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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2}}{d x ^{2}} (- sin (3 x))

\int x + 6 - x^{2} d x

\frac{d}{d x} (- \frac{12}{x ^{4}})

\int \frac{1}{( x + 4 ) x ^{2} + 8 x + 15} d x

y = 2 x^{2}, (- 1, 2)

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