What is the solution for 10y^{''}+6y^'+10.9y=0 ?

The solution for 10y^{''}+6y^'+10.9y=0 is y=e^{-(3t)/(10)}(c_{1}cos(t)+c_{2}sin(t))

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10y^{''}+6y^'+10.9y=0

10 y^{^{''}} + 6 y^{^{'}} + 10.9 y = 0

y = e^{- \frac{3 t}{10}} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

10 y^{''} (t) + 6 y^{'} (t) + 10.9 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{10}} (c_{1} cos (t) + c_{2} sin (t))

y = e^{- \frac{3 t}{10}} (c_{1} cos (t) + c_{2} sin (t))

y^{^{''}} (s) - 3 y^{^{'}} (s) + y (s) 3 = 0 es

y^{^{''''}} - k^{2} y^{^{''}} = 0

y^{^{''}} + \frac{2}{5} y^{^{'}} + \frac{1}{25} y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} + \frac{2}{5} y^{^{'}} + \frac{1}{25} y = 0, y (0) = 0, y^{^{'}} (0) = 4

y^{^{''''}} - y^{^{''}} = 2 (4 y^{^{'''}})

What is the solution for 10y^{''}+6y^'+10.9y=0 ?
The solution for 10y^{''}+6y^'+10.9y=0 is y=e^{-(3t)/(10)}(c_{1}cos(t)+c_{2}sin(t))