What is the solution for y^{''}+y^'-2y=3e^{-2t} ?

The solution for y^{''}+y^'-2y=3e^{-2t} is y=c_{1}e^t+c_{2}e^{-2t}-e^{-2t}t

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

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

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

Solution steps

y^{''} (t) + y^{'} (t) - 2 y = 3 e^{- 2 t}

Solve linear ODE:

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

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

(D^{3} - 3 D^{2} + 3 D - 1) y = e^{x} + 1

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

y^{^{''}} + 25 y = 2 sec (5 t)

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

y^{^{''}} + 5 y^{^{'}} + 6 y = 3

What is the solution for y^{''}+y^'-2y=3e^{-2t} ?
The solution for y^{''}+y^'-2y=3e^{-2t} is y=c_{1}e^t+c_{2}e^{-2t}-e^{-2t}t