What is the solution for f^{''}(t)+2f^'(t)+4f(t)=0 ?

The solution for f^{''}(t)+2f^'(t)+4f(t)=0 is f(t)=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 7 y^{^{'}} + 15 y = 0

y^{^{''''}} - 4 y^{^{'''}} + 5 y^{^{''}} = 0

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

y^{^{''}} + 4 y^{^{'}} + 2.5 y = 0

y^{^{'''}} - 3 y^{^{'}} - 2 y = 0

What is the solution for f^{''}(t)+2f^'(t)+4f(t)=0 ?
The solution for f^{''}(t)+2f^'(t)+4f(t)=0 is f(t)=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))