Let's solve the initial value problem using Laplace transforms step-by-step: Given: \[ y'' + 16y = 32t^2 - 48t + 68 \] with initial conditions: \[ y(0) = 0, \quad y'(0) = 21 \] --- ### **Step 1: Take the Laplace Transform of Both Sides** Recall: - \( \mathcal{L}\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0) \) - \( \mathcal{L}\{y(t)\} = Y(s) \) - \( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \) Apply Laplace transform: \[ \mathcal{L}\{y''\} + 16\mathcal{L}\{y\} = \mathcal{L}\{32t^2 - 48t + 68\} \] \[ (s^2 Y(s) - s y(0) - y'(0)) + 16 Y(s) = 32\mathcal{L}\{t^2\} - 48\mathcal{L}\{t\} + 68\mathcal{L}\{1\} \] Plug in initial conditions \( y(0) = 0, y'(0) = 21 \): \[ (s^2 Y(s) - 21) + 16 Y(s) = 32 \frac{2}{s^3} - 48 \frac{1}{s^2} + 68 \frac{1}{s} \] \[ (s^2 + 16)Y(s) - 21 = \frac{64}{s^3} - \frac{48}{s^2} + \frac{68}{s} \] --- ### **Step 2: Solve for \( Y(s) \)** \[ (s^2 + 16)Y(s) = \frac{64}{s^3} - \frac{48}{s^2} + \frac{68}{s} + 21 \] \[ Y(s) = \frac{64}{s^3(s^2 + 16)} - \frac{48}{s^2(s^2 + 16)} + \frac{68}{s(s^2 + 16)} + \frac{21}{s^2 + 16} \] --- ### **Step 3: Break into Partial Fractions** Let’s split each term as needed. #### (a) \( \frac{68}{s(s^2+16)} \): \[ \frac{68}{s(s^2+16)} = \frac{A}{s} + \frac{Bs+C}{s^2+16} \] Solve: \[ 68 = A(s^2+16) + (Bs+C)s \] Set \( s=0 \): \( 68 = 16A \implies A = \frac{68}{16} = \frac{17}{4} \) Now expand: \[ 68 = A s^2 + 16A + Bs^2 + Cs \] \[ 68 = (A+B)s^2 + Cs + 16A \] So, - \( A+B = 0 \implies B = -A = -\frac{17}{4} \) - \( C = 0 \) So, \[ \frac{68}{s(s^2+16)} = \frac{17}{4}\left(\frac{1}{s}\right) - \frac{17}{4}\left(\frac{s}{s^2+16}\right) \] #### (b) \( \frac{48}{s^2(s^2+16)} \): \[ \frac{48}{s^2(s^2+16)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs+D}{s^2+16} \] \[ 48 = A s(s^2+16) + B(s^2+16) + (Cs+D)s^2 \] Set \( s = 0 \): \( 48 = 16B \implies B = 3 \) Now, \[ 48 = A s^3 + 16A s + B s^2 + 16B + Cs^3 + D s^2 \] \[ 48 = (A+C) s^3 + (B+D) s^2 + 16A s + 16B \] So, - \( A+C = 0 \implies C = -A \) - \( B+D = 0 \implies D = -B = -3 \) - \( 16A = 0 \implies A = 0 \implies C = 0 \) Thus, \[ \frac{48}{s^2(s^2+16)} = \frac{3}{s^2} - \frac{3}{s^2+16} \] #### (c) \( \frac{64}{s^3(s^2+16)} \): \[ \frac{64}{s^3(s^2+16)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{Ds+E}{s^2+16} \] \[ 64 = A s^2(s^2+16) + B s (s^2+16) + C(s^2+16) + (Ds+E)s^3 \] Set \( s=0 \): \( 64 = 16C \implies C = 4 \) Expand: \[ 64 = A s^4 + 16A s^2 + B s^3 + 16B s + C s^2 + 16C + D s^4 + E s^3 \] \[ 64 = (A+D) s^4 + (B+E) s^3 + (16A+C) s^2 + 16B s + 16C \] So, - \( 16C = 64 \implies C = 4 \) - \( 16A + C = 0 \implies 