Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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}

 

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 10, 12 compartment 2024 exams today for 2.5 lakh students; exam timings, guidelines
anam-khan

CBSE compartment admit card 2024 out at cbse.gov.in; exams from July 15
anam-khan

Reports saying CBSE cannot conduct board exams twice a year currently ‘totally incorrect’; board denies claims

Latest Question