RD Sharma Class 12 Exercise 21.6 Differential Equation Solutions Maths

Lovekush kumar sainiUpdated on 08 Jan 2022, 11:38 AM IST

The Class 12 students need a proper guide to clarify their doubts on any concepts at their home. To make this process simpler, especially for the students who work out mathematical sums on their own for chapter 21, the RD Sharma Class 12th Exercise 21.6 solution book is handy. It is pretty standard that Differential Equation is a challenging portion. Yet with the help of authorized solution books, the students can face it easily.

Differential Equations Exercise: 21.6

Differential equations exercise 21.6 question 1

Answer: $x+\frac{1}{2} \log \left|1+y^{2}\right|=C$
Given: $\frac{d y}{d x}+\frac{1+y^{2}}{y}=0$
Hint: Separate y and x and then integrate both sides.
Solution:
$\begin{aligned} &\begin{aligned} \frac{d y}{d x}+\frac{1+y^{2}}{y} &=0 \\ \frac{d y}{d x} &=-\left(\frac{1+y^{2}}{y}\right) \end{aligned}\\ &\left(\frac{y}{1+y^{2}}\right) d y=-d x\\ &\text { [integrate both side] }\\ &\int\left(\frac{y}{1+y^{2}}\right) d y=-\int d x \end{aligned}$
$\begin{aligned} &\text { Let } 1+y^{2}=t\\ &\text { [differentiate with respect to } y \text { ] }\\ &\begin{aligned} 2 y d y &=d t \\ & \frac{1}{2} \int \frac{1}{t} d t=-\int d x \\ & \frac{1}{2} \log |t|=-x+C \\ & \frac{1}{2} \log \left|1+y^{2}\right|+x=C \end{aligned} \end{aligned}$

Differential equations exercise 21.6 question 2

Answer: $x=\frac{y^{2}}{2}-\frac{1}{2} \log \left|y^{2}+1\right|+C$
Given: $\frac{d y}{d x}=\frac{1+y^{2}}{y^{3}}$
Hint: Use integration by parts method
Solution: $\frac{d y}{d x}=\frac{1+y^{2}}{y^{3}}$
$\frac{y^{3}}{1+y^{2}} d y=d x$
On dividing $\frac{y^{3}}{1+y^{2}}$ we will write it as $y-\frac{y}{y^{2}+1} \text { i.e., } \mathrm{Q}+\frac{R}{D}$

$\int\left(y-\frac{y}{y^{2}+1}\right) d y=\int d x \; \; \; \; \; \; \; \; [integrate \: both\: sides]$
$\begin{aligned} &\int y d y-\int \frac{y}{y^{2}+1} d y=\int d x\\ &\frac{y^{2}}{2}-\frac{1}{2} \int \frac{2 y}{y^{2}+1} d y=x \end{aligned}$
$\text { Let } y^{2}+1=t$ [ Diffrentiate with reference to y]
$\begin{aligned} & 2 y d y=d t \\ &\qquad \begin{array}{c} \frac{y^{2}}{2}-\frac{1}{2} \int \frac{1}{t} d t=x \\ \\\frac{y^{2}}{2}-\frac{1}{2} \log |t|=x+C \end{array} \\ \\&\qquad \begin{array}{l} \frac{y^{2}}{2}-\frac{1}{2} \log \left|y^{2}+1\right|=x+C \end{array} \end{aligned}$

Differential equations exercise 21.6 question 3

Answer: $x+\cot y=C$
Given: $\frac{d y}{d x}=\sin ^{2} y$
Hint: You must know trigonometric identities
Solution: $\frac{d y}{d x}=\sin ^{2} y$
$\begin{aligned} &\frac{1}{\sin ^{2} y} d y=d x\\ &\operatorname{cosec}^{2} y d y=d x \quad \text { [applying integration] }\\ &\int \operatorname{cosec}^{2} y d y=\int d x \quad\: \; \; \; \; \; \; \; \; \; \left[\int \operatorname{cosec}^{2} y d y=-\operatorname{coty}+c\right]\\ &-\cot y=x+C\\ &x+\cot y=C \end{aligned}$

Differential equations exercise 21.6 question 4

Answer: $x+\cot y+y=C$
Given: $\frac{d y}{d x}=\frac{1-\cos 2 y}{1+\cos 2 y}$
Hint: Use trigonometric identities to simplify and integrate
Solution
$\frac{d y}{d x}=\frac{1-\cos 2 y}{1+\cos 2 y}$ $\left[\cos 2 x=1-\sin ^{2} x=2 \cos ^{2} x-1\right]$
$\frac{d y}{d x}=\frac{2 \sin ^{2} y}{2 \cos ^{2} y}$ $\left[\frac{\sin x}{\cos x}=\tan x\right]$
$\frac{d y}{d x}=\tan ^{2} y$
$\frac{1}{\tan ^{2} y} d y=d x$ $\left[\frac{1}{\tan x}=\cot x\right]$
$\int \cot ^{2} y d y=\int d x$ $\left[\cot ^{2} y=\operatorname{cosec}^{2} y-1\right]$
$\int\left(\operatorname{cosec}^{2} y-1\right) d y=\int d x$ $\left[\int \operatorname{cosec}^{2} y d y=-\cot y+c\right]$
$\begin{aligned} &-\cot y-y+C=x \\ &-\cot y-y-x=-C \\ &C=\cot y+y+x \end{aligned}$

Also see,

RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.1
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.2
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.3
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.4
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.6
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.7
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.8
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.9
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.10
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise 21.11
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise FBQ
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise MCQ
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise CSBQ
RD Sharma Solutions Class 12 Mathematics Chapter 21 Exercise RE

The Class 12 syllabus consists of the Differential Equation concept as its 21st chapter. Among the eleven exercises given in this chapter, the sixth exercise, ex 21.6, has fewer sums given in the textbook. This is because there are only four sums to be solved under the concept of the differential equation. Even though the count of the sums is less, students get doubts in the middle of these sums. The RD Sharma Class 12 Chapter 21 Exercise 21.6 reference material will lend a helping hand in such circumstances.

Exercise 21.6 has only Level 1 sums; therefore, it would not be tough to solve. Yet, these questions are not enough for a student to understand the concept deeply. Consequently, you can find various other practice sums to work out on the same concept given in the RD Sharma Class 12th Exercise 21.6 book. This is because Differential Equation is a concept that will only be easier to solve with a lot of practice. Therefore, the NCERT pattern solutions in this book give another valid reason for the CBSE students to adapt it.

The students can trust the answers given in the Class 12 RD Sharma Chapter 21 Exercise 21.6 Solution as much as they trust their mathematics teacher. A committee of staff members provides all the solutions presented here with very high expertise in this domain. Therefore, not only does this book help the student in preparing for their exams, but it also lends a helping hand in solving their homework and assignments.

And the other benefit that excites most of the students is that the RD Sharma Class 12 Solutions Differential Equation Ex 21.6 book can be downloaded for free from the Career 360 website. Thus, every student can access the RD Sharma Class 12th Exercise 21.6 book without spending money as they usually do to buy other books.

The usage of RD Sharma Class 12 Solutions Chapter 21 Ex 21.6 reference material to prepare questionnaires by teachers has increased its value. In the presence of this book, you need not depend on a teacher or a tutor each time you get doubts in solving differentiation.