What is the solution for 10y^{''}+20y^'+6260y=0 ?

The solution for 10y^{''}+20y^'+6260y=0 is y=e^{-t}(c_{1}cos(25t)+c_{2}sin(25t))

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10y^{''}+20y^'+6260y=0

10 y^{^{''}} + 20 y^{^{'}} + 6260 y = 0

y = e^{- t} (c_{1} cos (25 t) + c_{2} sin (25 t))

Solution steps

10 y^{''} (t) + 20 y^{'} (t) + 6260 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (25 t) + c_{2} sin (25 t))

y = e^{- t} (c_{1} cos (25 t) + c_{2} sin (25 t))

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\frac{d}{d x} ((arcsin (x)) d x)

\int \frac{1}{9 x ^{2} + 18 x + 10} d x

What is the solution for 10y^{''}+20y^'+6260y=0 ?
The solution for 10y^{''}+20y^'+6260y=0 is y=e^{-t}(c_{1}cos(25t)+c_{2}sin(25t))