What is the solution for 2y^{''}-2y^'-4y=2e^3t ?

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

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{e ^{3} t}{2} + \frac{e ^{3}}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{e ^{3} t}{2} + \frac{e ^{3}}{4}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{e ^{3} t}{2} + \frac{e ^{3}}{4}

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

y^{^{''}} - 4 y^{^{'}} + 5 y = 2 cos (3 x)

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

y^{^{''''}} + y^{^{'''}} = 1 - x^{2} e^{- x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 8 x e^{x}

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