What is the solution for y^{''}+5y^'+16y=0 ?

The solution for y^{''}+5y^'+16y=0 is y=e^{-(5t)/2}(c_{1}cos((sqrt(39)t)/2)+c_{2}sin((sqrt(39)t)/2))

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y^{''}+5y^'+16y=0

y^{^{''}} + 5 y^{^{'}} + 16 y = 0

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{39 t}{2}) + c_{2} sin (\frac{39 t}{2}))

Solution steps

y^{''} (t) + 5 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{39 t}{2}) + c_{2} sin (\frac{39 t}{2}))

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{39 t}{2}) + c_{2} sin (\frac{39 t}{2}))

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

10 x^{^{''}} + 30 x^{^{''}} + 20 x = 0, x (0) = - 7, x^{^{'}} (0) = - 2

y^{^{'''}} + y^{^{''}} - 22 y^{^{'}} - 40 y = 0

y^{^{''}} + 11 y^{^{'}} + 10 y = 0

81 y^{^{''}} - 144 y^{^{'}} + 73 y = 0, y (1) = 6, y^{^{'}} (1) = 5

What is the solution for y^{''}+5y^'+16y=0 ?
The solution for y^{''}+5y^'+16y=0 is y=e^{-(5t)/2}(c_{1}cos((sqrt(39)t)/2)+c_{2}sin((sqrt(39)t)/2))