What is the solution for y^{''}+0.5y^'+y=0 ?

The solution for y^{''}+0.5y^'+y=0 is y=e^{-t/4}(c_{1}cos((sqrt(15)t)/4)+c_{2}sin((sqrt(15)t)/4))

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y^{''}+0.5y^'+y=0

y^{^{''}} + 0.5 y^{^{'}} + y = 0

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

Solution steps

y^{''} (t) + 0.5 y^{'} (t) + 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^{^{''}} + 64 π^{2} y = 0

y^{^{''}} + 16 y^{^{'}} + 73 y = 0, y^{^{'}} (0) = 8

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 7

x^{^{''}} - x^{^{'}} - 110 x = 0, x (0) = 14

What is the solution for y^{''}+0.5y^'+y=0 ?
The solution for y^{''}+0.5y^'+y=0 is y=e^{-t/4}(c_{1}cos((sqrt(15)t)/4)+c_{2}sin((sqrt(15)t)/4))