Let's solve the ODE using the Laplace Transform method: ### Given: \[ y''(t) + 4y'(t) + 4y(t) = 3u_2(t) - 5u_4(t) \] with initial conditions: \[ y(0) = 1, \quad y'(0) = 0 \] where \( u_a(t) \) is the unit step function at \( t = a \). --- ## **Step 1: Take Laplace Transform of Both Sides** Recall: - \( \mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) \) - \( \mathcal{L}\{y'(t)\} = sY(s) - y(0) \) - \( \mathcal{L}\{y(t)\} = Y(s) \) - \( \mathcal{L}\{u_a(t)\} = \frac{e^{-as}}{s} \) Apply Laplace Transform: \[ \mathcal{L}\{y''(t)\} + 4\mathcal{L}\{y'(t)\} + 4\mathcal{L}\{y(t)\} = 3\mathcal{L}\{u_2(t)\} - 5\mathcal{L}\{u_4(t)\} \] Plug in initial conditions: \[ (s^2Y(s) - s y(0) - y'(0)) + 4(sY(s) - y(0)) + 4Y(s) = 3 \cdot \frac{e^{-2s}}{s} - 5 \cdot \frac{e^{-4s}}{s} \] Given \( y(0) = 1,~ y'(0) = 0 \): \[ (s^2Y(s) - s) + 4(sY(s) - 1) + 4Y(s) = 3 \frac{e^{-2s}}{s} - 5 \frac{e^{-4s}}{s} \] Expand: \[ s^2Y(s) - s + 4sY(s) - 4 + 4Y(s) = 3 \frac{e^{-2s}}{s} - 5 \frac{e^{-4s}}{s} \] Group \( Y(s) \) terms: \[ (s^2 + 4s + 4)Y(s) - (s + 4) = 3 \frac{e^{-2s}}{s} - 5 \frac{e^{-4s}}{s} \] --- ## **Step 2: Solve for \( Y(s) \)** \[ (s^2 + 4s + 4)Y(s) = (s + 4) + 3 \frac{e^{-2s}}{s} - 5 \frac{e^{-4s}}{s} \] \[ Y(s) = \frac{s + 4}{s^2 + 4s + 4} + \frac{3 e^{-2s}/s - 5 e^{-4s}/s}{s^2 + 4s + 4} \] \[ Y(s) = \frac{s + 4}{(s + 2)^2} + \frac{3 e^{-2s}}{s (s + 2)^2} - \frac{5 e^{-4s}}{s (s + 2)^2} \] --- ## **Step 3: Inverse Laplace Transform Each Term** ### **First Term: \( \frac{s + 4}{(s + 2)^2} \)** Let’s rewrite \( s + 4 = (s + 2) + 2 \): \[ \frac{s + 4}{(s + 2)^2} = \frac{s + 2}{(s + 2)^2} + \frac{2}{(s + 2)^2} = \frac{1}{s + 2} + \frac{2}{(s + 2)^2} \] Inverse Laplace: - \( \mathcal{L}^{-1}\left\{ \frac{1}{s + a} \right\} = e^{-a t} \) - \( \mathcal{L}^{-1}\left\{ \frac{1}{(s + a)^2} \right\} = t e^{-a t} \) So, \[ \mathcal{L}^{-1}\left\{ \frac{1}{s + 2} \right\} = e^{-2t} \] \[ \mathcal{L}^{-1}\left\{ \frac{2}{(s + 2)^2} \right\} = 2t e^{-2t} \] ### **Second and Third Terms** These involve time shifts (using the shifting theorem): \[ \mathcal{L}^{-1} \left\{ \frac{e^{-a s} F(s)}{s} \right\} = u_a(t) * f(t-a) \] But first, let’s find the inverse Laplace of \( \frac{1}{s (s + 2)^2} \): Let’s use partial fractions: \[ \frac{1}{s(s + 2)^2} = \frac{A}{s} + \frac{B}{s + 2} + \frac{C}{(s + 2)^2} \] Multiply both sides by \( s(s + 2)^2 \): \[ 1 = A (s + 2)^2 + B s (s + 2) + C s \] Expand \( (s + 2)^2 = s^2 + 4s + 4 \): \[ 1 = A(s^2 + 4s + 4) + B s (s + 2) + C s = A(s^2 + 4s + 4) + B(s^2 + 2s) + Cs \] \[ = (A + B) s^2 + (4A + 2B + C) s + 4A \] Set coefficients equal: - \( s^2: \quad A + B \) - \( s: \quad 4A + 2B + C \) - constant: \( 4A \) So: - \( A + B = 0 \) - \( 4A + 2B + C = 0 \) - \( 4A = 1 \implies