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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2 \frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} - y = 0

y^{^{''}} + 10 y^{^{'}} + \frac{8}{3} y = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (\frac{π}{2}) = 0, y (π) = - 1

15 y^{^{''}} - 4 y^{^{'}} - 3 y = 0

2 y^{^{''}} + 3 y^{^{'}} + 3 y = 0

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