What is the solution for y^{'''}-y^{''}-8y^'+12y=2e^t ?

The solution for y^{'''}-y^{''}-8y^'+12y=2e^t is y=c_{1}e^{-3t}+c_{2}e^{2t}+c_{3}te^{2t}+1/2 e^t

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y^{'''}-y^{''}-8y^'+12y=2e^t

y^{^{'''}} - y^{^{''}} - 8 y^{^{'}} + 12 y = 2 e^{t}

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

Solution steps

y^{'''} (t) - y^{''} (t) - 8 y^{'} (t) + 12 y = 2 e^{t}

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 3 x^{3}

x^{^{''}} + 2 x^{^{'}} + x = e^{- t}

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 e^{2 x}

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

What is the solution for y^{'''}-y^{''}-8y^'+12y=2e^t ?
The solution for y^{'''}-y^{''}-8y^'+12y=2e^t is y=c_{1}e^{-3t}+c_{2}e^{2t}+c_{3}te^{2t}+1/2 e^t