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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(D^{3} - 2 D^{2} - D + 2) y = 0

2 y^{^{'''}} + y^{^{''}} - 8 y^{^{'}} + 5 y = 0

\frac{d ^{2} y}{d t ^{2}} = \frac{d y}{d t}

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

y^{^{'''}} + 9 y^{^{''}} + 20 y^{^{'}} = 0

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