What is the solution for y^{''}+0.25y^'+y=0 ?

The solution for y^{''}+0.25y^'+y=0 is y=e^{-t/8}(c_{1}cos((3sqrt(7)t)/8)+c_{2}sin((3sqrt(7)t)/8))

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y^{''}+0.25y^'+y=0

y^{^{''}} + 0.25 y^{^{'}} + y = 0

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{3 7 t}{8}) + c_{2} sin (\frac{3 7 t}{8}))

Solution steps

y^{''} (t) + 0.25 y^{'} (t) + y = 0

Solve linear ODE:

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{3 7 t}{8}) + c_{2} sin (\frac{3 7 t}{8}))

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{3 7 t}{8}) + c_{2} sin (\frac{3 7 t}{8}))

2 y^{^{'''}} + 5 y^{^{''}} - 28 y^{^{'}} - 15 y = 0

4 y^{^{''}} - 20 y^{^{'}} + 169 y = 0

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5.5

5 x^{^{''}} + 4.9 x^{^{'}} + 10 x = 0

y^{^{''}} + 0 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

What is the solution for y^{''}+0.25y^'+y=0 ?
The solution for y^{''}+0.25y^'+y=0 is y=e^{-t/8}(c_{1}cos((3sqrt(7)t)/8)+c_{2}sin((3sqrt(7)t)/8))