What is the solution for y^{''''}-y^{''}=2(4y^{''}) ?

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

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y^{''''}-y^{''}=2(4y^{''})

y^{^{''''}} - y^{^{''}} = 2 (4 y^{^{''}})

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 1 y^{^{'}} - 2 y = 0

y^{^{''''}} - (- 5) y^{^{''}} + 4 y = 0

y^{^{''}} = 2 y - y

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 1, y^{^{'}} (0) = 10

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

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