What is the solution for y^{''}+4y^'+5y=0 ?

The solution for y^{''}+4y^'+5y=0 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))

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

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

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

What is the solution for y^{''}+4y^'+5y=0 ?
The solution for y^{''}+4y^'+5y=0 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))