What is the solution for y^{''}-4y=4t-8e^{-2t} ?

The solution for y^{''}-4y=4t-8e^{-2t} is y=c_{1}e^{2t}+c_{2}e^{-2t}-t+2te^{-2t}

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y^{''}-4y=4t-8e^{-2t}

y^{^{''}} - 4 y = 4 t - 8 e^{- 2 t}

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

Solution steps

y^{''} (t) - 4 y = 4 t - 8 e^{- 2 t}

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} = cos (3 t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x ^{4}}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4, y^{^{'}} (0) = - 1

y^{^{''}} + 9 y = cos (3 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = - 18

What is the solution for y^{''}-4y=4t-8e^{-2t} ?
The solution for y^{''}-4y=4t-8e^{-2t} is y=c_{1}e^{2t}+c_{2}e^{-2t}-t+2te^{-2t}