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

The solution for x^{''}+x^'+2x=0 is x=e^{-t/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))

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

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

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

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

Solve linear ODE:

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

\frac{d y}{d x} = 2 y (3 - y)

y = 5 arctan (x - 1 + x^{2})

\int \frac{1}{y ^{2} - 3 y} d y

y = - x + 4, y = \frac{x ^{2}}{3}

\frac{d}{d x} (csc^{2} (x) - 4)

What is the solution for x^{''}+x^'+2x=0 ?
The solution for x^{''}+x^'+2x=0 is x=e^{-t/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))