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

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

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

linear

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 8 y^{^{'}} + 17 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

9 y - y^{^{''}} = 0

y^{^{''}} - 10 y^{^{'}} + 34 y = 0, y (\frac{π}{6}) = 1, y (0) = 4

10 y^{^{''}} + 30 y^{^{'}} + 20 y = 0, y (0) = - 7, y^{^{'}} (0) = - 2

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

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