Let's solve the problem step-by-step: --- ### **Step 1: Write the Zero-Input Response (ZIR) Equation** Given system: \[ y^{(4)}(t) + 4y^{(3)}(t) + 7y^{(2)}(t) + 6y^{(1)}(t) + 2y(t) = w(t) - w^{(1)}(t) \] **Zero-input response (ZIR):** Set \( w(t) = 0 \) \[ y^{(4)}(t) + 4y^{(3)}(t) + 7y^{(2)}(t) + 6y^{(1)}(t) + 2y(t) = 0 \] --- ### **Step 2: Initial Conditions** Given: - \( y^{(3)}(0^-) = 1 \) - \( y^{(2)}(0^-) = 0 \) - \( y^{(1)}(0^-) = 2 \) - \( y(0^-) = 0 \) --- ### **Step 3: Find the Characteristic Equation** Assume solution \( y(t) = e^{rt} \): \[ r^4 + 4r^3 + 7r^2 + 6r + 2 = 0 \] --- ### **Step 4: Find the Roots** Let's factor the polynomial, or use rational root theorem or synthetic division. #### **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**. Divide by \( (r+1) \): \[ r^4 + 4r^3 + 7r^2 + 6r + 2 = (r+1)(r^3 + 3r^2 + 4r + 2) \] #### **Try r = -1 again for cubic:** \[ (-1)^3 + 3(-1)^2 + 4(-1) + 2 = -1 + 3 - 4 + 2 = 0 \] So, \( r = -1 \) is a **repeated root**. Divide cubic by \( (r+1) \): \[ r^3 + 3r^2 + 4r + 2 = (r+1)(r^2 + 2r + 2) \] So, \[ r^4 + 4r^3 + 7r^2 + 6r + 2 = (r+1)^2(r^2 + 2r + 2) \] #### **Quadratic roots:** \[ r^2 + 2r + 2 = 0 \implies r = \frac{-2 \pm \sqrt{4-8}}{2} = -1 \pm j \] --- ### **Step 5: General Solution** \[ y(t) = (A + Bt)e^{-t} + Ce^{-t}\cos t + De^{-t}\sin t \] --- ### **Step 6: Apply Initial Conditions** Let’s find \( A, B, C, D \): #### **Compute derivatives:** - \( y(t) = (A + Bt)e^{-t} + Ce^{-t}\cos t + De^{-t}\sin t \) **At \( t = 0 \):** - \( y(0) = A + C \) - \( y'(t) = [B e^{-t} - (A+Bt)e^{-t}] + Ce^{-t}(-\sin t - \cos t) + De^{-t}(\cos t - \sin t) \) - \( y'(0) = B - A - C + D \) - \( y''(0) \) and \( y'''(0) \) (details below) ##### **Calculate derivatives at \( t = 0 \):** **First derivative:** \[ y'(t) = (B - A - Bt)e^{-t} + Ce^{-t}(-\cos t - \sin t) + De^{-t}(\cos t - \sin t) \] At \( t=0 \): - \( y'(0) = (B - A) + C(-1) + D(1) = B - A - C + D \) **Second derivative:** \[ y''(t) = \frac{d}{dt} y'(t) \] We can use product rule, but for \( t = 0 \), it's easier to just substitute \( t=0 \) and use the derivatives: - \( y''(0) = -B + (B-A)(-1) + C(-1 - 0) + D(0 - 1) \) But let's systematically compute: Let’s define: - \( f_1 = (A+Bt)e^{-t} \) - \( f_2 = Ce^{-t}\cos t \) - \( f_3 = De^{-t}\sin t \) \( f_1' = (B - (A+Bt))e^{-t} \) At \( t=0 \): \( f_1'(0) = B - A \) \( f_1'' = \frac{d}{dt}(B - (A+Bt))e^{-t} = (0 - B) e^{-t} - (B - (A+Bt))e^{-t} \) At \( t=0 \): \( f_1''(0) = -B - (B - A) = -2B + A \) \( f_2' = Ce^{-t}(-\cos t - \sin t) \) At \( t=0 \): \( f_2'(0) = C(-1 - 0) = -C \) \( f_2'' = C \left( \frac{d}{dt}(e^{-t}(-\cos t - \sin t)) \right) \) But for simplicity, let's use the known pattern for repeated roots and exponentials. Let's write the system (at \( t=0 \)): 1. \( y(0) = A + C = 0 \implies C = -A \) 2. \( y'(0) = B - A - C + D = 2 \) - Substitute \( C = -A \): - \( y'(0) = B - A - (-A) + D = B - A + A + D = B + D = 2 \) 3. \( y''(0) = \text{(compute below)} = 0 \) 4. \( y'''(0) = \text{(compute below)} = 1 \) Let's compute \( y''(0) \): #### **Second derivative:** \[ y''(t) = \frac{d}{dt} y'(t) \] But let's use a symbolic math tool, or calculate directly: We have: \[ y(t) = (A + Bt)e^{-t} + Ce^{-t}\cos t + De^{-t}\sin t \] Let’s use the