What is the solution for (dy)/(dx)= y/(x^2-4) ?

The solution for (dy)/(dx)= y/(x^2-4) is y=e^{-1/4 (ln(x/2+1)-ln(x/2-1))+c_{1}}

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(dy)/(dx)= y/(x^2-4)

\frac{d y}{d x} = \frac{y}{x ^{2} - 4}

y = e^{- \frac{1}{4} (l n (\frac{x}{2} + 1) - l n (\frac{x}{2} - 1)) + c_{1}}

Solution steps

\frac{d y}{d x} = \frac{y}{x ^{2} - 4}

Solve separable ODE:

y = e^{- \frac{1}{4} (l n (\frac{x}{2} + 1) - l n (\frac{x}{2} - 1)) + c_{1}}

y = e^{- \frac{1}{4} (l n (\frac{x}{2} + 1) - l n (\frac{x}{2} - 1)) + c_{1}}

\frac{d}{d x} (5 x^{2} - y^{3}) = \frac{d}{d x} (7)

\frac{d v}{d t} = 9.56 - 0.1 v

2 x d x - y d y = 0, y = f (x)

y^{^{'}} = \frac{y}{x ^{2} - 4}

d y y^{- 2} = \frac{d x}{- x ^{2} + 144}

What is the solution for (dy)/(dx)= y/(x^2-4) ?
The solution for (dy)/(dx)= y/(x^2-4) is y=e^{-1/4 (ln(x/2+1)-ln(x/2-1))+c_{1}}