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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + 6 \frac{d x}{d t} + 3 x = 0

(d^{6} - 4 d^{4} + 4 d^{2}) y = 0

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

\frac{d ^{2} y}{d x ^{2}} = - 9 y

y^{^{''}} = 7 y^{^{'}} - 12 y, y^{^{'}} (0) = 1

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