What is the solution for y^{''}+4y^'+4y=5e^{-2t} ?

The solution for y^{''}+4y^'+4y=5e^{-2t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}+(5e^{-2t}t^2)/2

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y^{''}+4y^'+4y=5e^{-2t}

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 e^{- 2 t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{5 e ^{- 2 t} t ^{2}}{2}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 5 e^{- 2 t}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{5 e ^{- 2 t} t ^{2}}{2}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{5 e ^{- 2 t} t ^{2}}{2}

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{x}

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

y^{^{''}} + 4 y = 2 cos (x) + 3 sin (x)

y^{^{''}} + y = 3 e^{- 7 x}, y (0) = 0, y^{^{'}} (0) = 0

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

What is the solution for y^{''}+4y^'+4y=5e^{-2t} ?
The solution for y^{''}+4y^'+4y=5e^{-2t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}+(5e^{-2t}t^2)/2