What is the solution for y^{''}-y^'-6y=2e^{4t} ?

The solution for y^{''}-y^'-6y=2e^{4t} is y=c_{1}e^{3t}+c_{2}e^{-2t}+1/3 e^{4t}

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y^{''}-y^'-6y=2e^{4t}

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{4 t}

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

Solution steps

y^{''} (t) - y^{'} (t) - 6 y = 2 e^{4 t}

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 3, y (0) = 1, y^{^{'}} (0) = 0

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

(4 d^{2} + 12 d + 9) y = 2022

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

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

What is the solution for y^{''}-y^'-6y=2e^{4t} ?
The solution for y^{''}-y^'-6y=2e^{4t} is y=c_{1}e^{3t}+c_{2}e^{-2t}+1/3 e^{4t}