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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} + y^{^{'}} - y = 3 x + 4

y^{^{''}} - 5 y^{^{'}} + 6 y = x^{2} - 4

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

y^{^{''}} + 2 y = - 8 x sin (3 x)

y^{^{''}} - 14 y^{^{'}} + 49 y = 10 e^{7 x}

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