What is the solution for y^{''}+2y^'+2y=5 ?

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

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y^{''}+2y^'+2y=5

y^{^{''}} + 2 y^{^{'}} + 2 y = 5

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = 5

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 6 x^{2} + 4 x + 1

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

y^{^{''}} - 8 y^{^{'}} - 4 y - 2 = 2 x e^{x} sin (x)

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 6

y^{^{''}} - 6 y^{^{'}} + 9 y = 5 t^{2}, y (0) = 0, y^{^{'}} (0) = 1

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