What is the solution for 4y^{''}+20y^'+169y=0 ?

The solution for 4y^{''}+20y^'+169y=0 is y=e^{-(5t)/2}(c_{1}cos(6t)+c_{2}sin(6t))

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4y^{''}+20y^'+169y=0

4 y^{^{''}} + 20 y^{^{'}} + 169 y = 0

y = e^{- \frac{5 t}{2}} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

4 y^{''} (t) + 20 y^{'} (t) + 169 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{- \frac{5 t}{2}} (c_{1} cos (6 t) + c_{2} sin (6 t))

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 5, y^{^{'}} (0) = - 1

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

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

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

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

What is the solution for 4y^{''}+20y^'+169y=0 ?
The solution for 4y^{''}+20y^'+169y=0 is y=e^{-(5t)/2}(c_{1}cos(6t)+c_{2}sin(6t))