What is the solution for (D^2+3D+2)y=sin(2x) ?

The solution for (D^2+3D+2)y=sin(2x) is y=c_{1}e^{-x}+c_{2}e^{-2x}-1/20 sin(2x)-3/20 cos(2x)

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(D^2+3D+2)y=sin(2x)

(D^{2} + 3 D + 2) y = sin (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - \frac{1}{20} sin (2 x) - \frac{3}{20} cos (2 x)

Solution steps

(D^{2} + 3 D + 2) y = sin (2 x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - \frac{1}{20} sin (2 x) - \frac{3}{20} cos (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - \frac{1}{20} sin (2 x) - \frac{3}{20} cos (2 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 11 e^{3 x}, y (0) = e^{x}

y^{^{''}} + 16 y = e^{4 t} + 9 cos (2 t)

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{(- x)} cos (x)

\frac{d ^{2} y}{d x ^{2}} - 4 y^{^{'}} - 12 y = 2 x + 6

y^{^{'''}} - 3 y^{^{''}} + 4 y = 4 - e^{- 3 x}

What is the solution for (D^2+3D+2)y=sin(2x) ?
The solution for (D^2+3D+2)y=sin(2x) is y=c_{1}e^{-x}+c_{2}e^{-2x}-1/20 sin(2x)-3/20 cos(2x)