Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 42

Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 42

Answers (1)

Answer:   \begin{aligned} & x+\cot ^{-1} y=1+C e^{\tan ^{-1} y}\\ & \end{aligned}

Give:  \begin{aligned} & \left(\cot ^{-1} y+x\right) d y=\left(1+y^{2}\right) d x\\ \end{aligned}

Hint: Using integration by parts and  \int \frac{1}{1+x^{2}} d x

Explanation:   \left(\cot ^{-1} y+x\right) d y=\left(1+y^{2}\right) d x

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

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

This is a linear differential equation of the form

 \begin{aligned} &\frac{d x}{d y}+P x=Q \\ &P=\frac{-1}{1+y^{2}} \text { and } Q=\frac{\cot ^{-1} y}{1+y^{2}} \end{aligned}

The integrating factor  If  of this differential equation is

 \begin{aligned} &I f=e^{\int P d y} \\\\ &=e^{\int \frac{-1}{1+y^{2}} d y} \\\\ &=e^{-\int \frac{1}{1+y^{2}} d y} \end{aligned}

\begin{array}{ll} =e^{-\tan ^{-1} y} &\qquad\qquad {\left[\int \frac{1}{1+\mathrm{y}^{2}} d y=\tan ^{-1} y+C\right]} \\\\ =e^{\cot ^{-1} y} &\qquad {\left[\tan ^{-1} y=\cot ^{-1}\left(\frac{1}{y}\right)\right]} \end{array}

Hence, the solution is

x\: \text{If}=\int Q \: \text{If} \: d y+C

 =x\left(e^{\cot ^{-1} y}\right)=\int \frac{\cot ^{-1} y}{1+y^{2}} e^{\cot ^{-1} y} d y

Put   \cot ^{-1} y=t \\

\begin{aligned} & &\qquad \begin{aligned} =&-\frac{d y}{1+y^{2}}=d t \\ =& \frac{d y}{1+y^{2}}=-d t \end{aligned} \end{aligned} 

 

So,

=-\int t e^{t} d t  

Using integration by parts

\begin{aligned} &=-\left[t e^{t}-\int e^{t} d t\right] \\\\ &=-\left[t e^{t}-e^{t}-C\right] \\\\ &=-t e^{t}+e^{t}+C \\ \\&=x e^{\cot ^{-1} y}=-\cot ^{-1} y e^{\cot ^{-1} y}+e^{\cot ^{-1} y}+C \end{aligned}

 

Divide by  e^{cot^{-1}}y

 \begin{aligned} &=x=-\cot ^{-1} y+1+\frac{c}{e^{\cot ^{-1} y}} \\ &=x+\cot ^{-1} y=1+C e^{\tan ^{-1} y} \quad\left[\cot ^{-1} y=\tan ^{-1}\left(\frac{1}{y}\right)\right] \end{aligned}

Posted by

infoexpert27

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