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

The solution for y^{''}+2y^'+2y=e^{-t}sin(3t) is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))-1/8 e^{-t}sin(3t)

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

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

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) - \frac{1}{8} e^{- t} sin (3 t)

Solution steps

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

Solve linear ODE:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) - \frac{1}{8} e^{- t} sin (3 t)

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) - \frac{1}{8} e^{- t} sin (3 t)

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2 x} csc (x)

y^{^{''}} + 4 y = u_{π} (t) - u_{3 π} (t)

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

y^{^{''}} - 4 y = 4 x^{(2)} e^{(2)} x

y^{^{''}} - 4 y = e^{x} cos (x), y (0) = 1, y^{^{'}} (0) = 2

What is the solution for y^{''}+2y^'+2y=e^{-t}sin(3t) ?
The solution for y^{''}+2y^'+2y=e^{-t}sin(3t) is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))-1/8 e^{-t}sin(3t)