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}

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