What is the solution for linear y^{'''}+3y^{''}-8y^'-30y=0 ?

The solution for linear y^{'''}+3y^{''}-8y^'-30y=0 is y=c_{1}e^{3t}+e^{-3t}(c_{2}cos(t)+c_{3}sin(t))

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linear y^{'''}+3y^{''}-8y^'-30y=0

linear

y^{^{'''}} + 3 y^{^{''}} - 8 y^{^{'}} - 30 y = 0

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

Solution steps

y^{'''} (t) + 3 y^{''} (t) - 8 y^{'} (t) - 30 y = 0

Solve linear ODE:

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

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

y^{^{'''}} - 7 y^{^{''}} - 8 y^{^{'}} + 16 y = 0

y^{^{''}} + y + y^{^{'}} = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{39}{2}

y^{^{''}} + 18 y^{^{'}} + 80 y = 0, y (0) = 9, y^{^{'}} (0) = - 76

(D^{3} - 3 D) y = 0

What is the solution for linear y^{'''}+3y^{''}-8y^'-30y=0 ?
The solution for linear y^{'''}+3y^{''}-8y^'-30y=0 is y=c_{1}e^{3t}+e^{-3t}(c_{2}cos(t)+c_{3}sin(t))