Let's solve the problem **step by step**. --- ## **Step 1: Write the Homogeneous Equation (Zero-Input Response)** For **zero-input response** (ZIR), we set the input to zero: \[ y^{(4)}(t) + 4y^{(3)}(t) + 7y^{(2)}(t) + 6y^{(1)}(t) + 2y(t) = 0 \] --- ## **Step 2: Write the Initial Conditions** \[ y^{(3)}(0^-) = 1, \quad y^{(2)}(0^-) = 0, \quad y^{(1)}(0^-) = 2, \quad y(0^-) = 0 \] --- ## **Step 3: Assume a Solution and Find the Characteristic Equation** Assume \( y(t) = e^{rt} \): \[ r^4 + 4r^3 + 7r^2 + 6r + 2 = 0 \] --- ## **Step 4: Factor or Solve the Characteristic Equation** Let's attempt to factor or use the **Rational Root Theorem** to find real roots. ### Try \( r = -1 \): \[ (-1)^4 + 4(-1)^3 + 7(-1)^2 + 6(-1) + 2 = 1 - 4 + 7 - 6 + 2 = 0 \] So, \( r = -1 \) is a root. Now, divide the polynomial by \( (r + 1) \): ### Polynomial Division Divide \( r^4 + 4r^3 + 7r^2 + 6r + 2 \) by \( (r + 1) \): Using synthetic or long division, we get: \[ r^4 + 4r^3 + 7r^2 + 6r + 2 = (r + 1)(r^3 + 3r^2 + 4r + 2) \] Now, factor \( r^3 + 3r^2 + 4r + 2 \). Try \( r = -1 \) again: \[ (-1)^3 + 3(-1)^2 + 4(-1) + 2 = -1 + 3 - 4 + 2 = 0 \] So, \( r = -1 \) is a repeated root. Now, divide \( r^3 + 3r^2 + 4r + 2 \) by \( (r + 1) \): \[ r^3 + 3r^2 + 4r + 2 = (r + 1)(r^2 + 2r + 2) \] So, the factorization is: \[ r^4 + 4r^3 + 7r^2 + 6r + 2 = (r + 1)^2(r^2 + 2r + 2) \] Now, \( r^2 + 2r + 2 = 0 \) has roots: \[ r = \frac{-2 \pm \sqrt{4 - 8}}{2} = -1 \pm j \] --- ## **Step 5: Write the General Solution** \[ y(t) = (A + Bt) e^{-t} + Ce^{-t} \cos t + D e^{-t} \sin t \] --- ## **Step 6: Apply Initial Conditions** First, compute derivatives. ### \( y(t) = (A + Bt)e^{-t} + Ce^{-t}\cos t + D e^{-t}\sin t \) Let's expand derivatives: #### \( y(0) \): \[ y(0) = (A + 0) \cdot 1 + C \cdot 1 \cdot 1 + D \cdot 1 \cdot 0 = A + C \] Set \( y(0) = 0 \) (from \( y(0^-) = 0 \)): \[ A + C = 0 \tag{1} \] #### \( y'(t) \): \[ y'(t) = \frac{d}{dt} \left[ (A + Bt)e^{-t} \right] + \frac{d}{dt}\left[ Ce^{-t}\cos t \right] + \frac{d}{dt}\left[ D e^{-t}\sin t \right] \] - \( \frac{d}{dt}\left[(A + Bt) e^{-t}\right] = B e^{-t} + (A + Bt)(-e^{-t}) = B e^{-t} - (A + Bt) e^{-t} = (B - A - Bt) e^{-t} \) - \( \frac{d}{dt}[C e^{-t}\cos t] = C(-e^{-t}\cos t + e^{-t}(-\sin t)) = C e^{-t}(-\cos t - \sin t) \) - \( \frac{d}{dt}[D e^{-t}\sin t] = D(-e^{-t}\sin t + e^{-t}\cos t) = D e^{-t}(-\sin t + \cos t) \) Add all together: \[ y'(t) = (B - A - Bt) e^{-t} + C e^{-t}(-\cos t - \sin t) + D e^{-t}(-\sin t + \cos t) \] At \( t = 0 \): - \( e^{-0} = 1 \) - \( \cos 0 = 1 \) - \( \sin 0 = 0 \) So, \[ y'(0) = (B - A) + C(-1 - 0) + D(0 + 1) = (B - A) - C + D \] Set \( y'(0) = 2 \): \[ B - A - C + D = 2 \tag{2} \] #### \( y''(t) \): Differentiate \( y'(t) \): \[ y'(t) = (B - A - Bt) e^{-t} + C e^{-t}(-\cos t - \sin t) + D e^{-t}(-\sin t + \cos t) \] Let’s compute each term: - \( \frac{d}{dt}[(B - A - Bt)e^{-t}] = -B e^{-t} + (B - A - Bt)(-e^{-t}) = -B e^{-t} - (B - A - Bt)e^{-t} = (-B - B + A + Bt) e^{-t} = (A + Bt - 2B) e^{-t} \) Now, \( C e^{-t}(-\cos t - \sin t) \): - \( \frac{d}{dt}[C e^{-t}(-\cos t - \sin t)] \) - \( = C \frac{d}{dt} [e^{-t}(-\cos t - \sin t)] \) - \( \frac{d}{dt}[e^{-t}(-\cos t - \sin t)] = -e^{-t}(-\cos t - \sin t) + e^{-t}(\sin t - \cos t) \) - \( = e^{-t}(\cos t + \sin t) + e^{-t}(\sin t - \cos t) = e^{-t}(2\sin t) \) - So the derivative is \( C \cdot 2 e^{-t} \sin t \) Now, \( D e^{-t}(-\sin t + \cos t) \): - \( \frac{d}{dt}[D e^{-t}(-\sin t + \cos t)] \) - \( = D \frac{d}{dt}[e^{-t}(-\sin t + \cos t)] \) - \( = D(-e^{-t}(-\sin t + \cos t) + e^{-t}(-\cos t - \sin t)) \) - \( = D( e^{-t}(\sin t - \cos t) + e^{-t}(-\cos t - \sin t) ) \) - \( = D e^{-t}((\sin t - \cos t) + (-\cos t - \sin t)) \) - \( = D e^{-t}(-2\cos t) \) Now sum up: \[ y''(t) = (A + Bt - 2B) e^{-t} + 2 C e^{-t} \sin t - 2 D e^{-t} \cos t \] At \( t = 0 \): - \( e^{-0} = 1 \) - \( \sin 0 = 0 \) - \( \cos 0 = 1 \) So, \[ y''(0) = (A + 0 - 2B) + 2C \cdot 0 - 2D \cdot 1 = (A - 2B) - 2D \] Set \( y''(0) = 0 \): \[ A - 2B - 2D = 0 \tag{3} \] #### \( y'''(t) \): Differentiate \( y''(t) \): - \( \frac{d}{dt}[(A + Bt - 2B) e^{-t}] = (B) e^{-t} + (A + Bt - 2B)(-e^{-t}) = B e^{-t} - (A + Bt - 2B) e^{-t} = (B - A - Bt + 2B) e^{-t} = (3B - A - Bt) e^{-t} \) - \( \frac{d}{dt}[2C e^{-t} \sin t] = 2C(-e^{-t} \sin t + e^{-t} \cos t) = 2C e^{-t}(\cos t - \sin t) \) - \( \frac{d}{dt}[-2D e^{-t} \cos t] = -2D(-e^{-t} \cos t + e^{-t}(-\sin t)) = -2D(e^{-t} \cos t + e^{-t} \sin t) = -2D e^{-t} (\cos t + \sin t) \) Sum up: \[ y'''(t) = (3B - A - Bt) e^{-t} + 2C e^{-t}(\cos t - \sin t) - 2D e^{-t}(\cos t + \sin t) \] At \( t = 0 \): - \( e^{-0} = 1 \) - \( \cos 0 = 1 \) - \( \sin 0 = 0 \) So, \[ y'''(0) = (3B - A - 0) + 2C (1-0) - 2D (1+0) = 3B - A + 2C - 2D \] Set \( y'''(0) = 1 \): \[ 3B - A + 2C - 2D = 1 \tag{4} \] --- ## **Step 7: Solve the System of Equations** From above, we have: 1. \( A + C = 0 \) 2. \( B - A - C + D = 2 \) 3. \( A - 2B - 2D = 0 \) 4. \( 3B - A + 2C - 2D = 1 \) Let's solve step by step. ### From (1): \( C = -A \) Plug into the rest: 2. \( B - A - (-A) + D = 2 \implies B - A + A + D = 2 \implies B + D = 2 \) 3. \( A - 2B - 2D = 0 \) 4. \( 3B - A + 2(-A) - 2D = 1 \implies 3B - A - 2A - 2D = 1 \implies 3B - 3A - 2D = 1 \) Now, we have: - \( C = -A \) - \( B + D = 2 \) (a) - \( A - 2B - 2D = 0 \) (b) - \( 3B - 3A - 2D = 1 \) (c) From (a): \( D = 2 - B \) Plug into (b): \[ A - 2B - 2(2 - B) = 0 \\ A - 2B - 4 + 2B = 0 \\ A - 4 = 0 \implies A = 4 \] So, \( C = -A = -4 \) Now, from (a): \( D = 2 - B \) Plug these into (c): \[ 3B - 3A - 2D = 1 \\ 3B - 3(4) - 2(2 - B) = 1 \\ 3B - 12 - 4 + 2B = 1 \\ (3B + 2B) - 16 = 1 \\ 5B = 17 \\ B = \frac{17}{5} \] Now, \( D = 2 - B = 2 - \frac{17}{5} = \frac{10 - 17}{5} = -\frac{7}{5} \) So, summarizing: - \(