Step 1: Separate the variables.

The given differential equation is:

$x(x^2-1)\frac{dy}{dx}=1$

Rewriting this, we get:

$dy = \frac{dx}{x(x^2-1)}$

Step 2: Integrate both sides.

Integrating the LHS gives $y$. For the RHS, use partial fractions for $\frac{1}{x(x^2-1)}$.

$\frac{1}{x(x^2-1)} = \frac{1}{x(x-1)(x+1)} = \frac{-1}{x} + \frac{1}{2(x-1)} + \frac{1}{2(x+1)}$

Now, integrate:

$\int dy = \int \left( \frac{-1}{x} + \frac{1}{2(x-1)} + \frac{1}{2(x+1)} \right) dx$

Step 3: Find the general solution.

$y = -\ln|x| + \frac{1}{2}\ln|x-1| + \frac{1}{2}\ln|x+1| + C$

Combining the logarithmic terms:

$y = \frac{1}{2}[\ln|x-1| + \ln|x+1|] - \ln|x| + C$

$y = \frac{1}{2}\ln|x^2-1| - \ln|x| + C$

$y = \ln\left( \frac{\sqrt{x^2-1}}{|x|} \right) + C$

Step 4: Apply the given condition to find C.

Given condition: $y=0$ when $x=2$.

Substituting these values:

$0 = \ln\left( \frac{\sqrt{2^2-1}}{2} \right) + C$

$0 = \ln\left( \frac{\sqrt{3}}{2} \right) + C$

$C = -\ln\left( \frac{\sqrt{3}}{2} \right) = \ln\left( \frac{2}{\sqrt{3}} \right)$

Step 5: Write the particular solution.

Substitute $C$ back into the general solution:

$y = \ln\left( \frac{\sqrt{x^2-1}}{x} \right) + \ln\left( \frac{2}{\sqrt{3}} \right)$

$y = \ln\left( \frac{2\sqrt{x^2-1}}{\sqrt{3}x} \right)$

∴ $y = \ln\left( \dfrac{2\sqrt{x^2-1}}{\sqrt{3}\,x} \right)$