What is the solution for y^{''}+10y^'+288y=0 ?

The solution for y^{''}+10y^'+288y=0 is y=e^{-5t}(c_{1}cos(sqrt(263)t)+c_{2}sin(sqrt(263)t))

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y^{''}+10y^'+288y=0

y^{^{''}} + 10 y^{^{'}} + 288 y = 0

y = e^{- 5 t} (c_{1} cos (263 t) + c_{2} sin (263 t))

Solution steps

y^{''} (t) + 10 y^{'} (t) + 288 y = 0

Solve linear ODE:

y = e^{- 5 t} (c_{1} cos (263 t) + c_{2} sin (263 t))

y = e^{- 5 t} (c_{1} cos (263 t) + c_{2} sin (263 t))

(d^{2} + 6) y = 0

y^{^{''}} + 9.82 \cdot y^{^{'}} = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 6, y^{^{'}} (0) = - 6

2 y^{^{''}} + 23 y^{^{'}} - 12 y = 0

a \frac{d ^{2} y}{d x ^{2}} - y b^{2} = 0

What is the solution for y^{''}+10y^'+288y=0 ?
The solution for y^{''}+10y^'+288y=0 is y=e^{-5t}(c_{1}cos(sqrt(263)t)+c_{2}sin(sqrt(263)t))