Solve y''+3y'+2y=e^(2x) sin x

The equation is (D² + 3D + 2)y = e^(2x) sin x, with D = d/dx.

Complementary function (CF)

Auxiliary equation m² + 3m + 2 = 0, so m = −1, −2.

CF = c₁e^(−x) + c₂e^(−2x)

Particular integral (PI)

PI = 1/(D² + 3D + 2) · e^(2x) sin x

Use the shift rule: replace D by D + 2 and take e^(2x) out:

PI = e^(2x) · 1/[(D+2)² + 3(D+2) + 2] · sin x = e^(2x) · 1/(D² + 7D + 12) · sin x

Put D² = −1:

= e^(2x) · 1/(7D + 11) · sin x = e^(2x) · (11 − 7D)/(121 − 49D²) · sin x

With D² = −1 the denominator is 121 + 49 = 170:

= e^(2x)/170 · (11 sin x − 7 cos x)

General solution

y = c₁e^(−x) + c₂e^(−2x) + (e^(2x)/170)(11 sin x − 7 cos x)