The given equation is an initial-value problem with a Dirac delta function on the right-hand side. The differential equation is:

$y' + y = \delta(t-3), \quad y(0) = 2$

This leads to the equation involving an integral of both sides:

$\frac{d}{ds}(e^s y) \, ds = \int_0^t e^s \delta(s-3) \, ds$

Now, let's proceed step by step.

Step 1: Take the integral of both sides

On the left-hand side, you're integrating:

$\int_0^t \frac{d}{ds}(e^s y(s)) \, ds = e^t y(t) - e^0 y(0) = e^t y(t) - 2$

On the right-hand side, you're integrating a Dirac delta function:

$\int_0^t e^s \delta(s-3) \, ds$

The delta function $\delta(s-3)$ picks out the value of the integrand at $s = 3$, but only if $t > 3$. Thus:

\begin{cases} e^3, & \text{if } t \geq 3, \\ 0, & \text{if } t < 3. \end{cases}$$ ### Step 2: Write the expression after integrating both sides We now have: $$e^t y(t) - 2 = \begin{cases} e^3, & \text{if } t \geq 3, \\ 0, & \text{if } t < 3. \end{cases}$$ This is the next step after integrating both sides of the equation. Now, you can solve for $$y(t)$$ by isolating $$y(t)$$: $$y(t) = \begin{cases} e^{-t}(e^3 + 2), & \text{if } t \geq 3, \\ 2 e^{-t}, & \text{if } t < 3. \end{cases}$$ This gives you the solution to the differential equation. Would you like further details or clarification on this? --- Here are five related questions to extend your understanding: 1. How does the Dirac delta function affect solutions to differential equations? 2. What are the key properties of the exponential function when solving linear differential equations? 3. How would the solution change if the initial condition $$y(0)$$ were different? 4. What is the physical interpretation of a delta function in a differential equation? 5. Can you solve the homogeneous version of the differential equation, $$y' + y = 0$$? **Tip:** The Dirac delta function is often used to model impulses or sudden forces in physical systems. Its integral over any interval that includes its argument is always 1.