Given: \[ y'' - 3y' - 4y = 2e^{-x} \] with a known solution to the **homogeneous equation** \(y_1(x) = e^{-x}\). Let's solve the nonhomogeneous equation using reduction of order. --- ## **Step 1: Find the general solution to the homogeneous equation** The homogeneous equation is: \[ y'' - 3y' - 4y = 0 \] Assume \(y = e^{rx}\): \[ r^2 - 3r - 4 = 0 \] \[ (r-4)(r+1)=0 \implies r=4,\, r=-1 \] So, the general solution to the homogeneous equation is: \[ y_h(x) = C_1 e^{4x} + C_2 e^{-x} \] --- ## **Step 2: Find a particular solution using reduction of order** Since \(y_1(x) = e^{-x}\) is a solution, let's look for a particular solution of the form \[ y_p(x) = v(x) e^{-x} \] where \(v(x)\) is to be determined. Compute derivatives: \[ y_p' = v' e^{-x} - v e^{-x} \] \[ y_p'' = v'' e^{-x} - 2 v' e^{-x} + v e^{-x} \] Plug into the original equation: \[ y_p'' - 3y_p' - 4 y_p = 2e^{-x} \] \[ [v'' e^{-x} - 2 v' e^{-x} + v e^{-x}] - 3 [v' e^{-x} - v e^{-x}] - 4 [v e^{-x}] = 2e^{-x} \] Expand: \[ v'' e^{-x} - 2 v' e^{-x} + v e^{-x} - 3 v' e^{-x} + 3 v e^{-x} - 4 v e^{-x} = 2e^{-x} \] Combine like terms: - \(v'' e^{-x}\) - \(-2v' e^{-x} - 3v' e^{-x} = -5v' e^{-x}\) - \(v e^{-x} + 3 v e^{-x} - 4 v e^{-x} = (1+3-4) v e^{-x} = 0\) So: \[ v'' e^{-x} - 5 v' e^{-x} = 2 e^{-x} \] Divide both sides by \(e^{-x}\): \[ v'' - 5 v' = 2 \] --- ## **Step 3: Solve for \(v(x)\)** The equation: \[ v'' - 5 v' = 2 \] This is a linear ODE with constant coefficients. First, solve the homogeneous part: \[ v'' - 5 v' = 0 \] Characteristic equation: \[ r^2 - 5r = 0 \implies r(r-5) = 0 \implies r = 0,\, r = 5 \] So, \[ v_h(x) = A + B e^{5x} \] Now, find a particular solution to \(v'' - 5 v' = 2\): Guess a particular solution of the form \(v_p(x) = Cx\): \[ v_p' = C,\quad v_p'' = 0 \] \[ v_p'' - 5 v_p' = 0 - 5C = 2 \implies C = -\frac{2}{5} \] So, \[ v_p(x) = -\frac{2}{5} x \] Thus, \[ v(x) = A + B e^{5x} - \frac{2}{5}x \] --- ## **Step 4: Write the particular solution for \(y(x)\)** Recall: \[ y_p(x) = v(x) e^{-x} = \left( A + B e^{5x} - \frac{2}{5}x \right) e^{-x} \] But since \(A e^{-x}\) and \(B e^{4x}\) are already part of the homogeneous solution, the new particular solution is just: \[ y_p(x) = \left( -\frac{2}{5}x \right) e^{-x} = -\frac{2}{5} x e^{-x} \] --- ## **Step 5: General Solution** Combine with the homogeneous solution: \[ \boxed{ y(x) = C_1 e^{4x} + C_2 e^{-x} - \frac{2}{5} x e^{-x} } \]