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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = sin (3 t)

y^{^{''}} - 2 y^{^{'}} - 8 y = 6 + 3 x

2 y^{^{''}} + 4 y = 6 t

y^{^{''}} + 6 y^{^{'}} + 8 y = 8 t + 22, y (0) = - 7, y^{^{'}} (0) = 20

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 x \cdot e^{2 x}

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