Solve the following differential equation: x(x² - 1)(dy)/(dx) = 1, y = 0, given x = 2

$\frac{dy}{dx} = \frac{1}{x(x^2-1)}$, so $dy = \frac{dx}{x(x-1)(x+1)}$. Partial fractions: $\frac{1}{x(x-1)(x+1)} = -\frac1x + \frac{1}{2(x-1)} + \frac{1}{2(x+1)}$ Integrating: $y = -\log|x| + \frac12\log|x-1| + \frac12\log|x+1| + C = \frac12\log\left|\frac{x^2-1}{x^2}\right| + C$ Put $x = 2$, $y = 0$: $0 = \frac12\log\frac34 + C \Rightarrow C = -\frac12\log\frac34 = \frac12\log\frac43$ $y = \frac12\log\frac{x^2-1}{x^2} + \frac12\log\frac43$ Hence $y = \frac12\log\left|\frac{4(x^2-1)}{3x^2}\right|$.