What is the solution for y^{'''}+4y^{''}-128y=0 ?

The solution for y^{'''}+4y^{''}-128y=0 is y=c_{1}e^{4t}+e^{-4t}(c_{2}cos(4t)+c_{3}sin(4t))

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y^{'''}+4y^{''}-128y=0

y^{^{'''}} + 4 y^{^{''}} - 128 y = 0

y = c_{1} e^{4 t} + e^{- 4 t} (c_{2} cos (4 t) + c_{3} sin (4 t))

Solution steps

y^{'''} (t) + 4 y^{''} (t) - 128 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + e^{- 4 t} (c_{2} cos (4 t) + c_{3} sin (4 t))

y = c_{1} e^{4 t} + e^{- 4 t} (c_{2} cos (4 t) + c_{3} sin (4 t))

y^{^{''''}} - 2 y^{^{'''}} + 4 y^{^{''}} + 2 y^{^{'}} - 5 y = 0

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 2

(D^{3} + 6 D^{2} + 12 D + 8) y = 0

576 y^{^{''}} - 528 y^{^{'}} + 121 y = 0, y (0) = 9, y^{^{'}} (0) = 3

y^{^{''}} - 6 y^{^{'}} + 12 y^{^{'}} - 8 y = 0

What is the solution for y^{'''}+4y^{''}-128y=0 ?
The solution for y^{'''}+4y^{''}-128y=0 is y=c_{1}e^{4t}+e^{-4t}(c_{2}cos(4t)+c_{3}sin(4t))