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}+1/4 e^{3t}

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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{1}{4} e^{3 t}

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{1}{4} e^{3 t}

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

2 x^{^{''}} + 3 x^{^{'}} + 4 x = - 5 cos (t) + sin (t)

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- x} \cdot cos (x)

y^{^{''}} + 6 y^{^{'}} + 5 y = 30

4 y^{^{''}} - 4 y^{^{'}} + y = 80 e^{\frac{t}{2}}

- 2 y^{^{''}} + y = e^{- 0.2 x}, y (0) = 1, y^{^{'}} (1) = - y (1)

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}+1/4 e^{3t}