RD Sharma Class 12 Exercise 21.6 Differential Equation Solutions Maths

Lovekush kumar sainiUpdated on 24 Jan 2022, 04:20 PM IST

The CBSE board school students use the RD Sharma books to clear their doubts in various subjects. Regarding mathematics, especially chapter 21, the students find it challenging to work out the differential equations. Solution books like the RD Sharma Class 12th Exercise 21.5 lend them a helping hand to solve those sums without any difficulties.

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise: 21.5

Differential equation exercise 21.5 question 1

Answer:$y =\frac{x^3}{3}+\frac{x^2}{2}+\log \left | x \right |+C$
Hint: You have to integrate by applying integration of xⁿ.
Given: $\frac{dy}{dx}=x^2+x-\frac{1}{x},x\neq 0$
Solution: $\frac{dy}{dx}=x^2+x-\frac{1}{x}$
$dy=\left (x^2+x-\frac{1}{x} \right )dx$
Integrating both sides,
$\begin{aligned} &\int d y=\int\left(x^{2}+x-\frac{1}{x}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \frac{1}{x} d x=\log |x|+C\right]\\ &y=\frac{x^{3}}{3}+\frac{x^{2}}{2}+\log |x|+C \mid \end{aligned}$

Differential equation exercise 21.5 question 2

Answer: $\frac{x^6}{6}+\frac{x^3}{3}-2 \log \left | x \right |+C=y$
Hint: You have to integrate by applying integration of xⁿ.
Given:$\frac{dy}{dx}=x^5+x^2-\frac{2}{x},x\neq 0$
Solution:$\frac{dy}{dx}=x^5+x^2-\frac{2}{x}$
$dy=\left (x^5+x^2-\frac{2}{x} \right )dx$
Integrating both sides,
$\begin{aligned} &\int d y=\int\left(x^{5}+x^{2}-\frac{2}{x}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \frac{1}{x} d x=\log |x|+C\right]\\ &y=\frac{x^{6}}{6}+\frac{x^{3}}{3}-2 \log |x|+C \end{aligned}$

Differential equation exercise 21.5 question 3

Answer: $y+x^2=\frac{e^{3x}}{3}+C$
Hint: You have to integrate by applying integration of xⁿ.
Given: $\frac{dy}{dx}+2x=e^{3x}$
Solution:$\frac{dy}{dx}+2x=e^{3x}$
$\frac{dy}{dx}=-2x +e^{3x}$
$dy=\left (-2x +e^{3x} \right )dx$
Integrating both sides,
$\begin{aligned} &\int d y=\int\left(-2 x+e^{3 x}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \frac{1}{x} d x=\log |x|+C\right] \\ &y=-\frac{2 x^{2}}{2}+\frac{e^{a x}}{3} \\ &\quad=-x^{2}+\frac{e^{3 x}}{3} \\ &y+x^{2}=\frac{e^{3 x}}{3}+C \end{aligned}$

Differential equation exercise 21.5 question 4

Answer:$y=\tan^{-1}x+C$

Hint: You have to integrate by applying integration of xⁿ.

Given: $(x^2+1)\frac{dy}{dx}=1$

Solution:$\frac{dy}{dx}=\frac{1}{(x^2+1)}$

$dy=\frac{1}{(x^2+1)}dx$

Integrating both sides,

$\begin{array}{ll} \int d y=\int \frac{1}{1+x^{2}} d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad & {\left[\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+C\right]} \\\\ y=\tan ^{-1} x+C \end{array}$

Differential equation exercise 21.5 question 5

Answer:$y=2\tan ^ 2\frac{x}{2}-x+C$
Hint: You have to integrate by applying integration of xⁿ.
Given: $\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$
Solution: We know
$\begin{aligned} \cos 2 \theta &=2 \cos ^{2} \theta-1 \\ &=>\cos 2 \theta+1=2 \cos ^{2} \theta \end{aligned}$
Replace $\theta$ by $\frac{x}{2}$
$\begin{aligned} &=\cos 2\left(\frac{x}{2}\right)+1=2 \cos ^{2} \frac{x}{2} \\ &=>\cos x+1=2 \cos ^{2} \frac{x}{2} \ldots(i) \\ \cos 2 \theta &=1-2 \sin ^{2} \theta \\ &=>1-\cos 2 \theta=2 \sin ^{2} \theta \end{aligned}$
Replace $\theta$ by $\frac{x}{2}$
$\begin{aligned} &\Rightarrow 2 \sin ^{2} \frac{x}{2}=1-\cos 2\left(\frac{x}{2}\right)\\ &=>2 \sin ^{2} \frac{x}{2}=1-\cos x \end{aligned}$
From (i) and (ii)
$\begin{aligned} &\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}=>\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=>\tan ^{2} \frac{x}{2} \\ &\frac{d y}{d x}=\tan ^{2} \frac{x}{2} \end{aligned}$
Integrating both sides,
$\begin{aligned} &\int \frac{d y}{d x}=\int \tan ^{2} \frac{x}{2} \\ &y=\int \sec ^{2} \frac{x}{2}-1\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{2} x=\sec ^{2} x-1\right] \\ &\quad=\int \sec ^{2} \frac{x}{2} d x-\int 1 d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \sec ^{2} x d x=\tan x+c\right] \\ &=\frac{\tan ^{2} \frac{x}{2}}{\frac{1}{2}}-x+C \\ &y=2 \tan ^{2} \frac{x}{2}-x+C \end{aligned}$

The syllabus of class 12, mathematics, chapter 21, Differential Equation plays a vital role in making the students score marks. There are eleven exercises in this chapter. Exercise 21.5 revolves around the concept of differentiating the given equations. There are 26 questions in this exercise to be solved under the same topic. It consists of only the Level 1 part, and hence, the sums would be a bit easier. It is better to have a copy of the RD Sharma Class 12 Chapter 21 Exercise 21.5 book to clarify their doubts.

The CBSE syllabus school students generally use NCERT based books to cover their concepts. As the RD Sharma Class 12th Exercise 21.5 book also follows the NCERT pattern, it is given preference. Differential Equation might be a bit of a complex concept for the students to work out, but if it is practiced and well-understood, no one can prevent students from scoring high marks. Therefore, the Class 12 RD Sharma Chapter 21 Exercise 21.5 Solution material plays a significant role as it has numerous practice questions.

The students can refer to this book while encountering any doubts or unclarity in homework sums and assignments. Using the RD Sharma Class 12 Solutions Differential Equation Ex 21.5 book for homework would help them use the same to prepare for their exams too. A random person has not given the solutions provided in the RD Sharma books but a group of experts with skill and experience in the respective subject. Hence, the students need not worry about the accuracy of each answer.

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