What is the solution for y^{''}+y^'+1.25y=0 ?

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

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y^{''}+y^'+1.25y=0

y^{^{''}} + y^{^{'}} + 1.25 y = 0

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

Solution steps

y^{''} (t) + y^{'} (t) + 1.25 y = 0

Solve linear ODE:

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

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

121 y^{^{''}} + 88 y^{^{'}} + 16 y = 0

y^{^{'''}} - 4 y^{^{''}} + 3 y^{^{'}} = 0

y^{^{''}} - y^{^{'}} - 56 y = 0, y (0) = - 5, y^{^{'}} (0) = 5

y^{^{''}} + 36 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{24}{5}

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

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