What is the solution for 4y^{''}+4y^'+6y=0 ?

The solution for 4y^{''}+4y^'+6y=0 is y=e^{-t/2}(c_{1}cos((sqrt(5)t)/2)+c_{2}sin((sqrt(5)t)/2))

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4y^{''}+4y^'+6y=0

4 y^{^{''}} + 4 y^{^{'}} + 6 y = 0

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}))

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}))

\frac{d ^{2} y}{d x ^{2}} - 12 \frac{d y}{d x} + 35 y = 0

y^{^{''}} + y^{^{'}} + 3.25 y = 0

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

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

33 y^{^{''''}} + 323 y^{^{''}} + 400 y = 0

What is the solution for 4y^{''}+4y^'+6y=0 ?
The solution for 4y^{''}+4y^'+6y=0 is y=e^{-t/2}(c_{1}cos((sqrt(5)t)/2)+c_{2}sin((sqrt(5)t)/2))