Let's break down the solution into two main steps as requested: --- ## **Step 1: Find the Exact Solution** Given ODE: \[ \frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = \sin(t) \] ### **Step 1a: Solve the Homogeneous Equation** First, solve: \[ \frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = \] Characteristic equation: \[ r^2 + 4r + 4 = \implies (r+2)^2 = \implies r = -2 \text{ (double root)} \] Homogeneous solution: \[ y_h(t) = (A + Bt) e^{-2t} \] **Explanation:** The characteristic equation yields a repeated root at \( r = -2 \), giving the standard solution form for repeated real roots. --- ### **Step 1b: Find a Particular Solution** Assume a particular solution of the form: \[ y_p(t) = a\sin(t) + b\cos(t) \] Compute derivatives: \[ y_p' = a\cos(t) - b\sin(t) \] \[ y_p'' = -a\sin(t) - b\cos(t) \] Substitute into the ODE: \[ y_p'' + 4y_p' + 4y_p = \sin(t) \] \[ (-a\sin t - b\cos t) + 4(a\cos t - b\sin t) + 4(a\sin t + b\cos t) = \sin t \] Simplify: \[ (-a + 4a)\sin t + (-b + 4b)\cos t + 4(a\cos t - b\sin t) = \sin t \] \[ (3a - 4b)\sin t + (3b + 4a)\cos t = \sin t \] Set coefficients equal: \[ 3a - 4b = 1 \\ 3b + 4a = \] From the second equation: \( 3b + 4a = \implies b = -\frac{4}{3}a \) Substitute into the first: \[ 3a - 4(-\frac{4}{3}a) = 1 \implies 3a + \frac{16}{3}a = 1 \implies \frac{25}{3}a = 1 \implies a = \frac{3}{25} \] \[ b = -\frac{4}{3} \cdot \frac{3}{25} = -\frac{4}{25} \] So, the particular solution is: \[ y_p = \frac{3}{25}\sin t - \frac{4}{25}\cos t \] **Explanation:** The method of undetermined coefficients is used for the right-hand side \(\sin t\). Solving the system gives the constants for the particular solution. --- ### **Final Exact Solution** \[ \boxed{ y(t) = (A + Bt) e^{-2t} + \frac{3}{25}\sin t - \frac{4}{25}\cos t } \] --- ## **Step 2: Euler’s Method Setup** To use Euler’s method, convert the second-order ODE into a system of first-order ODEs: Let \( y_1 = y \), \( y_2 = \frac{dy}{dt} \): \[ \begin{cases} y_1' = y_2 \\ y_2' = \sin(t) - 4y_2 - 4y_1 \end{cases} \] Euler’s method update (with \( h=.1 \)): \[ y_1^{n+1} = y_1^n + h y_2^n \] \[ y_2^{n+1} = y_2^n + h [\sin(t_n) - 4y_2^n - 4y_1^n] \] **Explanation:** Euler’s method is applied to the system of first-order ODEs derived from the original second-order equation, allowing step-by-step approximation. --- ## **Final Answers (Summary):** 1. **Exact Solution:** \[ \boxed{ y(t) = (A + Bt) e^{-2t} + \frac{3}{25}\sin t - \frac{4}{25}\cos t } \] 2. **Euler’s Method System:** \[ \begin{cases} y_1' = y_2 \\ y_2' = \sin(t) - 4y_2 - 4y_1 \end{cases} \] with update steps: \[ y_1^{n+1} = y_1^n + h y_2^n \] \[ y_2^{n+1} = y_2^n + h [\sin(t_n) - 4y_2^n - 4y_1^n] \] --- **Continue with:** - Assign initial conditions to solve for \(A, B\) if given. - Implement Euler’s method iteratively for \( \leq t \leq 3\). - Compare the numerical (Euler’s) and exact solutions. - Plot both solutions for visualization.

