What is the solution for y^{''}+4y^'+4y=7e^{-2t} ?

The solution for y^{''}+4y^'+4y=7e^{-2t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}+(7e^{-2t}t^2)/2

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y^{''}+4y^'+4y=7e^{-2t}

y^{^{''}} + 4 y^{^{'}} + 4 y = 7 e^{- 2 t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{7 e ^{- 2 t} t ^{2}}{2}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 7 e^{- 2 t}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{7 e ^{- 2 t} t ^{2}}{2}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{7 e ^{- 2 t} t ^{2}}{2}

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8 e^{- t}

y^{^{''}} - 5 y = x^{2} - 2 x

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 e^{- t}

y^{^{''}} + 8 y^{^{'}} + 2 y = cos (x), y (0) = 0, y^{^{'}} (0) = 1

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

What is the solution for y^{''}+4y^'+4y=7e^{-2t} ?
The solution for y^{''}+4y^'+4y=7e^{-2t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}+(7e^{-2t}t^2)/2