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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

y^{^{''}} + 0 y^{^{'}} + 16 y = 0

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

y^{^{'''}} - 2 y^{^{''}} - 63 y^{^{'}} = 0

\frac{3}{4} x^{^{''}} + 72 x = 0, x (0) = - \frac{1}{4}, x^{^{'}} (0) = 0

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