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

The solution for y^{'''}-2y^{''}-y^'+2y=e^{4t} is y=c_{1}e^{2t}+c_{2}e^{-t}+c_{3}e^t+1/30 e^{4t}

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{30} e^{4 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{30} e^{4 t}

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{30} e^{4 t}

y^{^{''}} - y^{^{'}} - 12 y = e^{2 x} + e^{x} - 2

y^{^{'''}} + 64 y^{^{'}} = sec (8 x)

y^{^{''}} + 9 y = 18 x^{2} + 9 x - 6

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

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

What is the solution for y^{'''}-2y^{''}-y^'+2y=e^{4t} ?
The solution for y^{'''}-2y^{''}-y^'+2y=e^{4t} is y=c_{1}e^{2t}+c_{2}e^{-t}+c_{3}e^t+1/30 e^{4t}