RD Sharma Class 12 Exercise 21.1 Differential Equation Solutions Maths-Download PDF Online

# RD Sharma Class 12 Exercise 21.1 Differential Equation Solutions Maths-Download PDF Online

Edited By Satyajeet Kumar | Updated on Jan 24, 2022 04:19 PM IST

RD Sharma books are the gold standard for class 12 maths. They are the best material that students can refer to get in detail knowledge about the subject and get familiar with a lot of concepts. A majority of schools all over the country use RD Sharma books to set up question papers as well as to follow in their class lectures. This is why it remains the number one choice among students for maths.

Class 12 RD Sharma chapter 21 exercise 21.1 covers the chapter Differential Equations. This exercise contains 42 questions that are of Level-1 difficulty. Students can easily get through the solutions with a basic understanding of the chapter. With the help of RD Sharma Solutions Differential Equations Ex 21.1. Students can quickly finish the exercise and save time for their revision.

## Differential Equations Excercise: 21.1

Order=3 , Degree=1 , Non-linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\frac{d^{3}x}{dt^{3}}+\frac{d^{2}x}{dt^{2}}+\left ( \frac{dx}{dt} \right )^{2}=e^{t}$
Solution:
So, in this question, the order of the differential equation is 3 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
So, in this question. The dependent variable is x and the term $\frac{dx}{dt}$ is multiplied by itself. So the given equation is non-linear.
Therefore, Order=3 , Degree=1 , Non-linear

Differential Equation exercise 21.1 question 2

Order=2 , Degree=1 , linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\frac{d^{2}y}{dx^{2}}+4y=0$
Solution:
So, in this question, the order of the differential equation is 2 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
So, in this question. The dependent variable is y and its derivatives are multiplied with a constant or independent variable only. So this equation is linear differential equation.
Therefore, Order=2 , Degree=1 , linear

Differential Equation exercise 21.1 question 3

Order=1 , Degree=3 , Non-linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\left [ \frac{dy}{dx} \right ]^{2}+\frac{1}{\frac{dy}{dx}}=2$
Solution:
So, in this question, first we need to remove the term $\frac{1}{\frac{dy}{dx}}$ because this can be written as $\left (\frac{dy}{dx} \right )^{-1}$ which means a negative power.
So, the above equation becomes $\left [ \frac{dy}{dx} \right ]^{3}+1=2 \frac{dy}{dx}$
So, in this question, the order of the differential equation is 1 and the degree of the differential equation is 3.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
So, in this question. The dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So the given equation is non-linear.
Therefore, Order=1 , Degree=3 , Non-linear

Differential Equation exercise 21.1 question 4

Order=2 , Degree=2 , Non-linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\sqrt{1+\left [ \frac{dy}{dx} \right ]^{2}}=\left [ c\frac{d^{2}y}{dx^{2}} \right ]^{\frac{1}{3}}$
Solution:
In this question, we will raising both the sides the power 6. So we remove the fractional powers of derivatives of the dependent variable y.
So, the above equation becomes
$\left [1+\left [ \frac{dy}{dx} \right ]^{2} \right ]^{3}=\left [ c\frac{d^{2}y}{dx^{2}} \right ]^{2}$
$1+\left ( \frac{dy}{dx} \right )^{2}+3\left ( \frac{dy}{dx} \right )^{2}+3\left ( \frac{dy}{dx} \right )^{4}=c^{2}\left ( \frac{d^{2}y}{dx} \right )^{2}$
So, in this question, the order of the differential equation is 2 and the degree of the differential equation is 2.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
So, in this question. The dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself and many other are also. So the given equation is non-linear.
Therefore, Order=2 , Degree=2 , Non-linear

Differential Equation exercise 21.1 question 5

Order=2 , Degree=1, Non-linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\frac{d^{2}y}{dx^{2}}+\left [ \frac{dy}{dx} \right ]^{2}+xy=0$
Solution:
So, in this question, the order of the differential equation is 2 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
So, in this question. The dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So the given equation is non-linear.
Therefore, Order=2 , Degree=1 , Non-linear

Differential Equation exercise 21.1 question 6

Order=2 , Degree=2 , Non-linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\sqrt[3]{\frac{d^{2}y}{dx^{2}}}=\sqrt{\frac{dy}{dx}}$
Solution:
Squaring on both sides, we get
$\left (\sqrt[3]{\frac{d^{2}y}{dx^{2}}} \right )^{2}=\frac{dy}{dx}$
Cubing on both sides,
$\left (\frac{d^{2}y}{dx^{2}} \right )^{2}=\left (\frac{dy}{dx} \right )^{3}$
So, in this question, the order of the differential equation is 2 and the degree of the differential equation is 2.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
So, in this question. The dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So the given equation is non-linear.
Therefore, Order=2 , Degree=2 , Non-linear

