse Laplace transform of $\frac{1}{s-1}$ is $e^t$.

Therefore, the solution to the differential equation is: $x(t) = e^t$

Let me know if you'd like further clarification or additional steps! Also, here are some questions related to this solution:

1. What is the significance of the initial conditions in solving differential equations with the Laplace transform?
2. How does the Laplace transform simplify solving linear differential equations compared to traditional methods?
3. Can you generalize the Laplace transform method for higher-order differential equations?
4. How would the solution change if the non-homogeneous term was $\sin(t)$ instead of $e^t$?
5. What are the key properties of the Laplace transform that make it suitable for solving differential equations?

Tip: Always check the form of the differential equation and initial conditions before taking the Laplace transform to avoid mistakes!