What is the solution for y^{''}+2y^'+7y=0 ?

The solution for y^{''}+2y^'+7y=0 is y=e^{-t}(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))

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y^{''}+2y^'+7y=0

y^{^{''}} + 2 y^{^{'}} + 7 y = 0

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 7 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 14 y^{^{'}} + 53 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

2 x^{^{''}} - 2 x = 0

y^{^{''}} - 7 y^{^{'}} - 10 y = 0

y^{^{''}} - 16 y^{^{'}} + 63 y = 0

What is the solution for y^{''}+2y^'+7y=0 ?
The solution for y^{''}+2y^'+7y=0 is y=e^{-t}(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))