What is the solution for 2x^{''}+2x^'+3x=0 ?

The solution for 2x^{''}+2x^'+3x=0 is x=e^{-t/2}(c_{1}cos((sqrt(5)t)/2)+c_{2}sin((sqrt(5)t)/2))

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2x^{''}+2x^'+3x=0

2 x^{^{''}} + 2 x^{^{'}} + 3 x = 0

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}))

Solution steps

2 x^{''} (t) + 2 x^{'} (t) + 3 x = 0

Solve linear ODE:

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}))

x = e^{- \frac{t}{2}} (c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}))

What is the solution for 2x^{''}+2x^'+3x=0 ?
The solution for 2x^{''}+2x^'+3x=0 is x=e^{-t/2}(c_{1}cos((sqrt(5)t)/2)+c_{2}sin((sqrt(5)t)/2))