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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2^{x}, ln (x), \frac{1}{2}, 2

x \to 4 lim (- x^{2} + 4 x)

\int 3 y e^{3 x} d x

f (x) = \frac{2}{3 x + 5}, at x = - 1

4 y^{^{''}} - 4 y^{^{'}} + 50 y = 0

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