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

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

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y^'= y/(x^2-4)

y^{^{'}} = \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

y^{'} (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}}

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

y = y^{^{'}} x

\frac{d y}{d x} = x^{5} (y - 5), y (0) = 8

(1 - x^{2}) y^{^{'}} - y^{2} = 0

\frac{d x}{d t} = 125 - \frac{x}{100}, x (0) = 4000

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