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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + (π^{2} + 4) y = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 7, y^{^{'}} (0) = 11

16 y^{^{''''}} + 48 y^{^{''}} + 36 y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 11

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

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