What is the solution for 4y^{''}+y^'+8y=0 ?

The solution for 4y^{''}+y^'+8y=0 is y=e^{-t/8}(c_{1}cos((sqrt(127)t)/8)+c_{2}sin((sqrt(127)t)/8))

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4y^{''}+y^'+8y=0

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

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{127 t}{8}) + c_{2} sin (\frac{127 t}{8}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{127 t}{8}) + c_{2} sin (\frac{127 t}{8}))

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{127 t}{8}) + c_{2} sin (\frac{127 t}{8}))

\frac{d}{d x} (x + \frac{1}{x ^{2}} + x^{3})

x \to 5 lim (5 + x^{4})

\int e^{- s t} (t) d t

\int \frac{x ^{\frac{3}{2}} + 2 x ^{6} + 1.2}{x} d x

cos (arctan (3 x) - arcsin (2 x))

What is the solution for 4y^{''}+y^'+8y=0 ?
The solution for 4y^{''}+y^'+8y=0 is y=e^{-t/8}(c_{1}cos((sqrt(127)t)/8)+c_{2}sin((sqrt(127)t)/8))