Solve (D²−2D+5)y=e^(2x) sin x

The equation is (D² − 2D + 5)y = e^(2x) sin x.

Complementary function

m² − 2m + 5 = 0, so m = 1 ± 2i.

CF = eˣ(c₁ cos 2x + c₂ sin 2x)

Particular integral

PI = 1/(D² − 2D + 5) · e^(2x) sin x

Replace D by D + 2:

(D+2)² − 2(D+2) + 5 = D² + 2D + 5

PI = e^(2x) · 1/(D² + 2D + 5) · sin x

Put D² = −1: denominator 2D + 4.

= e^(2x) · 1/(2D + 4) · sin x = e^(2x) · (4 − 2D)/(16 − 4D²) · sin x

With D² = −1, the denominator is 16 + 4 = 20:

= e^(2x)/20 · (4 sin x − 2 cos x) = (e^(2x)/10)(2 sin x − cos x)

General solution

y = eˣ(c₁ cos 2x + c₂ sin 2x) + (e^(2x)/10)(2 sin x − cos x)