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

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

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

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

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{1}{5} sin (t) - \frac{2}{5} cos (t)

Solution steps

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

Solve linear ODE:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{1}{5} sin (t) - \frac{2}{5} cos (t)

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{1}{5} sin (t) - \frac{2}{5} cos (t)

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