general formula for derivatives of \( e^{-t}\cos t \) and \( e^{-t}\sin t \): \[ \frac{d}{dt}(e^{-t}\cos t) = -e^{-t}\cos t - e^{-t}\sin t \] \[ \frac{d}{dt}(e^{-t}\sin t) = -e^{-t}\sin t + e^{-t}\cos t \] So, \[ y'(t) = (B - (A+Bt))e^{-t} + C[-e^{-t}\cos t - e^{-}\sin t] + D[-e^{-t}\sin t + e^{-t}\cos t] \] At \( t=0 \): - \( (A+Bt)e^{-t} \to A \) at \( t=0 \) - \( (B - (A+Bt))e^{-t} \to B - A \) - \( -e^{-t}\cos t \to -1 \) - \( -e^{-t}\sin t \to 0 \) - \( -e^{-t}\sin t \to 0 \) - \( e^{-t}\cos t \to 1 \) So, \[ y'(0) = B - A + C(-1 - 0) + D(0 + 1) = B - A - C + D \] #### **Second derivative:** \[ y''(t) = \frac{d}{dt} y'(t) \] Let's differentiate the terms: 1. \( (B - (A+Bt))e^{-t} \) derivative: - \( -B e^{-t} - (B - (A+Bt))e^{-t} \) - At \( t=0 \): \( -B - (B - A) = -2B + A \) 2. \( C[-e^{-t}\cos t - e^{-t}\sin t] \) derivative: - \( C \times \frac{d}{dt}(-e^{-t}\cos t - e^{-t}\sin t) \) - First term: \( -e^{-t}\cos t \) derivative: \( e^{-t}\cos t + e^{-t}\sin t \) - Second term: \( -e^{-t}\sin t \) derivative: \( e^{-t}\sin t - e^{-t}\cos t \) - So, \( C \times [ (e^{-t}\cos t + e^{-t}\sin t) + (e^{-t}\sin t - e^{-t}\cos t) ] = C \times [2e^{-t}\sin t] \) - At \( t=0 \): 0 3. \( D[-e^{-t}\sin t + e^{-t}\cos t] \) derivative: - \( -e^{-t}\sin t \) derivative: \( -e^{-t}\sin t - e^{-t}\cos t \) - \( e^{-t}\cos t \) derivative: \( -e^{-t}\cos t + e^{-t}\sin t \) - So, total: \( D \times [ -e^{-t}\sin t - e^{-t}\cos t - e^{-t}\cos t + e^{-t}\sin t ] = D \times [ -2e^{-t}\cos t ] \) - At \( t=0 \): \( -2D \) Thus, \[ y''(0) = -2B + A + 0 - 2D \] Set equal to the initial condition: \[ y''(0) = 0 \implies -2B + A - 2D = 0 \] #### **Third derivative:** Let’s differentiate again: - The terms involving \( (A+Bt)e^{-t} \) will become: - The derivative of \( -2B + A \) at \( t=0 \) is: - The term \( -2B \) is constant, its derivative is 0. - The term \( A \) is a constant, its derivative is 0. - The original function \( (A+Bt)e^{-t} \) after three derivatives is: - First derivative: \( (B - (A+Bt))e^{-t} \) - Second: \( -2B + A \) - Third: The derivative of \( -2B + A \) is 0. So the only terms that contribute at \( t=0 \) are from the cosine and sine terms. To save time, let's use a symbolic tool for these, but let's continue with: Let’s summarize: 1. \( y(0) = A + C = 0 \implies C = -A \) 2. \( y'(0) = B + D = 2 \implies D = 2 - B \) 3. \( y''(0) = -2B + A - 2D = 0 \) 4. \( y'''(0) = [ \text{to be computed} ] = 1 \) Let's solve for \( A, B, D \): From (3): \[ -2B + A - 2D = 0 \implies A = 2B + 2D \] From (1): \( C = -A \) From (2): \( D = 2 - B \) Plug into (3): \[ A = 2B + 2D = 2B + 2(2 - B) = 2B + 4 - 2B = 4 \] So \( A = 4 \), thus \( C = -4 \) From earlier, \( D = 2 - B \). Now, let's calculate \( y'''(0) \): #### **Third Derivative:** Let’s write out the third derivative at \( t=0 \): - The third derivative of \( (A+Bt)e^{-t} \) at \( t=0 \): Recall, - First: \( (B - (A+Bt))e^{-t} \) - Second: \( -2B + A \) - Third: 0 So, for the exponential-polynomial part, the third derivative at \( t=0 \) is 0. For the cosine and sine parts, use the derivatives at \( t=0 \): From earlier, after two derivatives, the third derivative at \( t=0 \) for each: - For \( Ce^{-t}\cos t \): - First: \( -C(e^{-t}(\cos t + \sin t))