1. Problem: Solve the differential operator equation $$y'' + 2y' - 3y = (3x + 2)e^{-2x}$$ using the differential operator $D = \frac{d}{dx}$. 2. Formula: The operator form is $$(D^2 + 2D - 3)y = (3x + 2)e^{-2x}$$. 3. Step 1: Find the complementary function (CF) by solving the homogeneous equation: $$D^2 + 2D - 3 = 0$$ Characteristic equation: $$m^2 + 2m - 3 = 0$$ Factor: $$(m + 3)(m - 1) = 0$$ So, $$m = -3, 1$$ CF: $$y_c = C_1 e^{-3x} + C_2 e^{x}$$ 4. Step 2: Find the particular solution (PS) using the method of undetermined coefficients. Since the right side is $(3x + 2)e^{-2x}$, try: $$y_p = e^{-2x}(Ax + B)$$ 5. Step 3: Compute derivatives: $$y_p' = e^{-2x}(A - 2Ax - 2B)$$ $$y_p'' = e^{-2x}(-4A + 4Ax + 4B)$$ 6. Step 4: Substitute into the left side: $$y_p'' + 2y_p' - 3y_p = e^{-2x}[-4A + 4Ax + 4B + 2(A - 2Ax - 2B) - 3(Ax + B)]$$ Simplify inside brackets: $$-4A + 4Ax + 4B + 2A - 4Ax - 4B - 3Ax - 3B = (-4A + 2A) + (4Ax - 4Ax - 3Ax) + (4B - 4B - 3B)$$ $$= -2A - 3Ax - 3B$$ 7. Step 5: Equate to right side: $$-3Ax - 2A - 3B = 3x + 2$$ Match coefficients: For $x$: $$-3A = 3 \Rightarrow A = -1$$ For constants: $$-2A - 3B = 2$$ Substitute $A = -1$: $$-2(-1) - 3B = 2 \Rightarrow 2 - 3B = 2 \Rightarrow -3B = 0 \Rightarrow B = 0$$ 8. Step 6: Particular solution: $$y_p = e^{-2x}(-x) = -x e^{-2x}$$ 9. Final solution: $$y = y_c + y_p = C_1 e^{-3x} + C_2 e^{x} - x e^{-2x}$$ Answer: $$\boxed{y = C_1 e^{-3x} + C_2 e^{x} - x e^{-2x}}$$