What is the solution for x^{''}+4x^'+5x=16sin(t) ?

The solution for x^{''}+4x^'+5x=16sin(t) is x=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+2sin(t)-2cos(t)

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x^{''}+4x^'+5x=16sin(t)

x^{^{''}} + 4 x^{^{'}} + 5 x = 16 sin (t)

x = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + 2 sin (t) - 2 cos (t)

Solution steps

x^{''} (t) + 4 x^{'} (t) + 5 x = 16 sin (t)

Solve linear ODE:

x = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + 2 sin (t) - 2 cos (t)

x = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + 2 sin (t) - 2 cos (t)

y^{^{''}} + 16 y = 64 t, y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- 4 t}, y^{^{'}} (0) = 0, y (0) = 1

y^{^{''}} + 4 y = 4 t^{2} - 4 t + 10

2 y^{^{''}} - 5 y^{^{'}} + 3 y = (14 x - 10) e^{3 x}

y^{^{''}} + 9 y = 12 cos (3 t), y (0) = - 1, y^{^{'}} (0) = 1

What is the solution for x^{''}+4x^'+5x=16sin(t) ?
The solution for x^{''}+4x^'+5x=16sin(t) is x=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+2sin(t)-2cos(t)