What is the solution for x^{''}+8x^'+25x=0 ?

The solution for x^{''}+8x^'+25x=0 is x=e^{-4t}(c_{1}cos(3t)+c_{2}sin(3t))

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x^{''}+8x^'+25x=0

x^{^{''}} + 8 x^{^{'}} + 25 x = 0

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

Solution steps

x^{''} (t) + 8 x^{'} (t) + 25 x = 0

Solve linear ODE:

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

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

(D^{2} + 9) y = 0, Dy = 0, y = 2, x = 30

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = e^{6 x}

y^{^{'''''}} - 3 y^{^{'''}} = 0

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

y^{^{'''}} + 2 y^{^{''}} - 63 y^{^{'}} = 0

What is the solution for x^{''}+8x^'+25x=0 ?
The solution for x^{''}+8x^'+25x=0 is x=e^{-4t}(c_{1}cos(3t)+c_{2}sin(3t))