**Step 1: Take the Laplace Transform of the differential equation** Given: \( y'' + 3y' = \delta(t-1) \) \( y(0) = 0, \quad y'(0) = 1 \) Recall Laplace transforms: - \( \mathcal{L}\{y''\} = s^2 Y(s) - s y(0) - y'(0) \) - \( \mathcal{L}\{y'\} = s Y(s) - y(0) \) - \( \mathcal{L}\{\delta(t-a)\} = e^{-as} \) Apply Laplace transform to both sides: \[ \mathcal{L}\{y'' + 3y'\} = \mathcal{L}\{\delta(t-1)\} \] \[ (s^2 Y(s) - s y(0) - y'(0)) + 3(s Y(s) - y(0)) = e^{-s} \] Substitute \( y(0) = 0 \), \( y'(0) = 1 \): \[ (s^2 Y(s) - 1) + 3s Y(s) = e^{-s} \] \[ (s^2 + 3s) Y(s) - 1 = e^{-s} \] \[ (s^2 + 3s) Y(s) = e^{-s} + 1 \] \[ Y(s) = \frac{e^{-s} + 1}{s^2 + 3s} \] **Explanation:** We used the Laplace transform properties and substituted the initial conditions to convert the ODE into an algebraic equation in the Laplace domain. --- **Step 2: Find the inverse Laplace transform to solve for \( y(t) \)** Rewrite \( Y(s) \): \[ Y(s) = \frac{1}{s^2 + 3s} + \frac{e^{-s}}{s^2 + 3s} \] First, decompose \( \frac{1}{s^2 + 3s} \): \[ s^2 + 3s = s(s+3) \] \[ \frac{1}{s(s+3)} = \frac{A}{s} + \frac{B}{s+3} \] \[ 1 = A(s+3) + B s \] Let \( s = 0 \): \( 1 = 3A \implies A = \frac{1}{3} \) Let \( s = -3 \): \( 1 = B(-3) \implies B = -\frac{1}{3} \) So, \[ \frac{1}{s(s+3)} = \frac{1}{3} \frac{1}{s} - \frac{1}{3} \frac{1}{s+3} \] By the shifting theorem, \( \mathcal{L}^{-1}\left[\frac{e^{-as}}{F(s)}\right] = f(t-a) u(t-a) \), where \( f(t) = \mathcal{L}^{-1}[F(s)] \). So, \[ Y(s) = \left( \frac{1}{3} \frac{1}{s} - \frac{1}{3} \frac{1}{s+3} \right) + e^{-s} \left( \frac{1}{3} \frac{1}{s} - \frac{1}{3} \frac{1}{s+3} \right) \] Inverse Laplace: - \( \mathcal{L}^{-1}\left[\frac{1}{s}\right] = 1 \) - \( \mathcal{L}^{-1}\left[\frac{1}{s+3}\right] = e^{-3t} \) Therefore, \[ y_1(t) = \frac{1}{3}(1) - \frac{1}{3} e^{-3t} \] \[ y_2(t) = u(t-1) \left[ \frac{1}{3}(1) - \frac{1}{3} e^{-3(t-1)} \right] \] **Combine:** \[ y(t) = \frac{1}{3}(1 - e^{-3t}) + u(t-1)\left[ \frac{1}{3}(1 - e^{-3(t-1)}) \right] \] **Explanation:** We decomposed the Laplace domain solution into partial fractions, applied the shifting theorem for the delta function (which shifts the response by \( t-1 \)), and then took inverse Laplace transforms to get the solution in the time domain. --- **Final Answer:** \[ \boxed{ y(t) = \frac{1}{3}(1 - e^{-3t}) + u(t-1)\left[ \frac{1}{3}(1 - e^{-3(t-1)}) \right] } \] where \( u(t-1) \) is the unit step function.