To find the Laplace transform of the solution to the initial value problem, let's proceed by taking the Laplace transform of both sides of the differential equation:

Given: $x' + x = 3 + \delta(t - 4), \quad x(0) = 0$

Step 1: Applying the Laplace Transform

Taking the Laplace transform of each term:

1. For $x'(t)$: \mathcal{L}\{x'(t)\} = sX(s) - x(0) = sX(s) \quad \text{(since x(0) = 0)}

2. For $x(t)$: $\mathcal{L}\{x(t)\} = X(s)$

3. For the constant term $3$: $\mathcal{L}\{3\} = \frac{3}{s}$

4. For the Dirac delta function $\delta(t - 4)$: $\mathcal{L}\{\delta(t - 4)\} = e^{-4s}$

Step 2: Setting Up the Equation in the Laplace Domain

Substituting into the Laplace-transformed equation: $sX(s) + X(s) = \frac{3}{s} + e^{-4s}$

Combine like terms: $(s + 1)X(s) = \frac{3}{s} + e^{-4s}$

Step 3: Solving for $X(s)$

Now, isolate $X(s)$: $X(s) = \frac{3}{s(s + 1)} + \frac{e^{-4s}}{s + 1}$

Thus, the Laplace transform of the solution $x(t)$ is: $\mathcal{L}\{x(t)\}(s) = \frac{3}{s(s + 1)} + \frac{e^{-4s}}{s + 1}$

Would you like a detailed breakdown of the inverse Laplace transform or further explanation of each step?

Follow-Up Questions

1. How do we handle the inverse Laplace transform of functions involving exponential shifts?
2. What is the role of the Heaviside function in solving differential equations with shifted delta functions?
3. How do we interpret the solution in the time domain after taking the inverse Laplace transform?
4. What happens if the initial condition is non-zero in problems involving delta functions?
5. How does the Laplace transform help solve differential equations with discontinuous inputs?

Tip

In problems with delta functions, the Laplace transform allows us to handle impulses (instantaneous inputs) effectively by transforming them into exponential terms, making the analysis in the Laplace domain simpler.