What is the solution for 2y^{''}+4y^'+6y=0 ?

The solution for 2y^{''}+4y^'+6y=0 is y=e^{-t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

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2y^{''}+4y^'+6y=0

2 y^{^{''}} + 4 y^{^{'}} + 6 y = 0

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

Solution steps

2 y^{''} (t) + 4 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d}{d x} (4) + 4 y = 0

- 0.0375 x^{^{''}} + 0.4535 x^{^{'}} + 3.34 x = 0

y^{^{'}} = 7 y^{^{''}}

6 y^{^{'''}} + 78 y^{^{'}} - 696 y = 0

y^{^{''}} - \frac{1}{a ^{2}} y = 0

What is the solution for 2y^{''}+4y^'+6y=0 ?
The solution for 2y^{''}+4y^'+6y=0 is y=e^{-t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))