What is the solution for y^{''}+2y^'+1.25y=0 ?

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

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y^{''}+2y^'+1.25y=0

y^{^{''}} + 2 y^{^{'}} + 1.25 y = 0

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 1.25 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

3 \frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 3 y = 0

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

y^{^{'''}} - 7 y^{^{''}} + y^{^{'}} + 12 y = 0

y^{^{''}} - \frac{1}{4} y = 0, y (0) = 4, y^{^{'}} (0) = 0

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