Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 38 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 38 Maths Textbook Solution.

Answers (1)

Answer: \frac{2}{3} \tan ^{-1}\left(\frac{2 y+x}{x \sqrt{3}}\right)-\frac{\pi}{\sqrt{3}}=\log \left(x^{2}+x y+y^{2}\right)

Given: \left ( x-y \right )\frac{dy}{dx}=x+2y

To find: we have to find the solution of given differential equation.

Hint: we will put y=vx \: and\: \frac{dy}{dx}=v+x\frac{dv}{dx}

Solution: we have,

\left ( x-y \right )\frac{dy}{dx}=x+2y            ..(i)

It is homogeneous equation.

put y=vx \: and\: \frac{dy}{dx}=v+x\frac{dv}{dx}

So,

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

Separating the variables and integrating both side we get

\begin{aligned} &\int \frac{v-1}{1+2 v+v^{2}} d v=-\int \frac{d x}{x} \\ &\Rightarrow \int \frac{\frac{1}{2}(2 v+1)-\frac{3}{2}}{1+v+v^{2}} d v=-\log |x| \\ &\Rightarrow \frac{1}{2} \int \frac{(2 v+1)}{1+v+v^{2}}-\frac{3}{2} \int \frac{1}{1+v+v^{2}} d v=-\log |x| \\ &\Rightarrow \log \left|1+v+v^{2}\right|-3 \int \frac{1}{v^{2}+v+1+\left(\frac{1}{2}\right)^{4}-\left(\frac{1}{2}\right)^{2}}=-2 \log |x| \end{aligned}

\begin{aligned} &\left.\Rightarrow \log \left|v^{2}+v+1\right|-3 \frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1} \mid \frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=-2 \log |x|\\ &\text { Putting } \mathrm{v}=\frac{y}{x}\\ &\Rightarrow \log \left|\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right|-2 \sqrt{3} \tan ^{-1}\left(\frac{\frac{2 y}{x}+1}{\sqrt{3}}\right)=-\log |x|+c\\ &\Rightarrow \log \left|y^{2}+x y+x^{2}\right|-\log x^{2}-2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{x \sqrt{3}}\right)=-\log x^{2}+c\\ &\Rightarrow \log \left|y^{2}+x y+x^{2}\right|=c+2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{x \sqrt{3}}\right) \end{aligned}

It is given that y=0 when x=1

Putting y=0, x=1 in equation (ii) we get

\begin{aligned} &\Rightarrow \log |1|=c+2 \sqrt{3} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ &\Rightarrow 0=c+2 \sqrt{3} \times \frac{\pi}{6} \\ &\Rightarrow c=-\frac{\pi}{\sqrt{3}} \\ \end{aligned}

Putting value of c in equation (ii) we get

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

This is required solution.

 

Posted by

infoexpert21

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