What is the solution for y^{''}+y^'-2y=-260sin(3t) ?

The solution for y^{''}+y^'-2y=-260sin(3t) is y=c_{1}e^t+c_{2}e^{-2t}+22sin(3t)+6cos(3t)

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y^{''}+y^'-2y=-260sin(3t)

y^{^{''}} + y^{^{'}} - 2 y = - 260 sin (3 t)

y = c_{1} e^{t} + c_{2} e^{- 2 t} + 22 sin (3 t) + 6 cos (3 t)

Solution steps

y^{''} (t) + y^{'} (t) - 2 y = - 260 sin (3 t)

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 2 t} + 22 sin (3 t) + 6 cos (3 t)

y = c_{1} e^{t} + c_{2} e^{- 2 t} + 22 sin (3 t) + 6 cos (3 t)

y^{^{''}} - 9 y = 81 x^{3}

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

y^{^{''}} + 2 y^{^{'}} + y = 2 x + 5 - e^{- 2 x}

y^{^{''}} - y^{^{'}} - 2 y = 4 t^{2}, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} + 25 y = 2 csc^{2} (5 t)

What is the solution for y^{''}+y^'-2y=-260sin(3t) ?
The solution for y^{''}+y^'-2y=-260sin(3t) is y=c_{1}e^t+c_{2}e^{-2t}+22sin(3t)+6cos(3t)