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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - c^{2} x = 0

y^{^{''}} - 8 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 9 y^{^{'}} + 22 y = 0

y^{^{''}} + 6 y + 25 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

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

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