What is the solution for x^{''}+2x^'+26x=0 ?

The solution for x^{''}+2x^'+26x=0 is x=e^{-t}(c_{1}cos(5t)+c_{2}sin(5t))

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x^{''}+2x^'+26x=0

x^{^{''}} + 2 x^{^{'}} + 26 x = 0

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

Solution steps

x^{''} (t) + 2 x^{'} (t) + 26 x = 0

Solve linear ODE:

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

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

\frac{d ^{2} f}{d x ^{2}} + k^{2} f (x) = 0

2 y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

x^{^{''}} + 8 x^{^{'}} = 0, x (0) = - 3, x^{^{'}} (0) = 5

y^{^{''}} - 6 y^{^{''}} + 8 y = 0

y^{^{''}} - 14 y^{^{'}} + 113 y = 0

What is the solution for x^{''}+2x^'+26x=0 ?
The solution for x^{''}+2x^'+26x=0 is x=e^{-t}(c_{1}cos(5t)+c_{2}sin(5t))