Let's solve each problem step by step using the **annihilator method**: --- ## **19. Solve using the annihilator method: \( y'' + y = \cos(x) \)** ### **Step 1: Solve the Homogeneous Equation** Solve \( y'' + y = 0 \): The characteristic equation is \( r^2 + 1 = 0 \implies r = \pm i \). General solution: \( y_h = C_1 \cos(x) + C_2 \sin(x) \) **Explanation:** We always solve the associated homogeneous equation to find the complementary solution. --- ### **Step 2: Find a Particular Solution using Annihilator** The right side is \( \cos(x) \). - The annihilator for \( \cos(x) \) is \( (D^2 + 1) \). - Apply the annihilator to both sides: \[ (D^2 + 1)[y'' + y] = (D^2 + 1)[\cos(x)] \implies (D^2 + 1)(D^2 + 1) y = 0 \] Guess a particular solution: \( y_p = A x \cos(x) + B x \sin(x) \). Plug \( y_p \) into the original equation and solve for \( A \) and \( B \): \[ y_p'' + y_p = \cos(x) \] After substitution and simplification, \( y_p = 0 \cdot x \cos(x) + \frac{1}{2} x \sin(x) \). So, \( y_p = \frac{1}{2} x \sin(x) \). **Explanation:** The annihilator turns the nonhomogeneous equation into a higher-order homogeneous equation, allowing us to guess the particular solution and solve for constants. --- #### **Final Answer:** \[ \boxed{ y(x) = C_1 \cos(x) + C_2 \sin(x) + \frac{1}{2} x \sin(x) } \] --- ## **20. Use the annihilator method to solve: \( y'' - y = x e^x \)** ### **Step 1: Solve the Homogeneous Equation** Solve \( y'' - y = 0 \): Characteristic equation: \( r^2 - 1 = 0 \implies r = \pm 1 \). General solution: \( y_h = C_1 e^x + C_2 e^{-x} \) **Explanation:** Finding the general solution for the homogeneous part is always the first step. --- ### **Step 2: Find a Particular Solution using Annihilator** Right side: \( x e^x \). - The annihilator for \( x e^x \) is \( (D-1)^2 \). - Apply to both sides: \[ (D-1)^2 [y'' - y] = (D-1)^2 [x e^x] \implies (D-1)^2(D^2 - 1) y = 0 \] Guess: \( y_p = (A x + B) e^x \). Plug into the original equation: \[ y_p'' - y_p = [A x + B]'' e^x - [A x + B] e^x = x e^x \] This leads to \( y_p = \frac{1}{2} x^2 e^x \). **Explanation:** We use the annihilator to determine the correct form for the particular solution, then solve for coefficients. --- #### **Final Answer:** \[ \boxed{ y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{2} x^2 e^x } \] --- ## **21. Solve: \( y'' - 2y' + y = e^x + x \) using the annihilator method** ### **Step 1: Solve the Homogeneous Equation** Solve \( y'' - 2y' + y = 0 \): Characteristic equation: \( (r - 1)^2 = 0 \implies r = 1 \) (double root). General solution: \( y_h = (C_1 + C_2 x) e^{x} \) **Explanation:** Double root at \( r = 1 \) gives the form above. --- ### **Step 2: Find a Particular Solution using Annihilator** Right side: \( e^x + x \). - Annihilator for \( e^x \) is \( (D-1) \). - Annihilator for \( x \) is \( D^2 \). - The combined annihilator is \( (D-1)D^2 \). Apply to both sides: \[ (D-1)D^2 [y'' - 2y' + y] = (D-1)D^2 [e^x + x] \] Solve particular solution for each term: - For \( e^x \): Since \( e^x \) is a solution to the homogeneous equation (repeated root), guess \( y_{p,1} = A x^2 e^x \). - For \( x \): Guess \( y_{p,2} = B x + C \). Plug both into the original equation, solve for \( A, B, C \): - For \( e^x \) term, \( y_{p,1} = \frac{1}{2} x^2 e^x \). - For \( x \) term, \( y_{p,2} = x + 2 \). Total particular solution: \( y_p = \frac{1}{2} x^2 e^x + x + 2 \). **Explanation:** We use the annihilator for each right-hand term and sum up the particular solutions. --- #### **Final Answer:** \[ \boxed{ y(x) = (C_1 + C_2 x) e^{x} + \frac{1}{2} x^2 e^x + x + 2 } \] --- Let me know if you need step-by-step substitution for the particular solutions!

Certainly! Here are the final solutions for each problem: --- ### **19. Solve \( y'' + y = \cos(x) \)** **Final Answer:** \[ \boxed{ y(x) = C_1 \cos(x) + C_2 \sin(x) + \frac{1}{2} x \sin(x) } \] --- ### **20. Solve \( y'' - y = x e^x \)** **Final Answer:** \[ \boxed{ y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{2} x^2 e^x } \] --- ### **21. Solve \( y'' - 2 y' + y = e^x + x \)** **Final Answer:** \[ \boxed{ y(x) = (C_1 + C_2 x) e^{x} + \frac{1}{2} x^2 e^x + x + 2 } \] --- If you need detailed steps for each particular solution, feel free to ask!