Let's solve **Question 1.1** step by step using the **D-operator method**. --- ### **Q1.1** \[(D^2 - 2D + 5)y = x + 5\] #### **Step 1: Write the Operator Equation** Let \( D = \frac{d}{dx} \): \[(D^2 - 2D + 5)y = x + 5\] --- #### **Step 2: Find the Complementary Function (CF)** Solve the homogeneous part: \[(D^2 - 2D + 5)y = 0\] The auxiliary equation: \[ m^2 - 2m + 5 = 0 \] Solve for \( m \): \[ m = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i \] So the complementary function is: \[ y_{CF} = e^{x} \left( A \cos 2x + B \sin 2x \right) \] --- #### **Step 3: Find a Particular Integral (PI)** The PI is: \[ y_{PI} = \frac{1}{D^2 - 2D + 5}(x + 5) \] Let's find the PI **term by term**. ##### **A. For 5:** \[ \frac{1}{D^2 - 2D + 5}(5) \] For a constant, try \( y_p = C \). Apply \( D^2 - 2D + 5 \) to \( C \): \[ (D^2 - 2D + 5)C = 5C \] Set equal to 5: \[ 5C = 5 \implies C = 1 \] ##### **B. For \( x \):** Try \( y_p = ax + b \). \[ D(y_p) = a, \quad D^2(y_p) = 0 \] \[ (D^2 - 2D + 5)(ax + b) = 0 - 2a + 5(ax + b) = 5ax + 5b - 2a \] Set equal to \( x \): \[ 5ax + 5b - 2a = x \] Equate coefficients: - Coefficient of \( x \): \( 5a = 1 \implies a = \frac{1}{5} \) - Constant: \( 5b - 2a = 0 \implies b = \frac{2a}{5} = \frac{2}{25} \) So the PI for \( x \) is \( \frac{1}{5}x + \frac{2}{25} \). ##### **Combine PI terms:** Sum both PIs: \[ y_{PI} = \frac{1}{5}x + \frac{2}{25} + 1 = \frac{1}{5}x + \frac{27}{25} \] --- #### **Step 4: General Solution** Combine CF and PI: \[ \boxed{ y(x) = e^{x} \left[ A \cos 2x + B \sin 2x \right] + \frac{1}{5}x + \frac{27}{25} } \] --- ### **Final Answer:** \[ \boxed{ y(x) = e^{x} \left[ A \cos 2x + B \sin 2x \right] + \frac{1}{5}x + \frac{27}{25} } \] --- Let me know if you want a step-by-step for the next part!