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

The solution for 2y^{''}+3y^'+3y=0 is y=e^{-(3t)/4}(c_{1}cos((sqrt(15)t)/4)+c_{2}sin((sqrt(15)t)/4))

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

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

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{15 t}{4}) + c_{2} sin (\frac{15 t}{4}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{15 t}{4}) + c_{2} sin (\frac{15 t}{4}))

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{15 t}{4}) + c_{2} sin (\frac{15 t}{4}))

y^{^{'''}} - 8 y^{^{''}} + 16 y^{^{'}} = 0

y^{^{''}} + 5 y^{^{'}} - 6 y = 0, y (0) = - 5, y^{^{'}} (0) = 37

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

y^{^{''}} = y^{^{'}} + y

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

What is the solution for 2y^{''}+3y^'+3y=0 ?
The solution for 2y^{''}+3y^'+3y=0 is y=e^{-(3t)/4}(c_{1}cos((sqrt(15)t)/4)+c_{2}sin((sqrt(15)t)/4))