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}))

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)))