RD Sharma Class 12 Exercise 21.5 Differential Equation Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 21.5 Differential Equation Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 24, 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.

## 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.

Above all, the RD Sharma Class 12th Exercise 21.5 reference material is available for free of cost at the Career 360 website. This allows the students to download the best mathematics guide without any payment. RD Sharma solutions Many students have already benefitted with the help of RD Sharma books. Now, it’s your turn to use these books wisely and increase your performance in the tests and examinations.

Another benefit of using the RD Sharma Class 12 Solutions Chapter 21 Ex 21.5 book is that you might face questions asked from it at your public exams. Therefore, using this reference material will help you get exam-ready every day.

RD Sharma Chapter wise Solutions

1. Are the RD Sharma solution books affordable?

This RD Sharma books can be downloaded in PDF format from the Career 360 website. Hence, you need not worry about the cost and if you can afford it.

2. Which mathematics guide has the best and easy solutions for the Differential Equations chapter?

The RD Sharma Class 12th Exercise 21.5 book has the best and easy solutions for the people who find it difficult to solve the Differential Equation sums.

3. What advantages do the students get when they use the RD Sharma solution books?
• Accurate solutions by experts.

• Various practice sums are present to work out.

4. How many questions are answered in the RD Sharma solution book for exercise 21.5?

The RD Sharma Class 12th Exercise 21.5 contains the solutions for all the 26 questions asked in the textbook.

5. Are our RD Sharma books worth for the class 12 students to follow it?

The RD Sharma solution books contain the in-depth answer that clarifies their doubts and makes them understand the concept clearly. Therefore, the Class 12 students can use this guide to score well.

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