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Question:

$x\left(x^{2}-1\right) \frac{d y}{d x}=1 ; y=0$ when $x=2$

Solution:

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

$\Rightarrow d y=\frac{d x}{x\left(x^{2}-1\right)}$

$\Rightarrow d y=\frac{1}{x(x-1)(x+1)} d x$

Integrating both sides, we get:

$\int d y=\int \frac{1}{x(x-1)(x+1)} d x$          …(1)

Let $\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$.                …(2)

$\Rightarrow \frac{1}{x(x-1)(x+1)}=\frac{A(x-1)(x+1)+B x(x+1)+C x(x-1)}{x(x-1)(x+1)}$

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

Comparing the coefficients of $x^{2}, x_{t}$ and constant, we get:

$A=-1$

$B-C=0$

$A+B+C=0$

Solving these equations, we get $B=\frac{1}{2}$ and $C=\frac{1}{2}$.

Substituting the values of AB, and C in equation (2), we get:

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

Therefore, equation (1) becomes:

$\int d y=-\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{1}{x+1} d x$

$\Rightarrow y=-\log x+\frac{1}{2} \log (x-1)+\frac{1}{2} \log (x+1)+\log k$

$\Rightarrow y=\frac{1}{2} \log \left[\frac{k^{2}(x-1)(x+1)}{x^{2}}\right]$                 …(3)

Now, $y=0$ when $x=2$.

$\Rightarrow 0=\frac{1}{2} \log \left[\frac{k^{2}(2-1)(2+1)}{4}\right]$

$\Rightarrow \log \left(\frac{3 k^{2}}{4}\right)=0$

$\Rightarrow \frac{3 k^{2}}{4}=1$

$\Rightarrow 3 k^{2}=4$

$\Rightarrow k^{2}=\frac{4}{3}$

Substituting the value of $k^{2}$ in equation (3), we get:

$y=\frac{1}{2} \log \left[\frac{4(x-1)(x+1)}{3 x^{2}}\right]$

$y=\frac{1}{2} \log \left[\frac{4\left(x^{2}-1\right)}{3 x^{2}}\right]$