Let's solve the given initial value problem step-by-step: Given: \[ \frac{d^2y}{dt^2} + 2\frac{dy}{dt} + y = 3\delta(t-3), \quad y(0) = 0, \; y'(0) = 3 \] --- ### **Step 1: Take the Laplace Transform** Take the Laplace Transform of both sides (let \( Y(s) = \mathcal{L}\{y(t)\} \)): \[ \mathcal{L}\left\{ \frac{d^2y}{dt^2} + 2\frac{dy}{dt} + y \right\} = \mathcal{L}\{3\delta(t-3)\} \] Recall properties: - \(\mathcal{L}\{y''\} = s^2 Y(s) - s y(0) - y'(0)\) - \(\mathcal{L}\{y'\} = s Y(s) - y(0)\) - \(\mathcal{L}\{y\} = Y(s)\) - \(\mathcal{L}\{\delta(t-a)\} = e^{-as}\) Plug in initial conditions (\(y(0) = 0, y'(0) = 3\)): \[ [s^2 Y(s) - 0 - 3] + 2[s Y(s) - 0] + Y(s) = 3e^{-3s} \] \[ s^2 Y(s) - 3 + 2s Y(s) + Y(s) = 3e^{-3s} \] \[ (s^2 + 2s + 1)Y(s) = 3 + 3e^{-3s} \] \[ (s+1)^2 Y(s) = 3 + 3e^{-3s} \] \[ Y(s) = \frac{3}{(s+1)^2} + \frac{3e^{-3s}}{(s+1)^2} \] **Explanation:** We've used the Laplace transform to convert the differential equation into an algebraic equation in terms of \( Y(s) \). --- ### **Step 2: Inverse Laplace Transform** Now, take the inverse Laplace transform to find \( y(t) \): We know: - \( \mathcal{L}^{-1}\left\{ \frac{1}{(s+a)^2} \right\} = t e^{-a t} \) - \( \mathcal{L}^{-1}\left\{ e^{-as}F(s) \right\} = f(t-a)u(t-a) \), where \( u(t-a) \) is the unit step function. So, \[ y(t) = 3 \cdot t e^{-t} + 3 \cdot (t-3)e^{-(t-3)} u(t-3) \] **Explanation:** We used the inverse Laplace transform and the shifting property to get the solution in the time domain. --- ### **Final Answer** \[ \boxed{ y(t) = 3 t e^{-t} + 3 (t-3) e^{-(t-3)} u(t-3) } \] Where \( u(t-3) \) is the unit step function.