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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{4} y}{d x ^{4}} - 23 \frac{d ^{2} y}{d x ^{2}} - 50 y = 0

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} + 8 y^{^{'}} + 12 y = 0

y^{^{'''}} - y^{^{''}} - 8 y^{^{'}} + 12 y = 0

y^{^{''}} + 8 y^{^{'}} + 20 y = 0

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