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}))

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

y^{^{'''}} + 6 y^{^{''}} + 11 y^{^{'}} + 6 y = 0

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + y = 0

y^{^{''}} - 6 y^{^{'}} + 34 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = - 4, y^{^{'}} (0) = - 5

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))