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

The solution for (D^2-3D+2)y=cos(3x) is y=c_{1}e^{2x}+c_{2}e^x-9/130 sin(3x)-7/130 cos(3x)

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(D^2-3D+2)y=cos(3x)

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

y = c_{1} e^{2 x} + c_{2} e^{x} - \frac{9}{130} sin (3 x) - \frac{7}{130} cos (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} - \frac{9}{130} sin (3 x) - \frac{7}{130} cos (3 x)

y = c_{1} e^{2 x} + c_{2} e^{x} - \frac{9}{130} sin (3 x) - \frac{7}{130} cos (3 x)

2 y^{^{''}} + y = - e^{3} + e^{2 x}

y^{^{'''}} + y^{^{''}} = 6

y^{^{''}} + 3 y^{^{'}} - 4 y = 1, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 cos^{2} (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 e^{- t}, y (0) = 2, y^{^{'}} (0) = 0

What is the solution for (D^2-3D+2)y=cos(3x) ?
The solution for (D^2-3D+2)y=cos(3x) is y=c_{1}e^{2x}+c_{2}e^x-9/130 sin(3x)-7/130 cos(3x)