What is the solution for 2y^{''}+2y^'+3y=0 ?

The solution for 2y^{''}+2y^'+3y=0 is y=e^{-t/2}(c_{1}cos((sqrt(5)t)/2)+c_{2}sin((sqrt(5)t)/2))

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2y^{''}+2y^'+3y=0

2 y^{^{''}} + 2 y^{^{'}} + 3 y = 0

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

Solution steps

2 y^{''} (t) + 2 y^{'} (t) + 3 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}))

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

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

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 15, y^{^{'}} (0) = - 6

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 9

y^{^{''}} - 5 y^{^{'}} - 36 y = 0

What is the solution for 2y^{''}+2y^'+3y=0 ?
The solution for 2y^{''}+2y^'+3y=0 is y=e^{-t/2}(c_{1}cos((sqrt(5)t)/2)+c_{2}sin((sqrt(5)t)/2))