What is the solution for y^{''''}+8y^{'''}+16y^{''}=0 ?

The solution for y^{''''}+8y^{'''}+16y^{''}=0 is y=c_{1}+c_{2}t+c_{3}e^{-4t}+c_{4}te^{-4t}

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

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

y = c_{1} + c_{2} t + c_{3} e^{- 4 t} + c_{4} t e^{- 4 t}

Solution steps

y^{(4)} (t) + 8 y^{'''} (t) + 16 y^{''} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} t + c_{3} e^{- 4 t} + c_{4} t e^{- 4 t}

y = c_{1} + c_{2} t + c_{3} e^{- 4 t} + c_{4} t e^{- 4 t}

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 9

y^{^{''}} - 6 y^{^{'}} + 14 y = 0

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

What is the solution for y^{''''}+8y^{'''}+16y^{''}=0 ?
The solution for y^{''''}+8y^{'''}+16y^{''}=0 is y=c_{1}+c_{2}t+c_{3}e^{-4t}+c_{4}te^{-4t}