What is the solution for y^{''}+0.5y^'+4y=0 ?

The solution for y^{''}+0.5y^'+4y=0 is y=e^{-t/4}(c_{1}cos((3sqrt(7)t)/4)+c_{2}sin((3sqrt(7)t)/4))

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y^{''}+0.5y^'+4y=0

y^{^{''}} + 0.5 y^{^{'}} + 4 y = 0

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{3 7 t}{4}) + c_{2} sin (\frac{3 7 t}{4}))

Solution steps

y^{''} (t) + 0.5 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{3 7 t}{4}) + c_{2} sin (\frac{3 7 t}{4}))

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{3 7 t}{4}) + c_{2} sin (\frac{3 7 t}{4}))

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What is the solution for y^{''}+0.5y^'+4y=0 ?
The solution for y^{''}+0.5y^'+4y=0 is y=e^{-t/4}(c_{1}cos((3sqrt(7)t)/4)+c_{2}sin((3sqrt(7)t)/4))