Let's solve the given task in **2 steps** with explanations at each step: --- ### **Step 1: Solve the Differential Equation by Hand** Given: \[ \frac{d^2y}{dt^2} + y = 3\cos(4t) \] with initial conditions: \[ y(0) = 1,\quad y'(0) = 2 \] The general solution to this non-homogeneous equation is: \[ y(t) = y_h(t) + y_p(t) \] #### **Homogeneous solution \(y_h(t)\):** \[ \frac{d^2y}{dt^2} + y = 0 \] The characteristic equation is: \[ r^2 + 1 = 0 \implies r = \pm i \] So, \[ y_h(t) = A_c \cos(t) + A_s \sin(t) \] #### **Particular solution \(y_p(t)\):** Try a particular solution of the form: \[ y_p(t) = A_f \cos(4t) \] Plug into the equation: \[ \frac{d^2}{dt^2}[A_f \cos(4t)] + A_f \cos(4t) = 3\cos(4t) \] \[ -16A_f \cos(4t) + A_f \cos(4t) = 3\cos(4t) \] \[ (-16A_f + A_f)\cos(4t) = 3\cos(4t) \] \[ -15A_f = 3 \implies A_f = -\frac{1}{5} \] So, \[ y_p(t) = -\frac{1}{5} \cos(4t) \] Combine: \[ y(t) = A_c \cos(t) + A_s \sin(t) - \frac{1}{5} \cos(4t) \] Now, use initial conditions to find \(A_c\) and \(A_s\): \[ y(0) = A_c \cos(0) + A_s \sin(0) - \frac{1}{5} \cos(0) = 1 \] \[ A_c - \frac{1}{5} = 1 \implies A_c = 1 + \frac{1}{5} = \frac{6}{5} \] \[ y'(t) = -A_c \sin(t) + A_s \cos(t) + \frac{4}{5} \sin(4t) \] \[ y'(0) = -A_c \sin(0) + A_s \cos(0) + \frac{4}{5} \sin(0) = 2 \] \[ A_s = 2 \] The general solution is: \[ \boxed{y(t) = \frac{6}{5} \cos(t) + 2\sin(t) - \frac{1}{5} \cos(4t)} \] **Explanation:** We split the solution into the homogeneous (oscillatory) part and a particular part matching the driving force. The coefficients are found from the initial conditions. --- ### **Step 2: Identify the Coefficients for MATLAB** The solution is of the form: \[ A_c \cos(\omega t) + A_s \sin(\omega t) + A_f \cos(4t) \] Comparing with our answer: - \(A_c = \frac{6}{5}\) - \(A_s = 2\) - \(\omega = 1\) (since the homogeneous part uses \(\cos(t)\) and \(\sin(t)\)) - \(A_f = -\frac{1}{5}\) **Explanation:** We matched the solution form as required by the problem, reading off the constants for MATLAB. --- ## **Final Answer:** \[ A_c = \frac{6}{5},\quad A_s = 2,\quad \omega = 1,\quad A_f = -\frac{1}{5} \]

**Step 1: Solve the differential equation by hand** Given: \[ \frac{d^2 y}{dt^2} + y = 3 \cos(4t) \] with initial conditions: \[ y(0) = 1,\quad y'(0) = 2 \] **Homogeneous solution:** The homogeneous equation: \[ \frac{d^2 y}{dt^2} + y = 0 \] has characteristic roots: \[ r^2 + 1 = 0 \Rightarrow r = \pm i \] so the homogeneous solution: \[ y_h(t) = A_c \cos(t) + A_s \sin(t) \] **Particular solution:** Since the RHS is \(3 \cos(4t)\), try a particular solution of the form: \[ y_p(t) = A_f \cos(4t) \] Plug into the differential equation: \[ \frac{d^2 y_p}{dt^2} + y_p = 3 \cos(4t) \] \[ -16 A_f \cos(4t) + A_f \cos(4t) = 3 \cos(4t) \] \[ (-16 A_f + A_f) \cos(4t) = 3 \cos(4t) \] \[ -15 A_f = 3 \Rightarrow A_f = -\frac{1}{5} \] **Full solution:** \[ y(t) = A_c \cos(t) + A_s \sin(t) - \frac{1}{5} \cos(4t) \] **Use initial conditions:** At \(t=0\): \[ y(0) = A_c \cdot 1 + A_s \cdot 0 - \frac{1}{5} \cdot 1 = 1 \] \[ A_c - \frac{1}{5} = 1 \Rightarrow A_c = 1 + \frac{1}{5} = \frac{6}{5} \] Derivative: \[ y'(t) = -A_c \sin(t) + A_s \cos(t) + \frac{4}{5} \sin(4t) \] At \(t=0\): \[ y'(0) = -A_c \cdot 0 + A_s \cdot 1 + 0 = 2 \] \[ A_s = 2 \] --- ### **Summary of constants:** \[ \boxed{ \begin{aligned} A_c &= \frac{6}{5} \\ A_s &= 2 \\ A_f &= -\frac{1}{5} \\ \omega &= 1 \end{aligned} } \] --- **Step 2: Use MATLAB function to verify** The MATLAB function: ```matlab [Ac, As, omegas, Af] = secondUndampedSolver(B, A, omegaf, y0, v0) ``` is designed to find these coefficients for the equation: \[ \frac{d^2 y}{dt^2} + y = B \cos(\omega_f t) \] with initial conditions \(y(0)= y_0\), \(y'(0)=v_0\). Plugging in: ```matlab [B, A, omegaf, y0, v0] = deal(3, 1, 4, 1, 2); ``` Running the solver should produce coefficients close to: \[ A_c \approx \frac{6}{5},\quad A_s \approx 2,\quad \omega \approx 1,\quad A_f \approx -\frac{1}{5} \] which confirms the hand solution. --- ## **Final Answer:** \[ \boxed{ \begin{aligned} A_c &= \frac{6}{5} \\ A_s &= 2 \\ \text{omega} &= 1 \\ A_f &= -\frac{1}{5} \end{aligned} } \]