What is the solution for y^{''}-y^'-2y=3cos(3x) ?

The solution for y^{''}-y^'-2y=3cos(3x) is y=c_{1}e^{2x}+c_{2}e^{-x}-9/130 sin(3x)-33/130 cos(3x)

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{9}{130} sin (3 x) - \frac{33}{130} cos (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{9}{130} sin (3 x) - \frac{33}{130} cos (3 x)

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{9}{130} sin (3 x) - \frac{33}{130} cos (3 x)

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{a x}

25 y^{^{''}} - 10 y^{^{'}} - 2 y = 4 e^{t}

y^{^{''}} + 0.5 y^{^{'}} + y = cos (t)

y^{^{''}} + y = 2 cos (x), y (0) = 1, y^{^{'}} (0) = 0

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

What is the solution for y^{''}-y^'-2y=3cos(3x) ?
The solution for y^{''}-y^'-2y=3cos(3x) is y=c_{1}e^{2x}+c_{2}e^{-x}-9/130 sin(3x)-33/130 cos(3x)