What is the solution for y^{'''}-2y^{''}+4y^'-8=0 ?

The solution for y^{'''}-2y^{''}+4y^'-8=0 is y=c_{1}+e^t(c_{2}cos(sqrt(3)t)+c_{3}sin(sqrt(3)t))+2t

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y^{'''}-2y^{''}+4y^'-8=0

y^{^{'''}} - 2 y^{^{''}} + 4 y^{^{'}} - 8 = 0

y = c_{1} + e^{t} (c_{2} cos (3 t) + c_{3} sin (3 t)) + 2 t

Solution steps

y^{'''} (t) - 2 y^{''} (t) + 4 y^{'} (t) - 8 = 0

Solve linear ODE:

y = c_{1} + e^{t} (c_{2} cos (3 t) + c_{3} sin (3 t)) + 2 t

y = c_{1} + e^{t} (c_{2} cos (3 t) + c_{3} sin (3 t)) + 2 t

y^{^{''''}} - y = 12 t + cos (t)

y^{^{''}} + 8 y^{^{'}} = 8 x

y^{^{''}} + 2 y^{^{'}} + 2 y = x^{2} - 3 x e^{- 2 x} cos (5 x)

(D + 2)^{2} y = 12 x e^{- 2 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{2 x}

What is the solution for y^{'''}-2y^{''}+4y^'-8=0 ?
The solution for y^{'''}-2y^{''}+4y^'-8=0 is y=c_{1}+e^t(c_{2}cos(sqrt(3)t)+c_{3}sin(sqrt(3)t))+2t