What is the solution for y^{''}+4y^'+10y=0 ?

The solution for y^{''}+4y^'+10y=0 is y=e^{-2t}(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))

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y^{''}+4y^'+10y=0

y^{^{''}} + 4 y^{^{'}} + 10 y = 0

y = e^{- 2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{- 2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

2 y^{^{''}} + 6 y^{^{'}} + 8 y = 0

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

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

77 y^{^{''''}} + 106 y^{^{''}} + 9 y = 0

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

What is the solution for y^{''}+4y^'+10y=0 ?
The solution for y^{''}+4y^'+10y=0 is y=e^{-2t}(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))