What is the solution for 1/16 y^{''}+y^'+2y=0 ?

The solution for 1/16 y^{''}+y^'+2y=0 is y=c_{1}e^{4(sqrt(2)-2)t}+c_{2}e^{-4(2+sqrt(2))t}

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1/16 y^{''}+y^'+2y=0

\frac{1}{16} y^{^{''}} + y^{^{'}} + 2 y = 0

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

Solution steps

\frac{1}{16} y^{''} (t) + y^{'} (t) + 2 y = 0

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} + \frac{1}{3} y = 0

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

x^{^{''}} + 3 x^{^{'}} - 3 x = 0

y^{^{''}} + 8 y^{^{'}} + 160 y = 0

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

What is the solution for 1/16 y^{''}+y^'+2y=0 ?
The solution for 1/16 y^{''}+y^'+2y=0 is y=c_{1}e^{4(sqrt(2)-2)t}+c_{2}e^{-4(2+sqrt(2))t}