Let's solve the problem **step by step** as requested: --- ## **Step 1: Set up the Particular Solution Integral** The given differential equation is: \[ y'' + 4y' + 4y = e^{-2x} \] The *integral form* for a particular solution is: \[ y_p(x) = \int_{x_}^{x} G(x, t)\, f(t)\, dt \] where \( f(t) = e^{-2t} \) and \( G(x, t) \) is the Green's function for the corresponding homogeneous equation. **For a constant coefficient linear equation:** \[ y'' + 4y' + 4y = f(x) \] The Green's function is: \[ G(x,t) = (x-t)e^{-2(x-t)}\ \text{for}\ x > t \] **So, fill in the blank:** \[ y_p(x) = \int_{x_}^{x} (x-t) e^{-2(x-t)} e^{-2t} dt \] **Explanation:** We replaced \( G(x,t) \) with \( (x-t) e^{-2(x-t)} \) and \( f(t) \) with \( e^{-2t} \). --- ## **Step 2: Write the General Solution** The general solution is: \[ y(x) = y_h(x) + y_p(x) \] where \( y_h(x) \) is the general solution of the homogeneous equation: \[ y'' + 4y' + 4y = \] **Solve the characteristic equation:** \[ r^2 + 4r + 4 = \implies (r+2)^2 = \implies r = -2 \ (\text{double root}) \] So, \[ y_h(x) = (C_1 + C_2 x) e^{-2x} \] **So, fill in the blank:** \[ y(x) = (C_1 + C_2 x) e^{-2x} + y_p(x) \] **Explanation:** The general solution is a sum of the homogeneous solution (based on the characteristic equation) and the particular integral form. --- ## **Final Answers** ### **Step 1 Answer:** \[ \boxed{(x-t) e^{-2(x-t)} e^{-2t}} \] ### **Step 2 Answer:** \[ \boxed{(C_1 + C_2 x) e^{-2x}} \] --- **Final General Solution:** \[ y(x) = (C_1 + C_2 x) e^{-2x} + \int_{x_}^{x} (x-t) e^{-2(x-t)} e^{-2t} dt \]

Let's correct the solution step-by-step. --- ## **Step 1: Set up the Particular Solution Integral** The given differential equation is: \[ y'' + 4y' + 4y = e^{-2x} \] To find a particular solution \( y_p(x) \), we can use the method of undetermined coefficients since the right-hand side is an exponential function. ### **Assume a Form for the Particular Solution** Since the complementary solution will involve \( e^{-2x} \), we will assume a particular solution of the form: \[ y_p(x) = A x e^{-2x} \] where \( A \) is a constant to be determined. ### **Substitute into the Differential Equation** First, we compute the derivatives: - First derivative: \[ y_p' = A(e^{-2x} - 2xe^{-2x}) = A e^{-2x}(1 - 2x) \] - Second derivative: \[ y_p'' = A(-2e^{-2x} + 4xe^{-2x}) = A e^{-2x}(-2 + 4x) \] Now substitute \( y_p, y_p', \) and \( y_p'' \) into the original equation: \[ A e^{-2x}(-2 + 4x) + 4A e^{-2x}(1 - 2x) + 4Ax e^{-2x} = e^{-2x} \] This simplifies to: \[ A e^{-2x}(-2 + 4x + 4 - 8x + 4x) = e^{-2x} \] This reduces to: \[ A e^{-2x}(-2) = e^{-2x} \] Thus, setting coefficients equal gives: \[ -2A = 1 \implies A = -\frac{1}{2} \] **Answer for Step 1:** \[ y_p(x) = -\frac{1}{2} x e^{-2x} \] **Explanation:** We assumed a suitable form for the particular solution and found the constant \( A \) by substituting back into the original differential equation. --- ## **Step 2: Write the General Solution** The general solution is given by: \[ y(x) = y_h(x) + y_p(x) \] ### **Find the Homogeneous Solution** The homogeneous equation is: \[ y'' + 4y' + 4y = 0 \] ### **Solve the Characteristic Equation** The characteristic equation is: \[ r^2 + 4r + 4 = 0 \implies (r + 2)^2 = 0 \implies r = -2 \ (\text{double root}) \] ### **Form of the Homogeneous Solution** Thus, the general solution for the homogeneous part is: \[ y_h(x) = (C_1 + C_2 x) e^{-2x} \] ### **Combine the Solutions** The complete solution becomes: \[ y(x) = (C_1 + C_2 x) e^{-2x} - \frac{1}{2} x e^{-2x} \] This simplifies to: \[ y(x) = \left(C_1 + (C_2 - \frac{1}{2}) x\right) e^{-2x} \] **Answer for Step 2:** \[ y(x) = \left(C_1 + (C_2 - \frac{1}{2}) x\right) e^{-2x} \] **Explanation:** The general solution is a combination of the homogeneous solution and the particular solution, adapted for the double root case. --- ## **Final Answers** ### **Particular Solution:** \[ \boxed{y_p(x) = -\frac{1}{2} x e^{-2x}} \] ### **General Solution:** \[ \boxed{y(x) = \left(C_1 + (C_2 - \frac{1}{2}) x\right) e^{-2x}} \]