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

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

Answers (1)

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}

 

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