What is the solution for y^{''}+2y^'+2y=10sin(2t) ?

The solution for y^{''}+2y^'+2y=10sin(2t) is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))-sin(2t)-2cos(2t)

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y^{''}+2y^'+2y=10sin(2t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 10 sin (2 t)

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = 10 sin (2 t)

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} = 4 x

y^{^{''}} + 3 y = x^{3} - 1

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 e^{2 t}, y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} + y^{^{'}} = x, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = x^{2}

What is the solution for y^{''}+2y^'+2y=10sin(2t) ?
The solution for y^{''}+2y^'+2y=10sin(2t) is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))-sin(2t)-2cos(2t)