What is the solution for 1/5 y^{''}+2y^'+10y=0 ?

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

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1/5 y^{''}+2y^'+10y=0

\frac{1}{5} y^{^{''}} + 2 y^{^{'}} + 10 y = 0

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

Solution steps

\frac{1}{5} y^{''} (t) + 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

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

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

81 y^{^{''''}} = y

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} + \frac{5}{2} y^{^{'}} + 1.578125 y = 0

u^{^{''}} + \frac{2}{r} u^{^{'}} = 0

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

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