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

The solution for y^{'''}-2y^{''}-y^'+2y=e^{6t} is y=c_{1}e^{2t}+c_{2}e^{-t}+c_{3}e^t+1/140 e^{6t}

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{140} e^{6 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{140} e^{6 t}

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{140} e^{6 t}

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

y^{^{''}} + 4 y^{^{'}} = 10 sin (2 x)

y^{^{''}} - 4 y^{^{'}} + 3 y = 1, y (0) = 1, y^{^{'}} (0) = 0

(D^{2} + 1) y = sin (x)

y^{^{''}} + y = 16 cos (3 x)

What is the solution for y^{'''}-2y^{''}-y^'+2y=e^{6t} ?
The solution for y^{'''}-2y^{''}-y^'+2y=e^{6t} is y=c_{1}e^{2t}+c_{2}e^{-t}+c_{3}e^t+1/140 e^{6t}