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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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\int \frac{1}{( 6 x - 1 ) ^{2}} d x

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