A = \frac{1}{4} \) Then \( B = -A = -\frac{1}{4} \) Now \( 4A + 2B + C = 0 \): \[ 4 \cdot \frac{1}{4} + 2 \cdot \left( -\frac{1}{4} \right) + C = 0 \implies 1 - \frac{1}{2} + C = 0 \implies C = -\frac{1}{2} \] So: \[ \frac{1}{s (s + 2)^2} = \frac{1}{4} \frac{1}{s} - \frac{1}{4} \frac{1}{s + 2} - \frac{1}{2} \frac{1}{(s + 2)^2} \] Apply to both shifted terms. So, \[ Y(s) = \left[ \frac{1}{s + 2} + \frac{2}{(s + 2)^2} \right] + 3e^{-2s} \left[ \frac{1}{4} \frac{1}{s} - \frac{1}{4} \frac{1}{s + 2} - \frac{1}{2} \frac{1}{(s + 2)^2} \right] \] \[ -5e^{-4s} \left[ \frac{1}{4} \frac{1}{s} - \frac{1}{4} \frac{1}{s + 2} - \frac{1}{2} \frac{1}{(s + 2)^2} \right] \] Or, expand: \[ Y(s) = \frac{1}{s + 2} + \frac{2}{(s + 2)^2} + \frac{3}{4} e^{-2s} \frac{1}{s} - \frac{3}{4} e^{-2s} \frac{1}{s + 2} - \frac{3}{2} e^{-2s} \frac{1}{(s + 2)^2} \] \[ - \frac{5}{4} e^{-4s} \frac{1}{s} + \frac{5}{4} e^{-4s} \frac{1}{s + 2} + \frac{5}{2} e^{-4s} \frac{1}{(s + 2)^2} \] --- ### **Inverse Laplace of Shifted Terms** Recall: \[ \mathcal{L}^{-1} \{ e^{-a s} F(s) \} = u_a(t) f(t-a) \] So, for each term, the inverse Laplace is: - \( \mathcal{L}^{-1} \left\{ \frac{1}{s} \right\} = 1 \) - \( \mathcal{L}^{-1} \left\{ \frac{1}{s + 2} \right\} = e^{-2t} \) - \( \mathcal{L}^{-1} \left\{ \frac{1}{(s + 2)^2} \right\} = t e^{-2t} \) So, - \( e^{-a s} \frac{1}{s} \to u_a(t) \cdot 1 \) - \( e^{-a s} \frac{1}{s + 2} \to u_a(t) e^{-2(t-a)} \) - \( e^{-a s} \frac{1}{(s + 2)^2} \to u_a(t) (t-a) e^{-2(t-a)} \) --- ## **Step 4: Write the Final Solution** \[ y(t) = e^{-2t} + 2t e^{-2t} \] \[ + \frac{3}{4} u_2(t) \cdot 1 - \frac{3}{4} u_2(t) e^{-2(t-2)} - \frac{3}{2} u_2(t) (t-2) e^{-2(t-2)} \] \[ - \frac{5}{4} u_4(t) \cdot 1 + \frac{5}{4} u_4(t) e^{-2(t-4)} + \frac{5}{2} u_4(t) (t-4) e^{-2(t-4)} \] Or, grouped: \[ \boxed{ \begin{aligned} y(t) =\;& e^{-2t} + 2t e^{-2t} \\ &+ u_2(t) \left[ \frac{3}{4} - \frac{3}{4} e^{-2(t-2)} - \frac{3}{2} (t-2) e^{-2(t-2)} \right] \\ &+ u_4(t) \left[ -\frac{5}{4} + \frac{5}{4} e^{-2(t-4)} + \frac{5}{2} (t-4) e^{-2(t-4)} \right] \end{aligned} } \] --- ### **Summary of Steps** 1. **Laplace transform** the ODE and apply initial conditions. 2. **Solve for \( Y(s) \)** in the Laplace domain. 3. **Partial fraction decomposition** to simplify inverse transform. 4. **Inverse Laplace transform** each term, applying the time-shifting theorem for step functions. 5. **Combine** all terms for the final time-domain solution. --- **Final Answer:** \[ \boxed{ \begin{aligned} y(t) =\;& e^{-2t} + 2t e^{-2t} \\ &+ u_2(t) \left[ \frac{3}{4} - \frac{3}{4} e^{-2(t-2)} - \frac{3}{2} (t-2) e^{-2(t-2)} \right] \\ &+ u_4(t) \left[ -\frac{5}{4} + \frac{5}{4} e^{-2(t-4)} + \frac{5}{2} (t-4) e^{-2(t-4)} \right] \end{aligned} } \]