Differential Equation exercise 21.1 question 7

Order=4 , Degree=2 , Non-linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\frac{d^{4}y}{dx^{4}}=\left [ c+\left ( \frac{dy}{dx} \right )^{2} \right ]^{\frac{3}{2}}$
Solution:
Since this equation has fractional powers, we need to remove them.
So squaring on both sides, we get
$\left (\frac{d^{4}y}{dx^{4}} \right )^{2}=\left [ c+\left ( \frac{dy}{dx} \right )^{2} \right ]^{3}$
Solving both sides,
$\left (\frac{d^{4}y}{dx^{4}} \right )^{2}=c^{3}+\left ( \frac{dy}{dx} \right )^{6}+3c^{2}\left ( \frac{dy}{dx} \right )^{2}+3c\left ( \frac{dy}{dx} \right )^{4}$
So, in this question, the order of the differential equation is 4 and the degree of the differential equation is 2.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
So, in this question. The dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself, also the degree of the
equation is 2 which must be one for the equation to be linear. So the given equation is non-linear.
Therefore, Order=4 , Degree=2 , Non-linear

Differential Equation exercise 21.1 question 8

Order=1 , Degree=2 , linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$x+\frac{dy}{dx}=\sqrt{1+\left ( \frac{dy}{dx} \right )^{2}}$
Solution:
Since this equation has fractional powers, we need to remove them.
So squaring on both sides, we get
$\left (x+\frac{dy}{dx} \right )^{2}=1+\left ( \frac{dy}{dx} \right )^{2}\Rightarrow x^{2}+\left ( \frac{dy}{dx} \right )^{2}+2x\frac{dy}{dx}=1+\left ( \frac{dy}{dx} \right )^{2}$
$x^{2}+2x\frac{dy}{dx}-1=0$
So, in this question, the order of the differential equation is 1 and the degree of the differential equation is 2.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
So, in this question. The dependent variable is y and its derivatives are multiplied with a constant or independent variable only. So this equation is linear differential equation.
Therefore, Order=1 , Degree=2 , linear

Differential Equation exercise 21.1 question 9

Order=2 , Degree=1, linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.Given:$y\frac{d^{2}x}{dy^{2}}=y^{2}+1$
Solution:
Here in this question, the dependent variable is x and thus the order of the differential equation is 2 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
So, in this question. The dependent variable is x and its derivatives are multiplied with a constant or independent variable only. So this equation is linear differential equation.
Therefore, Order=2 , Degree=1 , linear

Differential Equation exercise 21.1 question 10

Order=2, Degree=1, Non- linear

Hint:

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

Given:

$s^{2}\frac{d^{2}t}{ds^{2}}+st\frac{dt}{ds}=s$

Solution:

Here in this question, the dependent variable is t and thus the order of the differential equation is 2 and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.

Here the dependent variable is t and its derivatives are multiplied together with$st\frac{dt}{ds}$. So this equation is non-linear differential equation.

Therefore, Order=2, Degree=1, Non-linear

Differential Equation exercise 21.1 question 11

Order=2, Degree=3 , Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$x^{2}\left [ \frac{d^{2}y}{dx^{2}} \right ]^{3}+y\left [ \frac{dy}{dx} \right ]^{4}+y^{4}=0$
Solution:
Here in this question, the order of the differential equation is 2 and the degree of the differential equation is .
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and its derivatives are multiplied together with $y \frac{dy}{dx}$ , also y is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=2, Degree=3 , Non-linear

Differential Equation exercise 21.1 question 12

Order=3 , Degree=1, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.Given:$\frac{d^{3}y}{dx^{3}}+\left ( \frac{d^{3}y}{dx^{2}} \right )+\frac{dy}{dx}+4y=\sin x$
Solution:
Here in this question, the order of the differential equation is 3 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and its derivatives are multiplied with itself $\frac{d^{3}y}{dx^{3}}$. So this equation is non-linear differential equation.
Therefore, Order=3 , Degree=1, Non-linear

Differential Equation exercise 21.1 question 13

Order=1, Degree=1, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\left ( xy^{2}+x \right )dx+\left ( y-x^{2}y \right )=0$
Solution:
The above equation can be written as
$x\left ( y^{2}+1 \right )dx=y\left ( x^{2}-1\right )dy\\ \frac{dy}{dx}\left ( x^{2}-1 \right )=x\left ( y^{2}+1 \right )\\ \frac{dy}{dx}x^{2}y-\frac{dy}{dx}y=xy^{2}+x$
Here in this question, the order of the differential equation is 1 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by y and also y is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=1, Degree=1, Non-linear

