What is the solution for y^{''}+5/2 y^'+1.578125y=0 ?

The solution for y^{''}+5/2 y^'+1.578125y=0 is y=e^{-1.25t}(c_{1}cos(0.125t)+c_{2}sin(0.125t))

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y^{''}+5/2 y^'+1.578125y=0

y^{^{''}} + \frac{5}{2} y^{^{'}} + 1.578125 y = 0

y = e^{- 1.25 t} (c_{1} cos (0.125 t) + c_{2} sin (0.125 t))

Solution steps

y^{''} (t) + \frac{5}{2} y^{'} (t) + 1.578125 y = 0

Solve linear ODE:

y = e^{- 1.25 t} (c_{1} cos (0.125 t) + c_{2} sin (0.125 t))

y = e^{- 1.25 t} (c_{1} cos (0.125 t) + c_{2} sin (0.125 t))

u^{^{''}} + \frac{2}{r} u^{^{'}} = 0

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

y^{^{''}} - 2 y^{^{'}} - 195 y = 0, y (0) = 10, y^{^{'}} (0) = - 46

y^{^{'''}} + 8 y^{^{''}} + 7 y^{^{'}} = 0

\frac{4}{9.8} y^{^{''}} + 0.4 y^{^{'}} + 4 y = 0

What is the solution for y^{''}+5/2 y^'+1.578125y=0 ?
The solution for y^{''}+5/2 y^'+1.578125y=0 is y=e^{-1.25t}(c_{1}cos(0.125t)+c_{2}sin(0.125t))