## **Step 1: Find the Exact Solution** ### **Step 1a: Solve the Homogeneous Equation** Given ODE: \[ \frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = 0 \] **Characteristic Equation:** \[ r^2 + 4r + 4 = 0 \implies (r+2)^2 = 0 \implies r = -2 \text{ (double root)} \] **Homogeneous Solution:** \[ y_h(t) = (A + Bt)e^{-2t} \] **Explanation:** The characteristic equation yields a repeated root at \( r = -2 \). This results in the standard solution form for repeated real roots, where \( A \) and \( B \) are constants determined by initial conditions. --- ### **Step 1b: Find a Particular Solution** Assume a particular solution of the form: \[ y_p(t) = a\sin(t) + b\cos(t) \] **Derivatives:** \[ y_p' = a\cos(t) - b\sin(t) \] \[ y_p'' = -a\sin(t) - b\cos(t) \] **Substituting into the ODE:** \[ y_p'' + 4y_p' + 4y_p = \sin(t) \] This leads to: \[ (-a\sin(t) - b\cos(t)) + 4(a\cos(t) - b\sin(t)) + 4(a\sin(t) + b\cos(t)) = \sin(t) \] **Combine Like Terms:** \[ (3a - 4b)\sin(t) + (3b + 4a)\cos(t) = \sin(t) \] **Set Coefficients Equal:** 1. \( 3a - 4b = 1 \) 2. \( 3b + 4a = 0 \) From the second equation, express \( b \): \[ b = -\frac{4}{3}a \] **Substituting into the first equation:** \[ 3a - 4(-\frac{4}{3}a) = 1 \implies 3a + \frac{16}{3}a = 1 \implies \frac{25}{3}a = 1 \implies a = \frac{3}{25} \] Calculating \( b \): \[ b = -\frac{4}{3} \cdot \frac{3}{25} = -\frac{4}{25} \] **Particular Solution:** \[ y_p = \frac{3}{25}\sin(t) - \frac{4}{25}\cos(t) \] **Explanation:** The method of undetermined coefficients is applied to find the particular solution. Solving the simultaneous equations gives constants \( a \) and \( b \). --- ### **Final Exact Solution** Combining both solutions: \[ \boxed{ y(t) = (A + Bt)e^{-2t} + \frac{3}{25}\sin(t) - \frac{4}{25}\cos(t) } \] --- ## **Step 2: Euler’s Method Setup** ### **Convert to First-Order System** Let \( y_1 = y \), \( y_2 = \frac{dy}{dt} \): \[ \begin{cases} y_1' = y_2 \\ y_2' = \sin(t) - 4y_2 - 4y_1 \end{cases} \] ### **Euler's Method Updates** Given step size \( h = 0.1 \): \[ y_1^{n+1} = y_1^n + h y_2^n \] \[ y_2^{n+1} = y_2^n + h \left[ \sin(t_n) - 4y_2^n - 4y_1^n \right] \] **Explanation:** Euler’s method approximates the solutions of the first-order system by iteratively updating the values of \( y_1 \) and \( y_2 \) based on their current values and the derivatives. --- ### **Final Answers (Summary)** 1. **Exact Solution:** \[ \boxed{ y(t) = (A + Bt)e^{-2t} + \frac{3}{25}\sin(t) - \frac{4}{25}\cos(t) } \] 2. **Euler’s Method System:** \[ \begin{cases} y_1' = y_2 \\ y_2' = \sin(t) - 4y_2 - 4y_1 \end{cases} \] Update steps: \[ y_1^{n+1} = y_1^n + h y_2^n \] \[ y_2^{n+1} = y_2^n + h \left[ \sin(t_n) - 4y_2^n - 4y_1^n \right] \] --- ### **Next Steps:** 1. Assign initial conditions to solve for \( A \) and \( B \). 2. Implement Euler’s method iteratively for \( 0 \leq t \leq 3 \). 3. Compare the numerical results with the exact solution. 4. Plot both solutions for visualization.