Solve (D²−3D+2)y=x²+x

The equation is (D² − 3D + 2)y = x² + x.

Complementary function

m² − 3m + 2 = 0, so (m − 1)(m − 2) = 0, m = 1, 2.

CF = c₁eˣ + c₂e^(2x)

Particular integral

PI = 1/(2 − 3D + D²) · (x² + x) = ½ [1 − (3D/2 − D²/2)]⁻¹ (x² + x) = ½ [1 + (3D/2 − D²/2) + (3D/2)² + …] (x² + x) = ½ [1 + 3D/2 + (9/4 − 1/2)D² + …] (x² + x) = ½ [1 + 3D/2 + (7/4)D²] (x² + x)

Now D(x² + x) = 2x + 1 and D²(x² + x) = 2:

= ½ [x² + x + (3/2)(2x + 1) + (7/4)(2)] = ½ [x² + x + 3x + 3/2 + 7/2] = ½ (x² + 4x + 5)

General solution

y = c₁eˣ + c₂e^(2x) + ½(x² + 4x + 5)

(Check: y_p = ½(x² + 4x + 5); y_p' = x + 2, y_p'' = 1; 1 − 3(x + 2) + (x² + 4x + 5) = x² + x ✓.)