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 21.10 Question 41 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 41 textbook solution.

Answers (1)

Answer : x=y^{2}-\frac{\pi^{2}}{4} \operatorname{cosec} y

Give : \frac{d x}{d y}+x \cot y=2 y+y^{2} \cot y, y \neq 0, x=0 \text { when } y=\frac{\pi}{2}

Hint : Using integration by parts and \int x^{n} d x

Explanation: \frac{d x}{d y}+x \cot y=2 y+y^{2} \cot y

                    =\frac{d x}{d y}+ (\cot y)\; x=2 y+y^{2} \cot y

This is a linear differential equation of the form

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

The integrating factor If  of this differential equation is

             \begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \cot y d x} \\ &=e^{\log |\sin y|} \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \cot x d x=\log |\sin x|+C\right] \\ &=\sin y\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[e^{\log e^{x}}=x\right] \end{aligned}

Hence, the solution is

          \begin{aligned} &x I f=\int Q I f d y+C \\ &=x(\sin y)=\int\left(2 y+y^{2} \cot y\right) \sin y d y+C \\ &=x \sin y=\int 2 y \sin y+y^{2} \cot y \sin y d y+C \\ &=x \sin y=\int 2 y \sin y+y^{2} \frac{\cos y}{\sin y} \sin y d y+C \end{aligned}

           \begin{aligned} &=x \sin y=\int 2 y \sin y d y+\int y^{2} \cos y d y+C \\ &=x \sin y=2 \int y \sin y d y+\int y^{2} \cos y d y+C \ldots(i) \end{aligned}

Using integration by parts

            \begin{aligned} &=2 \int y \sin y d y \\ &=2\left[\sin y \frac{y^{2}}{2}-\int \cos y \frac{y^{2}}{2} d y\right] \\ &=\sin y y^{2}-\int \cos y y^{2} d y \end{aligned}

Substituting on (i)

           \begin{aligned} &=x \sin y=\sin y y^{2}-\int \cos y y^{2} d y+\int y^{2} \cos y d y+C \\ &=x \sin y=\sin y y^{2}+C \end{aligned}

Divide by \sin y

             \begin{aligned} &=x=y^{2}+\frac{c}{\sin y} \\ &=x=y^{2}+C \operatorname{cosec} y \ldots(i i) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{1}{\sin y}=\operatorname{cosec} y\right] \end{aligned}

Now  x=0 \text { when } y=\frac{\pi}{2}

              =0=\left(\frac{\pi}{2}\right)^{2}+C \operatorname{cosec} \frac{\pi}{2}

              \begin{aligned} &=0=\frac{\pi^{2}}{4}+C(1)\; \; \; \; \; \; \; \; \quad\left[\operatorname{cosec} \frac{\pi}{2}=1\right] \\ &=C=-\frac{\pi^{2}}{4} \end{aligned}

Substituting in (ii)

            =x=y^{2}-\frac{\pi^{2}}{4} \operatorname{cosec} y

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 extends single girl child scholarship registration deadline till November 20
anam-khan

CBSE reminds schools to submit registration data of Class 9, 11 students by October 16
anam-khan

CBSE reopens online portal for submission of LOC for Class 10, 12 board exams 2026; submit forms by Oct 8

Latest Question