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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

4 y^{^{''}} - 2 y^{^{'}} - 2 y = 0, y (0) = 4, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 0

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

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