What is the solution for 4y^{''}+4y^'+7y=0 ?

The solution for 4y^{''}+4y^'+7y=0 is y=e^{-t/2}(c_{1}cos(sqrt(3/2)t)+c_{2}sin(sqrt(3/2)t))

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

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

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

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + 7 y = 0

Solve linear ODE:

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

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

5 y^{^{''}} - 13 y^{^{'}} - 6 y = 0

y^{^{''}} + 48 y^{^{'}} = 0

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

6 y^{^{''}} - 17 y^{^{'}} + 12 y = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 8, y^{^{'}} (0) = 18

What is the solution for 4y^{''}+4y^'+7y=0 ?
The solution for 4y^{''}+4y^'+7y=0 is y=e^{-t/2}(c_{1}cos(sqrt(3/2)t)+c_{2}sin(sqrt(3/2)t))