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}))

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = 1, y (0) = 22

20 y^{^{''}} + 60 y = 0

\frac{d ^{2} y}{d x ^{2}} - 36 y = 0

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} = 0

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))