What is the solution for y^{''}+6y^'+35y=0 ?

The solution for y^{''}+6y^'+35y=0 is y=e^{-3t}(c_{1}cos(sqrt(26)t)+c_{2}sin(sqrt(26)t))

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y^{''}+6y^'+35y=0

y^{^{''}} + 6 y^{^{'}} + 35 y = 0

y = e^{- 3 t} (c_{1} cos (26 t) + c_{2} sin (26 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 35 y = 0

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (26 t) + c_{2} sin (26 t))

y = e^{- 3 t} (c_{1} cos (26 t) + c_{2} sin (26 t))

(4 D^{3} - 15 D^{2} + 5 D + 6) y = 0

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

y^{^{''}} + 5 y^{^{'}} - 24 y = 0, y (0) = a, y^{^{'}} (0) = 80

44 y^{^{'''}} + 15 y^{^{''}} + y^{^{'}} = 0

16 x^{^{''}} + 8 x^{^{'}} + 145 x = 0, x (0) = - 2, x^{^{'}} (0) = 1

What is the solution for y^{''}+6y^'+35y=0 ?
The solution for y^{''}+6y^'+35y=0 is y=e^{-3t}(c_{1}cos(sqrt(26)t)+c_{2}sin(sqrt(26)t))