# Step 1: Solve the Homogeneous Equation First, solve the homogeneous part: \[ y'' + 16y = \] The characteristic equation is: \[ r^2 + 16 = \implies r = \pm4i \] So, the general solution to the homogeneous equation is: \[ y_h(t) = C_1 \cos(4t) + C_2 \sin(4t) \] **Explanation:** This gives us the complementary solution, representing the solution to the equation without the forcing term. --- # Step 2: Find a Particular Solution Assume a particular solution of the form: \[ y_p(t) = A\sin(3t) + B\cos(3t) \] Compute derivatives: \[ y_p' = 3A\cos(3t) - 3B\sin(3t) \] \[ y_p'' = -9A\sin(3t) - 9B\cos(3t) \] Plug into the original equation: \[ (-9A\sin(3t) - 9B\cos(3t)) + 16(A\sin(3t) + B\cos(3t)) = 7\sin(3t) \] \[ (-9A + 16A)\sin(3t) + (-9B + 16B)\cos(3t) = 7\sin(3t) \] \[ (7A)\sin(3t) + (7B)\cos(3t) = 7\sin(3t) \] So, \( 7A = 7 \implies A = 1 \) and \( 7B = \implies B = \). So, \( y_p(t) = \sin(3t) \). **Explanation:** We used the method of undetermined coefficients to guess a particular solution, substituted it, and solved for the coefficients. --- # Final Answer The general solution is: \[ \boxed{y(t) = C_1\cos(4t) + C_2\sin(4t) + \sin(3t)} \]

# Step 1: Solve the Homogeneous Equation First, we solve the homogeneous part of the equation: \[ y'' + 16y = 0 \] The characteristic equation is: \[ r^2 + 16 = 0 \] This simplifies to: \[ r^2 = -16 \implies r = \pm 4i \] So, the general solution to the homogeneous equation is: \[ y_h(t) = C_1 \cos(4t) + C_2 \sin(4t) \] **Explanation:** This solution represents the complementary function, where \(C_1\) and \(C_2\) are constants determined by initial conditions. The roots being purely imaginary indicates oscillatory behavior. --- # Step 2: Find a Particular Solution Next, we find a particular solution to the non-homogeneous equation: \[ y'' + 16y = 7\sin(3t) \] Assume a particular solution of the form: \[ y_p(t) = A\sin(3t) + B\cos(3t) \] Calculating the derivatives: \[ y_p' = 3A\cos(3t) - 3B\sin(3t) \] \[ y_p'' = -9A\sin(3t) - 9B\cos(3t) \] Substituting \(y_p\) and its derivatives into the original equation gives: \[ (-9A\sin(3t) - 9B\cos(3t)) + 16(A\sin(3t) + B\cos(3t)) = 7\sin(3t) \] Simplifying, we have: \[ (-9A + 16A)\sin(3t) + (-9B + 16B)\cos(3t) = 7\sin(3t) \] This results in: \[ (7A)\sin(3t) + (7B)\cos(3t) = 7\sin(3t) \] Setting coefficients equal, we find: \[ 7A = 7 \implies A = 1 \] \[ 7B = 0 \implies B = 0 \] Thus, the particular solution is: \[ y_p(t) = \sin(3t) \] **Explanation:** We used the method of undetermined coefficients to guess a form for the particular solution, calculated its derivatives, and substituted into the equation to solve for coefficients. --- # Final Answer The general solution is: \[ \boxed{y(t) = C_1\cos(4t) + C_2\sin(4t) + \sin(3t)} \]