Let's go step-by-step: ### 1. **Write the ODE and Initial Conditions** Given: \[ y'' + 4y = f(t), \quad y(0) = 1, \quad y'(0) = 0 \] where \[ f(t) = \begin{cases} 0 & 0 \leq t < 5 \\ e^{t} & t \geq 5 \end{cases} \] --- ### 2. **Express \( f(t) \) Using the Heaviside Function** The piecewise function can be written as: \[ f(t) = 0 \cdot u(t) + e^{t} u(t-5) = e^{t} u(t-5) \] But for \( t < 5 \), \( f(t) = 0 \), and for \( t \geq 5 \), \( f(t) = e^t \). So, \[ f(t) = e^{t} u(t-5) \] --- ### 3. **Take the Laplace Transform of Both Sides** Recall: - \( \mathcal{L}[y''] = s^2 Y(s) - s y(0) - y'(0) \) - \( \mathcal{L}[y] = Y(s) \) - \( \mathcal{L}[e^{at}u(t-c)] = \frac{e^{-cs}}{s-a} \) (shift theorem) Applying Laplace transform: \[ \mathcal{L}\{y'' + 4y\} = \mathcal{L}\{f(t)\} \] \[ s^2 Y(s) - s y(0) - y'(0) + 4Y(s) = \mathcal{L}\{e^{t}u(t-5)\} \] Plug in initial conditions \( y(0) = 1 \), \( y'(0) = 0 \): \[ s^2 Y(s) - s(1) + 0 + 4Y(s) = \mathcal{L}\{e^{t}u(t-5)\} \] \[ (s^2 + 4) Y(s) - s = \mathcal{L}\{e^{t}u(t-5)\} \] --- ### 4. **Laplace Transform of Right Side** \[ \mathcal{L}\{e^{t}u(t-5)\} = e^{-5s} \mathcal{L}\{e^{t+5}\} = e^{-5s} \mathcal{L}\{e^{t}\} \] But more generally, use the second shifting theorem: \[ \mathcal{L}\{e^{at}u(t-c)\} = \frac{e^{-cs}}{s - a} \] Here, \( a = 1, c = 5 \): \[ \mathcal{L}\{e^{t}u(t-5)\} = \frac{e^{-5s}}{s-1} \] --- ### 5. **Write the Laplace Equation** So, \[ (s^2 + 4)Y(s) - s = \frac{e^{-5s}}{s-1} \] --- ### 6. **Solve for \( Y(s) \)** \[ (s^2 + 4)Y(s) = s + \frac{e^{-5s}}{s-1} \] \[ Y(s) = \frac{s}{s^2 + 4} + \frac{e^{-5s}}{(s-1)(s^2 + 4)} \] --- ### 7. **Inverse Laplace of Each Term** #### First term: \[ \mathcal{L}^{-1}\left[\frac{s}{s^2 + 4}\right] = \cos(2t) \] #### Second term (using shifting theorem): \[ \mathcal{L}^{-1}\left[\frac{e^{-5s}}{(s-1)(s^2 + 4)}\right] = u(t-5) \cdot h(t-5) \] where \[ h(t) = \mathcal{L}^{-1}\left[\frac{1}{(s-1)((s-1)^2 + 4)}\right] \] So, shift \( t \to t-5 \) after finding the inverse Laplace of \( \frac{1}{(s-1)((s-1)^2 + 4)} \). --- ### 8. **Find \( \mathcal{L}^{-1}\left[\frac{1}{(s-1)((s-1)^2 + 4)}\right] \)** Let \( S = s-1 \): \[ \frac{1}{(s-1)((s-1)^2 + 4)} = \frac{1}{S(S^2+4)} \] Partial fractions: \[ \frac{1}{S(S^2+4)} = \frac{A}{S} + \frac{BS+C}{S^2+4} \] Multiply both sides by \( S(S^2+4) \): \[ 1 = A(S^2+4) + (BS+C)S \] \[ 1 = A S^2 + 4A + B S^2 + C S \] \[ 1 = (A+B)S^2 + C S + 4A \] So, - \( S^2 \): \( A + B = 0 \implies B = -A \) - \( S \): \( C = 0 \) - Constant: \( 4A = 1 \implies A = \frac{1}{4} \) So, - \( B = -\frac{1}{4} \) - \( C = 0 \) Thus, \[ \frac{1}{S(S^2+4)} = \frac{1}{4} \left( \frac{1}{S} - \frac{S}{S^2+4} \right) \] Back to \( s \): \[ \frac{1}{(s-1)((s-1)^2+4)} = \frac{1}{4}\left(\frac{1}{s-1} - \frac{s-1}{(s-1)^2+4}\right) \] --- ### 9. **Find the Inverse Laplace of Each Part** \[ \mathcal{L}^{-1}\left[\frac{1}{s-1}\right] = e^{t} \] \[ \mathcal{L}^{-1}\left[\frac{s-1}{(s-1)^2+4}\right] = e^{t} \cos(2t) \] So, \[ h(t) = \frac{1}{4} \left( e^{t} - e^{t} \cos(2t) \right) = \frac{1}{4} e^{t} (1 - \cos(2t)) \] --- ### 10. **Apply the Shift (Second Shifting Theorem)** \[ u(t-5) h(t-5) = u(t-5) \cdot \frac{1}{4} e^{t-5} [1 - \cos(2(t-5))] \] --- ### **Final Solution** \[ \boxed{ y(t) = \cos(2t) + u(t-5) \cdot \frac{1}{4} e^{t-5} \left[1 - \cos(2(t-5))\right] } \] --- ### **Summary of Steps Involving the Piecewise Function** - We rewrote the piecewise \( f(t) \) using the Heaviside function as \( e^t u(t-5) \). - Used the Laplace shift theorem to find its transform and its inverse. - Applied the second shifting theorem to write the solution in terms of \( u(t-5) \) and a shifted function. If you have any questions about any step, let me know!