Differential Equation exercise 21.1 question 14

Order=1, Degree=2, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\sqrt{1-y^{2}}dx+\sqrt{1-x^{2}}dy=0$
Solution:
The above equation can be written as
$\sqrt{1-x^{2}}\frac{dy}{dx}=-\sqrt{1-y^{2}}$
Since the power of y cannot be rational so squaring on both sides
$\left (\sqrt{1-x^{2}}\frac{dy}{dx} \right )^{2}=\left (-\sqrt{1-y^{2}} \right )^{2}$
$\left (1-x^{2} \right )\left (\frac{dy}{dx} \right )^{2}=1-y^{2}$
Here in this question, the order of the differential equation is 1 and the degree of the differential equation is 2.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=1, Degree=2, Non-linear

Differential Equation exercise 21.1 question 15

Order=2, Degree=3, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\frac{d^{2}y}{dx^{2}}=\left ( \frac{dy}{dx} \right )^{\frac{2}{3}}$
Solution:
Since the power of $\frac{dy}{dx}$ is not rational we need to make it rational.
So cubing on both sides, we get
$\left (\frac{d^{2}y}{dx^{2}} \right )^{3}=\left ( \frac{dy}{dx} \right )^{2}$
Here in this question, the order of the differential equation is 2 and the degree of the differential equation is 3.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=2, Degree=3, Non-linear

Differential Equation exercise 21.1 question 17

Order=2, Degree=2, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$5\frac{d^{2}y}{dx^{2}}=\left [ 1+\left ( \frac{dy}{dx} \right )^{2} \right ]^{\frac{3}{2}}$
Solution:
Since the equation has rational powers, we need to remove them.
So squaring both sides, we get
$25\left (\frac{d^{2}y}{dx^{2}} \right )^{2}=\left [ 1+\left ( \frac{dy}{dx} \right )^{2} \right ]^{3}$
$25\left (\frac{d^{2}y}{dx^{2}} \right )^{2}= 1+\left ( \frac{dy}{dx} \right )^{6} +3\left ( \frac{dy}{dx} \right )^{2}+3\left ( \frac{dy}{dx} \right )^{4}$
Here in this question, the order of the differential equation is 2 and the degree of the differential equation is 2 .
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=2, Degree=2, Non-linear

Differential Equation exercise 21.1 question 18

Order=1, Degree=2, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$y=x\frac{dy}{dx}+a\sqrt{1+\left ( \frac{dy}{dx} \right )^{2}}$
Solution:
First of all, we will rearrange the above equation as follows
$y-x\frac{dy}{dx}=a\sqrt{1+\left ( \frac{dy}{dx} \right )^{2}}$
Since the equation has rational powers, we need to remove them.
So squaring both sides, we get
$y^{2}-2xy\frac{dy}{dx}+x^{2}\left ( \frac{dy}{dx} \right )^{2}=a^{2}\left [ 1+\left ( \frac{dy}{dx} \right )^{2} \right ]$
Here in this question, the order of the differential equation is 1 and the degree of the differential equation is 2.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=1, Degree=2, Non-linear

Differential Equation exercise 21.1 question 19

Order=1, Degree=2, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$y=px+\sqrt{a^{2}p^{2}+b^{2}}$ where $p=\frac{dy}{dx}$
Solution:
First of all, we will rearrange the above equation as follows
$y-x\frac{dy}{dx}=a\sqrt{\frac{b^{2}}{a^{2}}+\left ( \frac{dy}{dx} \right )^{2}}$

Here we have substituted the value of $p$ and taken out $a^{2}$ from the root. Since the equation has rational powers, we need to remove them.

So squaring both sides, we get
$\left (y-x\frac{dy}{dx} \right )^{2}=a^{2}\left [\frac{b^{2}}{a^{2}} +\left ( \frac{dy}{dx} \right )^{2} \right ]\\ y^{2}-2xy\frac{dy}{dx}+x^{2}\left ( \frac{dy}{dx} \right )^{2}=a^{2}\left [\frac{b^{2}}{a^{2}} +\left ( \frac{dy}{dx} \right )^{2} \right ]$
Here in this question, the order of the differential equation is 1 and the degree of the differential equation is 2.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=1, Degree=2, Non-linear

Differential Equation exercise 21.1 question 20

Order=1, Degree=1, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\frac{dy}{dx}=e^{y}=0$
Solution:
Concept of the question
$e^{y}=1+y+\frac{y^{2}}{2!}+\frac{y^{3}}{3!}+......\frac{y^{n}}{n!}$
So the equation becomes as follows,
$\frac{dy}{dx}+1+y+\frac{y^{2}}{2!}+\frac{y^{3}}{3!}+......=0$
Here in this question, the order of the differential equation is 1 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term y is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=1, Degree=1, Non-linear

