What is the solution for x^{''}+6x^'+25x=0 ?

The solution for x^{''}+6x^'+25x=0 is x=e^{-3t}(c_{1}cos(4t)+c_{2}sin(4t))

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x^{''}+6x^'+25x=0

x^{^{''}} + 6 x^{^{'}} + 25 x = 0

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

Solution steps

x^{''} (t) + 6 x^{'} (t) + 25 x = 0

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{''}} - 4 y = 0

y^{^{''}} - 12 y^{^{'}} - 28 y = 0

y^{^{''''}} - 9 y^{^{'''}} + 30 y^{^{''}} - 44 y^{^{'}} + 24 y = 0

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

6 y^{^{''}} + y^{^{'}} - y = 0, y (0) = e^{\frac{x}{3}}

What is the solution for x^{''}+6x^'+25x=0 ?
The solution for x^{''}+6x^'+25x=0 is x=e^{-3t}(c_{1}cos(4t)+c_{2}sin(4t))