$x({x^2} - 1)\cfrac{{dy}}{{dx}} = 1;y = 0$ when $x = 2$

.: We have, $x({x^2} - 1)\cfrac{{dy}}{{dx}} = 1$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{x(x + 1)(x - 1)}} \Rightarrow dy = \cfrac{{dx}}{{x(x + 1)(x - 1)}}$

…(1)
Integrating (1) both sides,

we get
$\int d y = \int {\cfrac{{dx}}{{x(x + 1)(x - 1)}}}$

Let $\cfrac{1}{{x(x + 1)(x - 1)}} = \cfrac{A}{x} + \cfrac{B}{{x + 1}} + \cfrac{C}{{x - 1}}$

$\Rightarrow 1 = A(x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1)$

Putting $x = 0$,

we get $1 = A(0 + 1)(0 - 1) \Rightarrow A = - 1$

Putting $x + 1 = 0 \Rightarrow x = - 1$,

we get $1 = B( - 1)( - 1 - 1\rangle \Rightarrow B = \cfrac{1}{2}$

Putting$x - 1 = 0 \Rightarrow x = 1$ ,

we get $1 = C(1)(1 + 1) \Rightarrow C = \cfrac{1}{2}$

$\therefore$ $y = \int {\left[ {\cfrac{{ - 1}}{x} + \cfrac{{\cfrac{1}{2}}}{{x + 1}} + \cfrac{{\cfrac{1}{2}}}{{x - 1}}} \right]dx}$

$\Rightarrow y = - \log x + \cfrac{1}{2}\log (x + 1) + \cfrac{1}{2}\log (x - 1) + C$

$\Rightarrow y = \cfrac{1}{2}\log \left| {\left( {\cfrac{{{x^2} - 1}}{{{x^2}}}} \right)} \right| + C$

When, $x = 2,y = 0$

$0 = \cfrac{1}{2}\log \left| {\left( {\cfrac{3}{4}} \right)} \right| + C$

$\Rightarrow C = - \cfrac{1}{2}\log \cfrac{3}{4}$

Hence, particular solution is $y = \cfrac{1}{2}\log \left( {\cfrac{{{x^2} - 1}}{{{x^2}}}} \right) - \cfrac{1}{2}\log \cfrac{3}{4}$