Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 29 maths textbook solution

Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 29 maths textbook solution

Answers (1)

best_answer

Given: Slope of tangent  =\frac{x^{2}+y^{2}}{2 x y}.

To prove: The curves which slope at any point (x, y) \text { is } \frac{x^{2}+y^{2}}{2 x y}  are in rectangular hyperbola.

Hint: The homogeneous differential equation put y=v x \frac{d y}{d x}=v+x \frac{d v}{d x}

Solution: Shape of tangent  =\frac{x^{2}+y^{2}}{2 x y}

        =\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}

It is a homogeneous equation

Out y=vx

Then,

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

        \begin{aligned} &=x \frac{d v}{d x}=\frac{1+v^{2}}{2 v}-v \\\\ &=x \frac{d v}{d x}=\frac{1+v^{2}-2 v^{2}}{2 v} \\\\ &=x \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \end{aligned}

        =\frac{2 v}{1-v^{2}}=\frac{d x}{x}

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

        =\frac{2 v}{v^{2}-1}=-\frac{d x}{x}        [Separating variables]

Integrating on both sides

        \begin{aligned} &=\int \frac{2 v}{1-v^{2}} d v=-\int \frac{1}{x} d x \\\\ &=\int \frac{1}{t} d t=-\int \frac{1}{x} d x \quad\left[1-v^{2}=t, 2 v d v=d t\right] \end{aligned}

        \begin{aligned} &=\log t=-\log x+\log C \\\\ &=\log \left(1-v^{2}\right)=-\log x+\log C \\\\ &=-\log x=\log \left(1-v^{2}\right)-\log C \\\\ &=-\log x=\log \frac{\left(1-v^{2}\right)}{c} \end{aligned}

        \begin{aligned} &=\log x^{-1}=\log \frac{\left(1-v^{2}\right)}{c} \\\\ &=\log \frac{1}{x}=\log \frac{\left(1-v^{2}\right)}{c} \end{aligned}

Using antilogarithm

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

        \begin{aligned} &=\frac{x^{2}-y^{2}}{x^{2}}=\frac{c}{x} \\\\ &=x^{2}-y^{2}=\frac{c x^{2}}{x} \\\\ &=x^{2}-y^{2}=C x \end{aligned}

Hence, x^{2}-y^{2}=C x is the equation for a rectangular hyperbola.

 

Posted by

infoexpert26

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