What is the solution for 16x^{''}+40x^'+169x=0 ?

The solution for 16x^{''}+40x^'+169x=0 is x=e^{-(5t)/4}(c_{1}cos(3t)+c_{2}sin(3t))

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16x^{''}+40x^'+169x=0

16 x^{^{''}} + 40 x^{^{'}} + 169 x = 0

x = e^{- \frac{5 t}{4}} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

16 x^{''} (t) + 40 x^{'} (t) + 169 x = 0

Solve linear ODE:

x = e^{- \frac{5 t}{4}} (c_{1} cos (3 t) + c_{2} sin (3 t))

x = e^{- \frac{5 t}{4}} (c_{1} cos (3 t) + c_{2} sin (3 t))

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

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

What is the solution for 16x^{''}+40x^'+169x=0 ?
The solution for 16x^{''}+40x^'+169x=0 is x=e^{-(5t)/4}(c_{1}cos(3t)+c_{2}sin(3t))