What is the solution for y^{''}+4y^'+4y=4e^{-2t} ?

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

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y^{''}+4y^'+4y=4e^{-2t}

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 e^{- 2 t}

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 4 e^{- 2 t}

Solve linear ODE:

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

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

\int \frac{( x - 3 )}{( x + 4 ) ( x - 5 )} d x

\frac{d}{d x} (x^{3} - 5 x^{2} + 7 x - 3)

\frac{\partial}{\partial x} (x^{2} y - \frac{1}{x y})

\frac{d}{d x} (arccot (x))

f (x) = x^{2} 9 - x^{2}

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