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

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

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y^{''}+2y^'=-4y

y^{^{''}} + 2 y^{^{'}} = - 4 y

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

Solution steps

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

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))

y^{^{''}} = y^{^{'}} + y

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

y^{^{''}} - 9 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 2

u^{^{''}} + 2 u^{^{'}} + 2 H (t) = 0

55 x^{^{''}} + 220 x = 0, x (0) = 0, x^{^{'}} (0) = - 6

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