What is the solution for y^{''}+5y^'+14y=0 ?

The solution for y^{''}+5y^'+14y=0 is y=e^{-(5t)/2}(c_{1}cos((sqrt(31)t)/2)+c_{2}sin((sqrt(31)t)/2))

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y^{''}+5y^'+14y=0

y^{^{''}} + 5 y^{^{'}} + 14 y = 0

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{31 t}{2}) + c_{2} sin (\frac{31 t}{2}))

Solution steps

y^{''} (t) + 5 y^{'} (t) + 14 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{31 t}{2}) + c_{2} sin (\frac{31 t}{2}))

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{31 t}{2}) + c_{2} sin (\frac{31 t}{2}))

y^{^{'''}} + 6 y^{^{''}} + 5 y^{^{'}} = 0

y^{^{''}} + 50 y = 0

y^{^{''}} + a y = 0, y^{^{'}} (0) = 0, y^{^{'}} (1) = 0

2 x^{^{''}} + 8 x^{^{'}} + 52 x = 0

y^{^{'''}} + 4 y^{^{''}} - 7 y^{^{'}} - 10 y = 0

What is the solution for y^{''}+5y^'+14y=0 ?
The solution for y^{''}+5y^'+14y=0 is y=e^{-(5t)/2}(c_{1}cos((sqrt(31)t)/2)+c_{2}sin((sqrt(31)t)/2))