Let's solve the differential equation step by step: ## Given: \[ y'' + 4y' + 5y = 15x + 5e^{-2x} \] --- ### **a. Find a particular solution (\(y_p\))** The right side is \(15x + 5e^{-2x}\). We guess the particular solution as the sum for each part: #### For \(15x\) (polynomial part): Guess: \(y_{p1} = Ax + B\) Calculate derivatives: \[ y_{p1}' = A \\ y_{p1}'' = 0 \] Plug into the equation: \[ 0 + 4A + 5(Ax + B) = 15x \] \[ 5Ax + (4A + 5B) = 15x \] Matching coefficients: - \(5A = 15 \implies A = 3\) - \(4A + 5B = 0 \implies 4(3) + 5B = 0 \implies B = -\frac{12}{5}\) So, the particular solution for the polynomial part is: \[ y_{p1} = 3x - \frac{12}{5} \] --- #### For \(5e^{-2x}\) (exponential part): Guess: \(y_{p2} = Ce^{-2x}\) Calculate derivatives: \[ y_{p2}' = -2Ce^{-2x} \\ y_{p2}'' = 4Ce^{-2x} \] Plug into the equation: \[ 4Ce^{-2x} + 4(-2Ce^{-2x}) + 5Ce^{-2x} = 5e^{-2x} \] \[ [4C - 8C + 5C]e^{-2x} = 5e^{-2x} \] \[ (C)e^{-2x} = 5e^{-2x} \implies C = 5 \] So, the particular solution for the exponential part is: \[ y_{p2} = 5e^{-2x} \] --- #### Combine both parts: \[ \boxed{y_p = 3x - \frac{12}{5} + 5e^{-2x}} \] --- ### **b. General solution to the homogeneous equation (\(y_h\))** The homogeneous equation is: \[ y'' + 4y' + 5y = 0 \] The characteristic equation: \[ r^2 + 4r + 5 = 0 \] Solve using quadratic formula: \[ r = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i \] So, the general solution is: \[ \boxed{y_h = c_1 e^{-2x} \cos{x} + c_2 e^{-2x} \sin{x}} \] --- ### **c. General solution to the original nonhomogeneous equation** The general solution is the sum of the homogeneous and particular solutions: \[ \boxed{ y = c_1 e^{-2x} \cos{x} + c_2 e^{-2x} \sin{x} + 3x - \frac{12}{5} + 5e^{-2x} } \] --- ## **Summary of Answers** **a.** \( y_p = 3x - \frac{12}{5} + 5e^{-2x} \) **b.** \( y_h = c_1 e^{-2x} \cos{x} + c_2 e^{-2x} \sin{x} \) **c.** \( y = c_1 e^{-2x} \cos{x} + c_2 e^{-2x} \sin{x} + 3x - \frac{12}{5} + 5e^{-2x} \)