What is the solution for x^{''}+4x^'+5x=0 ?

The solution for x^{''}+4x^'+5x=0 is x=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))

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x^{''}+4x^'+5x=0

x^{^{''}} + 4 x^{^{'}} + 5 x = 0

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

Solution steps

x^{''} (t) + 4 x^{'} (t) + 5 x = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial n} (nm)

x \to 2 lim (\frac{3 x + 2}{x + 2})

\int_{0}^{2} (2 x + 2 - x^{2} - 2) d x

x \to \infty lim (\frac{ln ( x ^{2} - x + 1 )}{ln ( x ^{10} + x + 1 )})

\frac{\partial}{\partial x} (3 sin (2 x))

What is the solution for x^{''}+4x^'+5x=0 ?
The solution for x^{''}+4x^'+5x=0 is x=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))