What is the solution for 2y^{''}+4y^'+8y=0 ?

The solution for 2y^{''}+4y^'+8y=0 is y=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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2y^{''}+4y^'+8y=0

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

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

2 y^{^{''}} - 3 y^{^{'}} + 1 y = 0

x^{^{''}} + 12 x^{^{'}} + 27 x = 0, x (0) = 1, x^{^{'}} (0) = 0

(4 D^{4} - 45 D^{2} - 70 D - 24) y = 0

y^{^{''}} - 6 y^{^{'}} + 113 y = 0

10 \frac{d ^{2} y}{d x ^{2}} + 20 \frac{d y}{d x} + 10 y = 0

What is the solution for 2y^{''}+4y^'+8y=0 ?
The solution for 2y^{''}+4y^'+8y=0 is y=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))