#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 65 Subquestion (i) textbook solution.

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

Hint          : You must know the rules of solving differential equation and integration

Given       :    $x\left(x^{2}-1\right) \frac{d y}{d x}=1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad, y=0 \text { where } x=2$

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

\begin{aligned} &\frac{d y}{d x}=\frac{1}{x\left(x^{2}-1\right)} \\ &d y=\left\{\frac{1}{x\left(x^{2}-1\right)}\right\} d x \end{aligned}

Integrating both sides,

\begin{aligned} &\int d y=\int\left\{\frac{1}{x\left(x^{2}-1\right)}\right\} d x \\ &y=\int \frac{1}{x(x+1)(x-1)} d x+c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right] \end{aligned}

Let,

$\begin{gathered} \frac{1}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1} \\ 1=A(x+1)(x-1)+B x(x-1)+C x(x+1) \\ 1=A\left(x^{2}-1\right)+B\left(x^{2}-x\right)+C\left(x^{2}+x\right) \\ 1=x^{2}(A+B+C)+x(-B+C)-A \end{gathered}$

Comparing both sides,

$\begin{array}{r} -A=1 \\ -B+C=0 \\ A+B+C=0 \end{array}$

Therefore, $A=-1, B=\frac{1}{2}, C =\frac{1}{2}$

Therefore,

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

Now,

\begin{aligned} &y=\int\left(\frac{-1}{x}+\frac{1}{2(x+1)}+\frac{1}{2(x-1)}\right) d x+c \\ &y=-\log |x|+\frac{1}{2} \log |x+1|+\frac{1}{2} \log |x-1|+c \\ &y=\frac{1}{2} \log |x+1|+\frac{1}{2} \log |x-1|-\log |x|+c \end{aligned}

Given, $y(2)=0$

Therefore, $y=0,x=2$

\begin{aligned} &0=\frac{1}{2} \log |2+1|+\frac{1}{2} \log |2-1|-\log |2|+c \\ &c=\log |2|-\frac{1}{2} \log (3) \quad[\log 1=0] \end{aligned}

Therefore the solution is ,

\begin{aligned} &y=\frac{1}{2} \log |x-1|+\frac{1}{2} \log |x+1|-\log |x|+\log (2)-\frac{1}{2} \log (3) \\ &2 y=\log |x-1|+\log |x+1|-2 \log |x|+2 \log (2)-\log 3 \\ &2 y=\log |x-1|+\log |x+1|-\log x^{2}+\log \left(2^{2}\right)-\log 3 \end{aligned}

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