What is the solution for y^{''}+2y^'+2y=1 ?

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

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y^{''}+2y^'+2y=1

y^{^{''}} + 2 y^{^{'}} + 2 y = 1

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = 1

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{7 - sec ( x )}{tan ( x )})

x \to - \infty lim (\frac{x}{9 - x ^{2}})

x^{2} y^{^{''}} + 2 x y^{^{'}} - 72 y = 0

\int_{0}^{2} \frac{x + 1}{x ( x ^{2} + 1 )} d x

y^{^{'}} = \frac{( x - y sin ( 2 x ) )}{sin ^{2} ( x )}

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