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

The solution for 2y^{''}-2y^'-4y=2e^{2t} is y=c_{1}e^{2t}+c_{2}e^{-t}+(e^{2t}t)/3

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} + \frac{e ^{2 t} t}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} + \frac{e ^{2 t} t}{3}

y = c_{1} e^{2 t} + c_{2} e^{- t} + \frac{e ^{2 t} t}{3}

y^{^{'''}} - 6 y^{^{''}} - 6 = cos (x)

y^{^{'''}} + 6 y^{^{''}} + 11 y^{^{'}} + 6 y = 3 + 3 x

y^{^{'''}} + 4 y^{^{'}} = e^{x} + sin (x)

y^{^{''}} + y = 6 sin (2 x)

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

What is the solution for 2y^{''}-2y^'-4y=2e^{2t} ?
The solution for 2y^{''}-2y^'-4y=2e^{2t} is y=c_{1}e^{2t}+c_{2}e^{-t}+(e^{2t}t)/3