What is the solution for y^{''}+4y^'+4y=1 ?

The solution for y^{''}+4y^'+4y=1 is y=c_{1}e^{-2t}+c_{2}te^{-2t}+1/4

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y^{''}+4y^'+4y=1

y^{^{''}} + 4 y^{^{'}} + 4 y = 1

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{1}{4}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 1

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{1}{4}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{1}{4}

y^{^{''}} - 2 y^{^{'}} + y = x + e^{x}

y^{^{''}} + 4 y^{^{'}} = tan (2 x)

y^{^{''}} - 5 y^{^{'}} - 6 y = - sin (2 x)

y^{^{''}} - 4 y^{^{'}} + 5 y = \frac{2 e ^{x}}{sin ( x )}

3 y^{^{''}} - 6 y^{^{'}} + 6 y = e x sec (x)

What is the solution for y^{''}+4y^'+4y=1 ?
The solution for y^{''}+4y^'+4y=1 is y=c_{1}e^{-2t}+c_{2}te^{-2t}+1/4