Let's walk through this **step-by-step**: --- ### 1. ** the System Equation** Given: \[ y^{(4)}(t) + 4y^{(3)}(t) + 7y^{(2)}(t) + 6y^{(1)}(t) + 2y(t) = w(t) - w^{(1)}(t) \] **Zero-Input Response (ZIR):** - Set the input to zero: \( w(t) = 0 \) - The equation becomes: \[ y^{(4)}(t) + 4y^{(3)}(t) + 7y^{(2)}(t) + 6y^{(1)}(t) + 2y(t) = 0 \] --- ### 2. **Write Initial Conditions** \[ y^{(3)}(0^-) = 1, \quad y^{(2)}(0^-) = 0, \quad y^{(1)}(0^-) = 2, \quad y(0^-) = 0 \] --- ### 3. **Solve the Homogeneous Equation** #### a. **Find Characteristic Equation** Let \( y(t) = e^{st} \): \[ s^4 + 4s^3 + 7s^2 + 6s + 2 = 0 \] #### b. **Factor/Find Roots** Let's factor or use the rational root theorem or direct factoring. Let's try possible rational roots: ±1, ±2. Plug \( s = -1 \): \[ (-1)^4 + 4(-1)^3 + 7(-1)^2 + 6(-1) + 2 = 1 - 4 + 7 - 6 + 2 = 0 \] So \( s = -1 \) is a root. Divide \( s^4 + 4s^3 + 7s^2 + 6s + 2 \) by \( (s+1) \): Use synthetic or polynomial long division: \[ (s^4 + 4s^3 + 7s^2 + 6s + 2) \div (s+1) \] - \( s^4 / s = s^3 \rightarrow s^3(s+1) = s^4 + s^3 \) - Subtract: \( (4s^3 - s^3) = 3s^3 \) - \( 3s^3 / s = 3s^2 \rightarrow 3s^2(s+1) = 3s^3 + 3s^2 \) - Subtract: \( (7s^2 - 3s^2) = 4s^2 \) - \( 4s^2 / s = 4s \rightarrow 4s(s+1) = 4s^2 + 4s \) - Subtract: \( (6s - 4s) = 2s \) - \( 2s / s = 2 \rightarrow 2(s+1) = 2s + 2 \) - Subtract: \( (2 - 2) = 0 \) So: \[ s^4 + 4s^3 + 7s^2 + 6s + 2 = (s+1)(s^3 + 3s^2 + 4s + 2) \] Try \( s = -1 \) again in the cubic: \[ (-1)^3 + 3(-1)^2 + 4(-1) + 2 = -1 + 3 - 4 + 2 = 0 \] So \( (s+1) \) is a factor again. Divide: - \( s^3 / s = s^2 \rightarrow s^2(s+1) = s^3 + s^2 \) - Subtract: \( (3s^2 - s^2) = 2s^2 \) - \( 2s^2 / s = 2s \rightarrow 2s(s+1) = 2s^2 + 2s \) - Subtract: \( (4s - 2s) = 2s \) - \( 2s / s = 2 \rightarrow 2(s+1) = 2s + 2 \) - Subtract: \( (2 - 2) = 0 \) So: \[ s^3 + 3s^2 + 4s + 2 = (s+1)(s^2 + 2s + 2) \] Now: \[ s^4 + 4s^3 + 7s^2 + 6s + 2 = (s+1)^2(s^2 + 2s + 2) \] Find roots of \( s^2 + 2s + 2 = 0 \): \[ s = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i \] #### **Summary of Roots:** - \( s_1 = -1 \) (with multiplicity 2) - \( s_2 = -1 + i \) - \( s_3 = -1 - i \) --- ### 4. **Write the General Solution** \[ y(t) = (A + Bt)e^{-t} + Ce^{(-1 + i)t} + De^{(-1 - i)t} \] But, to avoid complex coefficients, express complex exponentials as sines and cosines: \[ Ce^{(-1 + i)t} + De^{(-1 - i)t} = e^{-t}[E \cos t + F \sin t] \] So, the general solution is: \[ \boxed{ y(t) = (A + Bt)e^{-t} + e^{-t}\left[ E \cos t + F \sin t \right] } \] Or, \[ \boxed{ y(t) = (A + Bt + E \cos t + F \sin t) e^{-t} } \] --- ### 5. **Apply Initial Conditions** Recall: - \( y(0^-) = 0 \) - \( y^{(1)}(0^-) = 2 \) - \( y^{(2)}(0^-) = 0 \) - \( y^{(3)}(0^-) = 1 \) First, we need to compute derivatives at \( t = 0 \): Let’s write: \[ y(t) = (A + Bt + E \cos t + F \sin t) e^{-t} \] Let \( f(t) = A + Bt + E \cos t + F \sin t \), so: \[ y(t) = f(t) e^{-t} \] #### First Derivative: \[ y'(t) = f'(t)e^{-t} - f(t)e^{-t} \] \[ = e^{-t}\left[ f'(t) - f(t) \right] \] #### Second Derivative: \[ y''(t) = \frac{d}{dt}\left( e^{-t}[f'(t) - f(t)] \right) \] Product rule again: \[ = -e^{-t}[f'(t) - f(t)] + e^{-t}[f''(t) - f'(t)] \] \[ = e^{-t} \left( f''(t) - 2f'(t) + f(t) \right) \] #### Third Derivative: \[ y'''(t) = \frac{d}{dt} \left( e^{-t}[f''(t) - 2f'(t) + f(t)] \right) \] \[ = -e^{-t}[f''(t) - 2f'(t) + f(t)] + e^{-t}[f'''(t) - 2f''(t) + f'(t)] \] \[ = e^{-t} [f'''(t) - 3f''(t) + 3f'(t) - f(t)] \] #### Fourth Derivative: \[ y^{(4)}(t) = \frac{d}{dt} ( e^{-t}[f'''(t) - 3f''(t) + 3f'(t) - f(t)] ) \] \[ = -e^{-t}[f'''(t) - 3f''(t) + 3f'(t) - f(t)] + e^{-t}[f^{(4)}(t) - 3f'''(t) + 3f''(t) - f'(t)] \] \[ = e^{-t}[f^{(4)}(t) - 4f'''(t) + 6f''(t) - 4f'(t) + f(t)] \] Now, evaluate everything at \( t = 0 \): - \( e^{-0} = 1 \) - \( f(0) = A + E \) - \( f'(t) = B - E \sin t + F \cos t \), so \( f'(0) = B + F \) - \( f''(t) = -E \cos t - F \sin t \), so at \( t=0 \): \( f''(0) = -E \) - \( f'''(t) = E \sin t - F \cos t \), so at \( t=0 \): \( f'''(0) = -F \) - \( f^{(4)}(t) = -E \cos t - F \sin t \), at \( t=0 \): \( f^{(4)}(0) = -E \) Now, let's fill in the values: #### At \( t = 0 \): - \( f(0) = A + E \) - \( f'(0) = B + F \) - \( f''(0) = -E \) - \( f'''(0) = -F \) - \( f^{(4)}(0) = -E \) So, \[ \begin{align*} y(0) &= f(0) = A + E \\ y'(0) &= f'(0) - f(0) = (B + F) - (A + E) = B + F - A - E \\ y''(0) &= f''(0) - 2f'(0) + f(0) = -E - 2(B + F) + (A + E) = -E - 2B - 2F + A + E = A - 2B - 2F \\ y'''(0) &= f'''(0) - 3f''(0) + 3f'(0) - f(0) \\ &= -F - 3(-E) + 3(B + F) - (A + E) \\ &= -F + 3E + 3B + 3F - A - E \\ &= 3B + 2E + 2F - A \end{align*} \] Now match to initial conditions: - \( y(0) = 0 \) - \( y'(0) = 2 \) - \( y''(0) = 0 \) - \( y'''(0) = 1 \) So, \[ \begin{cases} A + E = 0 \hspace{2cm} (1) \\ B + F - A - E = 2 \hspace{2cm} (2) \\ A - 2B - 2F = 0 \hspace{2cm} (3) \\ 3B + 2E + 2F - A = 1 \hspace{2cm} (4) \\ \end{cases} \] Let's solve step by step. ### 6. **Solve for Coefficients** From (1): \( A = -E \) Substitute \( A = -E \) into (2), (3), (4): (2): \( B + F - (-E) - E = 2 \rightarrow B + F + E - E = 2 \rightarrow B + F = 2 \) (3): \( -E - 2B - 2F = 0 \rightarrow -E = 2B + 2F \rightarrow -E = 2(B + F) \) But from above, \( B + F = 2 \rightarrow -E = 4 \rightarrow E = -4 \). Now \( A = -E = 4 \). Now, \( B + F = 2 \). Now, substitute \( A = 4 \), \( E = -4 \) into (4): \( 3B + 2E + 2F - A = 1 \) \[ 3B + 2(-4) + 2F - 4 = 1 \\ 3B - 8 + 2F - 4 = 1 \\ 3B + 2F - 12 = 1 \\ 3B + 2F = 13 \] Recall \( B + F = 2 \). Let’s solve: - \( B + F = 2 \) — (a) - \( 3B + 2F = 13 \) — (b) Multiply (a) by 2: \( 2B + 2F = 4 \) Subtract from (b): \( (3B + 2F) - (2B + 2F) = 13 -