What is the solution for y^{'''}-y^{''}-4y^'+4y=20e^{3t} ?

The solution for y^{'''}-y^{''}-4y^'+4y=20e^{3t} is y=c_{1}e^t+c_{2}e^{-2t}+c_{3}e^{2t}+2e^{3t}

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y^{'''}-y^{''}-4y^'+4y=20e^{3t}

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 20 e^{3 t}

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

Solution steps

y^{'''} (t) - y^{''} (t) - 4 y^{'} (t) + 4 y = 20 e^{3 t}

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} + 17 y = 10 e^{4 x} sec (x)

y^{^{''}} - 4 y = 12

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

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

y^{^{''}} + y^{^{'}} - 2 y = 4, y (0) = 2, y^{^{'}} (0) = 1

What is the solution for y^{'''}-y^{''}-4y^'+4y=20e^{3t} ?
The solution for y^{'''}-y^{''}-4y^'+4y=20e^{3t} is y=c_{1}e^t+c_{2}e^{-2t}+c_{3}e^{2t}+2e^{3t}