What is the solution for 4y^{''}+y^'+2y=0 ?

The solution for 4y^{''}+y^'+2y=0 is y=e^{-t/8}(c_{1}cos((sqrt(31)t)/8)+c_{2}sin((sqrt(31)t)/8))

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4y^{''}+y^'+2y=0

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

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{31 t}{8}) + c_{2} sin (\frac{31 t}{8}))

Solution steps

4 y^{''} (t) + y^{'} (t) + 2 y = 0

Solve linear ODE:

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{31 t}{8}) + c_{2} sin (\frac{31 t}{8}))

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{31 t}{8}) + c_{2} sin (\frac{31 t}{8}))

y^{^{''}} + 14 y^{^{'}} + 40 y = 0, y (0) = 9, y^{^{'}} (0) = - 48

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

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

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

What is the solution for 4y^{''}+y^'+2y=0 ?
The solution for 4y^{''}+y^'+2y=0 is y=e^{-t/8}(c_{1}cos((sqrt(31)t)/8)+c_{2}sin((sqrt(31)t)/8))