Let's solve the differential equation in **2 steps** with explanations: ### Given: \[ y'' + 9y = 1 - \cos 3x + 4\sin 3x \] --- ## **Step 1: Solve the Homogeneous Equation** Solve \( y'' + 9y = 0 \). **Characteristic equation:** \[ r^2 + 9 = 0 \implies r^2 = -9 \implies r = \pm 3i \] **General solution to the homogeneous part:** \[ y_h(x) = C_1 \cos 3x + C_2 \sin 3x \] **Explanation:** We find the complementary solution by solving the homogeneous equation. The characteristic equation gives complex roots, so the solution is a combination of sine and cosine. --- ## **Step 2: Find a Particular Solution** Guess a particular solution of the form: \[ y_p(x) = A + B \cos 3x + D \sin 3x \] Plug into the original equation: First, compute derivatives: \[ y_p' = -3B \sin 3x + 3D \cos 3x \] \[ y_p'' = -9B \cos 3x - 9D \sin 3x \] Substitute \( y_p \) and \( y_p'' \) into the equation: \[ (-9B \cos 3x - 9D \sin 3x) + 9(A + B \cos 3x + D \sin 3x) = 1 - \cos 3x + 4\sin 3x \] Simplify: \[ -9B \cos 3x - 9D \sin 3x + 9A + 9B \cos 3x + 9D \sin 3x = 1 - \cos 3x + 4\sin 3x \] \[ 9A = 1 \\ 0 \cdot \cos 3x = -\cos 3x \\ 0 \cdot \sin 3x = 4\sin 3x \] So, - \( 9A = 1 \implies A = \frac{1}{9} \) - For \( \cos 3x \): \( 0 = -1 \implies B = -\frac{1}{9} \) - For \( \sin 3x \): \( 0 = 4 \implies D = \frac{4}{9} \) So, the particular solution is: \[ y_p(x) = \frac{1}{9} - \frac{1}{9}\cos 3x + \frac{4}{9} \sin 3x \] **Explanation:** We use the method of undetermined coefficients. Substitute and equate coefficients to solve for unknowns. --- ## **Final General Solution** \[ \boxed{ y(x) = C_1 \cos 3x + C_2 \sin 3x + \frac{1}{9} - \frac{1}{9}\cos 3x + \frac{4}{9} \sin 3x } \] Or, grouping like terms: \[ \boxed{ y(x) = (C_1 - \frac{1}{9})\cos 3x + (C_2 + \frac{4}{9})\sin 3x + \frac{1}{9} } \]