16A + 4 = 0 \implies A = -\frac{1}{4} \) - \( 16B = 0 \implies B = 0 \) - \( D = -A = \frac{1}{4} \) - \( E = -B = 0 \) So, \[ \frac{64}{s^3(s^2+16)} = -\frac{1}{4}\frac{1}{s} + 4\frac{1}{s^3} + \frac{1}{4}\frac{s}{s^2+16} \] --- ### **Step 4: Combine All Terms** Now plug everything back into \( Y(s) \): \[ Y(s) = \left[-\frac{1}{4}\frac{1}{s} + 4\frac{1}{s^3} + \frac{1}{4}\frac{s}{s^2+16}\right] - \left[3\frac{1}{s^2} - 3\frac{1}{s^2+16}\right] + \left[\frac{17}{4}\frac{1}{s} - \frac{17}{4}\frac{s}{s^2+16}\right] + \frac{21}{s^2+16} \] Group similar terms: - \( \frac{1}{s} \) terms: \( -\frac{1}{4} + \frac{17}{4} = 4 \) - \( \frac{1}{s^3} \) term: \( 4 \) - \( \frac{1}{s^2} \) term: \( -3 \) - \( \frac{s}{s^2+16} \) terms: \( \frac{1}{4} - \frac{17}{4} = -4 \) - \( \frac{1}{s^2+16} \) terms: \( 3 + 21 = 24 \) So: \[ Y(s) = 4\frac{1}{s} + 4\frac{1}{s^3} - 3\frac{1}{s^2} - 4\frac{s}{s^2+16} + 24\frac{1}{s^2+16} \] --- ### **Step 5: Take the Inverse Laplace Transform** Recall: - \( \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1 \) - \( \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = t \) - \( \mathcal{L}^{-1}\left\{\frac{1}{s^3}\right\} = \frac{t^2}{2} \) - \( \mathcal{L}^{-1}\left\{\frac{s}{s^2+a^2}\right\} = \cos(at) \) - \( \mathcal{L}^{-1}\left\{\frac{1}{s^2+a^2}\right\} = \frac{1}{a}\sin(at) \) Apply these: \[ y(t) = 4 \cdot 1 + 4 \cdot \frac{t^2}{2} - 3t - 4\cos(4t) + 24 \cdot \frac{1}{4}\sin(4t) \] \[ = 4 + 2t^2 - 3t - 4\cos(4t) + 6\sin(4t) \] --- ## **Final Answer** \[ \boxed{ y(t) = 2t^2 - 3t + 4 + 6\sin(4t) - 4\cos(4t) } \]

Certainly! Let's go through the solution step-by-step in detail, including all calculations and reasoning. --- ## **Problem Restatement** Solve the initial value problem (IVP): \[ y'' + 16 y = 32 t^2 - 48 t + 68, \quad y(0) = 0, \quad y'(0) = 21 \] using Laplace transforms. --- ## **Step 1: Take Laplace Transform of Both Sides** Recall the Laplace transforms: - \( \mathcal{L}\{ y''(t) \} = s^2 Y(s) - s y(0) - y'(0) \) - \( \mathcal{L}\{ y(t) \} = Y(s) \) - \( \mathcal{L}\{ t^n \} = \frac{n!}{s^{n+1}} \) Apply to the differential equation: \[ \mathcal{L}\{ y'' \} + 16 \mathcal{L}\{ y \} = \mathcal{L}\{ 32 t^2 - 48 t + 68 \} \] Plugging in the initial conditions \( y(0) = 0 \), \( y'(0) = 21 \): \[ s^2 Y(s) - s \cdot 0 - 21 + 16 Y(s) = \mathcal{L}\{ 32 t^2 - 48 t + 68 \} \] Simplify the left side: \[ s^2 Y(s) - 21 + 16 Y(s) = \text{RHS} \] Calculate RHS: \[ \mathcal{L}\{32 t^2\} = 32 \cdot \frac{2!}{s^3} = 32 \cdot \frac{2}{s^3} = \frac{64}{s^3} \] \[ \mathcal{L}\{-48 t\} = -48 \cdot \frac{1!