Let's work **step by step** through the problem: --- ### **Step 1: Write the Zero-Input Response (ZIR) Differential Equation** Given: \[ y^{(4)}(t) + 4y^{(3)}(t) + 7y^{(2)}(t) + 6y^{(1)}(t) + 2y(t) = w(t) - w^{(1)}(t) \] For **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: \[ \begin{aligned} &y^{(3)}(0^-) = 1\\ &y^{(2)}(0^-) = 0\\ &y^{(1)}(0^-) = 2\\ &y(0^-) = 0 \end{aligned} \] --- ### **Step 3: Find the Characteristic Equation** Assume solution \( y(t) = e^{rt} \): \[ r^4 + 4r^3 + 7r^2 + 6r + 2 = 0 \] Let's factor or solve the quartic equation. --- #### **Step 3a: Try Rational Roots** Possible rational roots (±factors of 2 over factors of 1): ±1, ±2. Test \( r = -1 \): \[ (-1)^4 + 4(-1)^3 + 7(-1)^2 + 6(-1) + 2 = 1 - 4 + 7 - 6 + 2 = 0 \] So, \( r = -1 \) is a root. Divide the quartic by \( (r + 1) \): Perform synthetic division or polynomial division: Divide \( r^4 + 4r^3 + 7r^2 + 6r + 2 \) by \( r + 1 \): - Coefficients: 1 | 4 | 7 | 6 | 2 Bring down 1: -1 × 1 = -1 ⇒ 4 + (-1) = 3 -1 × 3 = -3 ⇒ 7 + (-3) = 4 -1 × 4 = -4 ⇒ 6 + (-4) = 2 -1 × 2 = -2 ⇒ 2 + (-2) = 0 So, \[ (r + 1)(r^3 + 3r^2 + 4r + 2) = 0 \] 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 also a root of the cubic. Now divide \( r^3 + 3r^2 + 4r + 2 \) by \( r + 1 \): Coefficients: 1 | 3 | 4 | 2 -1 × 1 = -1 ⇒ 3 + (-1) = 2 -1 × 2 = -2 ⇒ 4 + (-2) = 2 -1 × 2 = -2 ⇒ 2 + (-2) = 0 So, \[ (r + 1)(r^2 + 2r + 2) \] So, **full factorization:** \[ (r + 1)^2 (r^2 + 2r + 2) = 0 \] --- ### **Step 4: General Solution** Roots: - \( r_1 = -1 \) (multiplicity 2) - \( r_{2,3} = -1 \pm j \), since \( r^2 + 2r + 2 = 0 \implies r = -1 \pm j \) So, \[ y(t) = (A + Bt) e^{-t} + C e^{(-1 + j)t} + D e^{(-1 - j)t} \] But answer should have **no complex coefficients**. Use Euler's formula: \[ C e^{(-1 + j)t} + D e^{(-1 - j)t} = e^{-t} (E \cos t + F \sin t) \] So, \[ \boxed{ y(t) = (A + Bt) e^{-t} + e^{-t} (E \cos t + F \sin t) } \] or, more compactly, \[ y(t) = e^{-t} [A + Bt + E \cos t + F \sin t] \] --- ### **Step 5: Apply Initial Conditions** Compute derivatives at \( t = 0 \): Let’s write \[ y(t) = e^{-t} (A + Bt + E \cos t + F \sin t) \] Let’s first write all derivatives at \( t = 0 \): #### \( y(0) \): \[ y(0) = e^{0} (A + 0 + E \cdot 1 + F \cdot 0) = A + E = 0 \] #### \( y'(t) \): \[ y'(t) = \frac{d}{dt}\left[e^{-t}(A + Bt + E \cos t + F \sin t)\right] \] Using product rule: \[ = -e^{-t} (A + Bt + E \cos t + F \sin t) + e^{-t}(B + (-E) \sin t + F \cos t) \] So, \[ y'(t) = e^{-t}\left[ -A - Bt - E \cos t - F \sin t + B + F \cos t - E \sin t \right] \] At \( t = 0 \): \[ y'(0) = [ -A - 0 - E \cdot 1 - F \cdot 0 + B + F \cdot 1 - E \cdot 0 ] = -A - E + B + F \] Given \( y'(0^-) = 2 \): \[ y'(0) = -A - E + B + F = 2 \] #### \( y''(t) \): Let’s compute the second derivative at \( t = 0 \): \[ y''(t) = \frac{d}{dt} y'(t) \] First, expand \( y'(t) \): \[ y'(t) = e^{-t} \left[ -A - Bt - E \cos t - F \sin t + B + F \cos t - E \sin t \right] \] Group terms: \[ = e^{-t} \left[ -A + B - Bt + (-E + F) \cos t + (-F - E) \sin t \right] \] Now, \[ y''(t) = \frac{d}{dt} \left[ e^{-t} Q(t) \right] = -e^{-t} Q(t) + e^{-t} Q'(t) \] So, \[ y''(t) = e^{-t} [Q'(t) - Q(t)] \] Let’s compute \( Q(t) \): \[ Q(t) = -A + B - Bt + (-E + F) \cos t + (-F - E) \sin t \] So, \[ Q'(t) = -B + (-E + F)(-\sin t) + (-F - E) \cos t \] So, \[ Q'(t) = -B - (E - F)\sin t + (-F - E)\cos t \] Now, at \( t = 0 \), \(\sin 0 = 0, \cos 0 = 1\): \( Q(0) = -A + B + (-E + F)\cdot 1 + 0 = -A + B - E + F \) \( Q'(0) = -B + 0 + (-F - E) \cdot 1 = -B - F - E \) So, \[ y''(0) = [Q'(0) - Q(0)] = [-B - F - E] - [-A + B - E + F] = -B - F - E + A - B + E - F \] Simplify: - \( -B - B = -2B \) - \( -F - F = -2F \) - \( -E + E = 0 \) So, \[ y''(0) = A - 2B - 2F \] Given \( y''(0^-) = 0 \): \[ A - 2B - 2F = 0 \] #### \( y'''(t) \): Let’s differentiate again: \[ y'''(t) = \frac{d}{dt} y''(t) = -e^{-t}[Q'(t) - Q(t)] + e^{-t}[Q''(t) - Q'(t)] \] \[ = e^{-t}[Q''(t) - 2Q'(t) + Q(t)] \] Compute \( Q''(t) \): - \( Q'(t) = -B - (E - F)\sin t + (-F - E)\cos t \) - \( Q''(t) = -(E - F)\cos t + (-F - E)(-\sin t) \) - \( = -(E - F)\cos t + (F + E)\sin t \) At \( t = 0 \): - \( \cos 0 = 1, \sin 0 = 0 \) So, \[ Q''(0) = -(E - F) \cdot 1 + (F + E) \cdot 0 = -E + F \] Recall, - \( Q'(0) = -B - F - E \) - \( Q(0) = -A + B - E + F \) So, \[ y'''(0) = Q''(0) - 2Q'(0) + Q(0) \] Plug in values: \[ = [-E + F] - 2(-B - F - E) + [-A + B - E + F] \] \[ = -E + F + 2B + 2F + 2E -A + B - E + F \] Now combine like terms: - \( -A \) - \( 2B + B = 3B \) - \( -E + 2E - E = 0 \) - \( F + 2F + F = 4F \) So, \[ y'''(0) = -A + 3B + 4F \] Given \( y'''(0^-) = 1 \): \[ -A + 3B + 4F = 1 \] --- ### **Step 6: Solve for Coefficients Using Initial Conditions** We have the following system: \[ \begin{cases} A + E = 0 \\ -A - E + B + F = 2 \\ A - 2B - 2F = 0 \\ -A + 3B + 4F = 1 \end{cases} \] Let's solve step by step. #### From 1: \( A + E = 0 \implies E = -A \) Plug \( E = -A \) into the rest. #### 2: \( -A - E + B + F = 2 \implies -A - (-A) + B + F = 2 \implies 0 + B + F = 2 \implies B + F = 2 \) #### 3: \( A - 2B - 2F = 0 \implies A = 2B + 2F \) #### 4: \( -A + 3B + 4F = 1 \) Now substitute \( A = 2B + 2F \): So equation 4: \[ -(2B + 2F) + 3B + 4F = 1 \implies -2B - 2F + 3B + 4F = 1 \implies (3B - 2B) + (4F - 2F) = 1 \implies B + 2F = 1 \] Now, we have: - \( B + F = 2 \) (from step 2) - \( B + 2F = 1 \) (from above) Subtract the first from the second: \[ (B + 2F) - (B + F) = 1 - 2 \implies (B + 2F - B - F) = -1 \implies F = -1 \] Now plug \( F = -1 \) back into \( B + F = 2 \): \[ B + (-1) = 2 \implies B = 3 \] Now plug \( B = 3, F = -1 \) into \( A = 2B + 2F \): \[ A = 2(3) + 2(-1) = 6 - 2 = 4 \] Now \( E = -A = -4