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

The solution for 16y-8y^'=y^{''} is y=c_{1}e^{-4(1+sqrt(2))t}+c_{2}e^{4(sqrt(2)-1)t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = 9

6 y^{^{''}} - 7 y^{^{'}} + y = 0

y^{^{'''}} - 17 y^{^{''}} + 40 y^{^{'}} + 300 y = 0

y^{^{''}} + 4 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = - 1

\frac{24}{32} y^{^{''}} + 72 y = 0

What is the solution for 16y-8y^'=y^{''} ?
The solution for 16y-8y^'=y^{''} is y=c_{1}e^{-4(1+sqrt(2))t}+c_{2}e^{4(sqrt(2)-1)t}