Let's solve the given initial value problem in two steps, as requested: ### **Step 1: Solve for \( t < 2 \)** The differential equation is: \[ y'' + 4y = t, \quad y(0) = 0,\, y'(0) = 0 \] #### **Homogeneous Solution:** The associated homogeneous equation is \( y'' + 4y = 0 \). The characteristic equation is: \[ r^2 + 4 = 0 \implies r = \pm 2i \] So, the general solution to the homogeneous part is: \[ y_h(t) = C_1 \cos(2t) + C_2 \sin(2t) \] #### **Particular Solution:** Try a particular solution of the form \( y_p(t) = At + B \). Plug into the non-homogeneous equation: \[ y_p'' + 4y_p = 0 + 4(At + B) = 4At + 4B \] We want this to equal \( t \), so: \[ 4A t + 4B = t \implies 4A = 1 \implies A = \frac{1}{4}, \quad 4B = 0 \implies B = 0 \] Thus, the particular solution is \( y_p(t) = \frac{1}{4} t \). #### **General Solution for \( t < 2 \):** \[ y_1(t) = C_1 \cos(2t) + C_2 \sin(2t) + \frac{1}{4}t \] **Apply Initial Conditions:** - \( y(0) = 0 \): \( C_1 + 0 + 0 = 0 \implies C_1 = 0 \) - \( y'(t) = -2C_1\sin(2t) + 2C_2\cos(2t) + \frac{1}{4} \) - \( y'(0) = 2C_2 + \frac{1}{4} = 0 \implies C_2 = -\frac{1}{8} \) So, \[ y_1(t) = -\frac{1}{8} \sin(2t) + \frac{1}{4} t \qquad \text{for } t < 2 \] **Explanation:** We solved the equation for \( t < 2 \) using standard methods for nonhomogeneous ODEs, including finding the particular and homogeneous solutions, then applying initial conditions. --- ### **Step 2: Solve for \( t \geq 2 \) (and match conditions at \( t = 2 \))** Now, the equation is: \[ y'' + 4y = 0 \quad \text{for } t \geq 2 \] The general solution is: \[ y_2(t) = A \cos(2t) + B \sin(2t) \] We need to match \( y \) and \( y' \) at \( t = 2 \): From step 1: \[ y_1(2) = -\frac{1}{8} \sin(4) + \frac{1}{2} \] \[ y_1'(t) = -\frac{1}{8} \cdot 2\cos(2t) + \frac{1}{4} = -\frac{1}{4}\cos(2t) + \frac{1}{4} \] \[ y_1'(2) = -\frac{1}{4} \cos(4) + \frac{1}{4} \] At \( t = 2 \), set: \[ y_2(2) = y_1(2) \] \[ y_2'(2) = y_1'(2) \] Compute derivatives: \[ y_2'(t) = -2A \sin(2t) + 2B \cos(2t) \] So, \[ y_2(2) = A \cos(4) + B \sin(4) = -\frac{1}{8} \sin(4) + \frac{1}{2} \] \[ y_2'(2) = -2A \sin(4) + 2B \cos(4) = -\frac{1}{4} \cos(4) + \frac{1}{4} \] **Solve for \( A \) and \( B \):** System: \[ \begin{cases} A \cos(4) + B \sin(4) = -\frac{1}{8} \sin(4) + \frac{1}{2} \\ -2A \sin(4) + 2B \cos(4) = -\frac{1}{4} \cos(4) + \frac{1}{4} \end{cases} \] **Explanation:** We used continuity of \( y \) and \( y' \) at \( t = 2 \) to set up a system of equations for \( A \) and \( B \). --- ### **Final Answer:** \[ y(t) = \begin{cases} -\frac{1}{8} \sin(2t) + \frac{1}{4} t, & t < 2 \\ A \cos(2t) + B \sin(2t), & t \geq 2 \end{cases} \] Where \( A \) and \( B \) are found by solving: \[ \begin{cases} A \cos(4) + B \sin(4) = -\frac{1}{8} \sin(4) + \frac{1}{2} \\ -2A \sin(4) + 2B \cos(4) = -\frac{1}{4} \cos(4) + \frac{1}{4} \end{cases} \] You can leave your answer in this form, or solve explicitly for \( A \) and \( B \) if desired. Let me know if you want the explicit values!