What is the solution for 10y^{''}+2y^'+5y=0 ?

The solution for 10y^{''}+2y^'+5y=0 is y=e^{-t/(10)}(c_{1}cos((7t)/(10))+c_{2}sin((7t)/(10)))

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10y^{''}+2y^'+5y=0

10 y^{^{''}} + 2 y^{^{'}} + 5 y = 0

y = e^{- \frac{t}{10}} (c_{1} cos (\frac{7 t}{10}) + c_{2} sin (\frac{7 t}{10}))

Solution steps

10 y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- \frac{t}{10}} (c_{1} cos (\frac{7 t}{10}) + c_{2} sin (\frac{7 t}{10}))

y = e^{- \frac{t}{10}} (c_{1} cos (\frac{7 t}{10}) + c_{2} sin (\frac{7 t}{10}))

\frac{d y}{d x} = \frac{4 y}{x}

ln (x + 4)

(x y - y^{2}) d x + x^{3} d y = 0

f (x) = 3 x^{2} - 3 x

y = \frac{x}{1 + x}

What is the solution for 10y^{''}+2y^'+5y=0 ?
The solution for 10y^{''}+2y^'+5y=0 is y=e^{-t/(10)}(c_{1}cos((7t)/(10))+c_{2}sin((7t)/(10)))