What is the solution for 4y^{''}+4y^'+3y=0 ?

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

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4y^{''}+4y^'+3y=0

4 y^{^{''}} + 4 y^{^{'}} + 3 y = 0

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

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + 3 y = 0

Solve linear ODE:

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

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

2 y^{^{''}} + 5 y^{^{'}} + 3 y = 0, y (0) = 3, y^{^{'}} (0) = - 4

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

y^{^{''''}} + 6 y^{^{'''}} + 9 y^{^{''}} = 0

y^{^{'''}} + 2 y^{^{''}} - 7 y^{^{'}} + 4 y = 0

y^{^{''}} - 8 y^{^{'}} - 12 y = 0

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