}{s^2} = -48 \cdot \frac{1}{s^2} = -\frac{48}{s^2} \] \[ \mathcal{L}\{68\} = 68 \cdot \frac{1}{s} = \frac{68}{s} \] So, \[ s^2 Y(s) - 21 + 16 Y(s) = \frac{64}{s^3} - \frac{48}{s^2} + \frac{68}{s} \] Bring all \(Y(s)\) terms to one side: \[ (s^2 + 16) Y(s) = 21 + \frac{64}{s^3} - \frac{48}{s^2} + \frac{68}{s} \] --- ## **Step 2: Solve for \(Y(s)\)** \[ Y(s) = \frac{21 + \frac{64}{s^3} - \frac{48}{s^2} + \frac{68}{s}}{s^2 + 16} \] Express as a single fraction: \[ Y(s) = \frac{21 (s^2 + 16) + \frac{64}{s^3}(s^2 + 16) - \frac{48}{s^2}(s^2 + 16) + \frac{68}{s}(s^2 + 16)}{s^2 + 16} \] But it's easier to write directly as: \[ Y(s) = \frac{21}{s^2 + 16} + \frac{64}{s^3 (s^2 + 16)} - \frac{48}{s^2 (s^2 + 16)} + \frac{68}{s (s^2 + 16)} \] --- ## **Step 3: Partial Fraction Decomposition** We will individually decompose the four terms: ### Term 1: \[ Y_1(s) = \frac{21}{s^2 + 16} \] **Inverse Laplace:** \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2 + a^2}\right\} = \frac{1}{a} \sin(a t) \] Here, \( a=4 \), so: \[ \mathcal{L}^{-1}\left\{\frac{21}{s^2 + 16}\right\} = 21 \times \frac{1}{4} \sin(4 t) = \frac{21}{4} \sin(4 t) \] --- ### Term 2: \[ Y_2(s) = \frac{64}{s^3 (s^2 + 16)} \] This requires partial fractions: \[ \frac{64}{s^3 (s^2 + 16)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{D s + E}{s^2 + 16} \] Multiply both sides by the denominator: \[ 64 = A s^2 (s^2 + 16) + B s (s^2 + 16) + C (s^2 + 16) + (D s + E) s^3 \] Set \( s=0 \): \[ 64 = C \times 16 \Rightarrow C = \frac{64}{16} = 4 \] Now, expand the other parts: \[ A s^2 (s^2 + 16) = A s^4 + 16 A s^2 \] \[ B s (s^2 + 16) = B s^3 + 16 B s \] \[ C (s^2 + 16) = 4 s^2 + 64 \] \[ (D s + E) s^3 = D s^4 + E s^3 \] Combine all: \[ 64 = (A + D) s^4 + (B + E) s^3 + (16A + 4) s^2 + 16 B s + 64 \] Matching coefficients: - \( s^4 \): \[ A + D = 0 \Rightarrow D = -A \] - \( s^3 \): \[ B + E = 0 \Rightarrow E = -B \] - \( s^2 \): \[ 16A + 4 = 0 \Rightarrow 16A = -4 \Rightarrow A = -\frac{4}{16} = -\frac{1}{4} \] Then: \[ D = -A = \frac{1}{4} \] - \( s^1 \): \[ 16 B = 0 \Rightarrow B=0 \] - Constant term: \[ 64 = 64 \quad \text{(already satisfied)} \] And \( E = -B = 0 \). **Final partial fractions:** \[ \frac{64}{s^3 (s^2 + 16)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{D s + E}{s^2 + 16} \] \[ = -\frac{1}{4} \frac{1}{s} + 0 + \frac{4}{s^3} + \frac{\frac{1}{4}s + 0}{s^2 + 16} \] --- ### **Inverse Laplace of Term 2:** \[ \mathcal{L}^{-1}\left\{ - \frac{1}{4} \frac{1}{s} \right\} = -\frac{1}{4} \times 1 = -\frac{1}{4} \] \[ \mathcal{L}^{-1}\left\{ \frac{4}{s^3} \right\} = 4 \times \frac{t^2}{2} = 2 t^2 \] \[ \mathcal{L}^{-1}\left\{ \frac{\frac{1}{4} s}{s^2 + 16} \right\} = \frac{1}{4} \cos(4 t) \] --- ### **Term 3:** \[ Y_3(s) = - \frac{48}{s^2 (s^2 + 16)} \] Partial fractions: \[ \frac{-48}{s^2 (s^2 + 16)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^2 + 16} \] Multiply through: \[ -48 = A s (s^2 + 16) + B (s^2 + 16) + C s^2 \] Set \( s=0 \): \[ -48 = B \times 16 \Rightarrow B = -3 \] Compare coefficients: - \( s^3 \): \[ A s^3 + C s^2 + 16 A s \] - \( s^2 \): \[ 16A + B + C = \text{coefficient of } s^2 \] From the expression: \[ -48 = A s^3 + (C + 16A) s^2 + 16 A s + 16 B \] Matching coefficients: - \( s^3 \): \[ A = 0 \quad \text{(since no } s^3 \text{ term in RHS)} \] - \( s^2 \): \[ C + 16 A = C + 0 = -3 \Rightarrow C = -3 \] - \( s \): \[ 16 A = 0 \] - Constant: \[ 16 B = -48 \Rightarrow B = -3 \] All consistent. **Inverse Laplace:** \[ \mathcal{L}^{-1}\left\{ -\frac{48}{s^2 (s^2 + 16)} \right\} = -3 \times t + (-3) \times \frac{1}{s^2 + 16} \] Recall: \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2 + a^2}\right\} = \frac{1}{a} \sin(a t) \Rightarrow \text{for } a=4, \quad \frac{1}{4} \sin(4 t) \] So: \[ -3 t - 3 \times \frac{1}{4} \sin(4 t) = -3 t - \frac{3}{4} \sin(4 t) \] --- ### **Term 4:** \[ Y_4(s) = \frac{68}{s (s^2 + 16)} \] Partial fractions: \[ \frac{68}{s (s^2 + 16)} = \frac{A}{s} + \frac{B s + C}{s^2 + 16} \] Multiply both sides by denominator: \[ 68 = A (s^2 + 16) + (B s + C) s \] Set \( s=0 \): \[ 68 = A \times 16 \Rightarrow A = \frac{68}{16} = \frac{17}{4} \] Now, expand: \[ 68 = A s^2 + 16 A + B s^2 + C s \] Matching coefficients: - \( s^2 \): \[ A + B = 0 \Rightarrow B = -A = -\frac{17}{4} \] - \( s \): \[ C s \Rightarrow C \text{ is free (since no } s \text{ term on RHS), but to satisfy the full identity, equate coefficients: } C=0 \] - Constant: \[ 16 A = 16 \times \frac{17}{4} = 4 \times 17 = 68 \quad \text{matches RHS} \] Thus, \[ \frac{68}{s (s^2 + 16)} = \frac{17/4}{s} - \frac{17/4 s}{s^2 + 16} \] Inverse Laplace: \[ \mathcal{L}^{-1}\left\{\frac{17/4}{s}\right\} = \frac{17}{4} \] \[ \mathcal{L}^{-1}\left\{ - \frac{17/4 s}{s^2 + 16} \right\} = - \frac{17}{4} \cos(4 t) \] --- ## **Step 4: Write the total solution \( y(t) \)** Sum all inverse transforms: \[ y(t) = \text{Term 1} + \text{Term 2} + \text{Term 3} + \text{Term 4} \] \[ \boxed{ \begin{aligned} y(t) &= \frac{21}{4} \sin(4 t) + \left( -\frac{1}{4} + 2 t^2 + \frac{1}{4} \cos(4 t) \right) \\ &\quad + \left( -3 t - \frac{3}{4} \sin(4 t) \right) + \left( \frac{17}{4} - \frac{17}{4} \cos(4 t) \right) \end{aligned} } \] Combine like terms: - \( t^2 \): \[ 2 t^2 \]