Differential Equation exercise 21.1 question 21

Order=2, Degree=1, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$x^{2}\frac{d^{2}y}{dx^{2}}=\left [ 1+\left ( \frac{dy}{dx} \right )^{2} \right ]^{4}$
Solution:
On solving the equation,
$x^{2}\frac{d^{2}y}{dx^{2}}=\left [ 1+\left ( \frac{dy}{dx} \right )^{2} \right ]^{4}$
$x^{2}\frac{d^{2}y}{dx^{2}}= 1+4\left ( \frac{dy}{dx} \right )^{2} +6\left ( \frac{dy}{dx} \right )^{4} +4\left ( \frac{dy}{dx} \right )^{6} +\left ( \frac{dy}{dx} \right )^{8}$
Here in this question, the order of the differential equation is 2 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=2, Degree=1, Non-linear

Differential Equation exercise 21.1 question 22

Order=2, Degree = Not defined, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\left ( \frac{d^{2}y}{dx^{2}} \right )^{2}+\left ( \frac{dy}{dx} \right )^{2}=x \sin \left (\frac{d^{2}y}{dx^{2}} \right )$
Solution:
Concept of the question
$\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+......+\left ( \frac{(-1^{r}x^{2r+1})}{(2r+1)!} \right )$
So in this question, x of $\sin x$ is replaced by $\frac{d^{2}y}{dx^{2}}$ which means that the power of $\frac{d^{2}y}{dx^{2}}$ is not defined as it approaches to infinity by the above formula.
Here in this question, the order of the differential equation is 2 and the degree of the differential equation is not defined.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is 2 and the term $\frac{dy}{dx}$ is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=2, Degree = Not defined, Non- linear

Differential Equation exercise 21.1 question 23

Order=2, Degree=2 , Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\left ( {y}'' \right )^{2}+\left ( {y}' \right )^{3}+\sin y=0$
Solution:
Concept of the question
$\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+......+\left ( \frac{(-1)^{r}x^{2x+1}}{\left ( 2r+1 \right )!} \right )$
So in this question, x of $\sin x$ is replaced by y which means that the power of y is not defined as it approaches to infinity by the above formula.
Here in this question, the order of the differential equation is 2 and the degree of the differential equation is not defined.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term y is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=2, Degree=2, Non- linear

Differential Equation exercise 21.1 question 24

Order=2, Degree=1, linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\frac{d^{2}y}{dx^{x}}+5x\left ( \frac{dy}{dx} \right )-6y=\log x$
Solution:
Here in this question, the order of the differential equation is 2 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied with x and 5 both. So this equation is linear differential equation.
Therefore, Order=2, Degree=1, linear

Differential Equation exercise 21.1 question 25

Order=3, Degree=1 , Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\frac{d^{3}y}{dx^{3}}+\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}+y \sin y=0$
Solution:
Concept of the question
$\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+.......+\left ( \frac{(-1)^{r}x^{2r+1}}{(2r+1)!} \right )$
So in this question, x of $\sin x$ is replaced by y which means that the power of y is not defined as it approaches to infinity by the above formula.
Here in this question, the order of the differential equation is 3 and the degree of the differential equation is 1.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term y is multiplied by itself. So this equation is non-linear differential equation.
Therefore, Order=3, Degree=1 , Non- linear

Differential Equation exercise 21.1 question 26

Order=2, Degree-Not defined , Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\frac{d^{2}y}{dx^{2}}+3\left ( \frac{dy}{dx} \right )^{2}=x^{2} \log\left ( \frac{d^{2}y}{dx^{2}} \right )$
Solution:
Concept of the question
For the degree to be defined of any differential equation, the equation must be expressible in the form of a polynomial.
But in this question the degree of the differential equation is not defined because the term on the right hand side is not expressible in the term of a polynomial.
Here in this question, the order of the differential equation is 2 and the degree of the differential equation is not defined.
Since the degree of the equation is not defined, the equation is non-linear.
Therefore, Order=2, Degree=Not defined, Non- linear

Differential Equation exercise 21.1 question 27

Order=1, Degree=3, Non- linear
Hint:
The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.
Given:
$\left ( \frac{dy}{dx} \right )^{3}-4\left ( \frac{dy}{dx} \right )^{2}+7y=\sin x$
Solution:
Here in this question, the order of the differential equation is 1 and the degree of the differential equation is 3.
In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.
Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied with itself. So this equation is non-linear differential equation.
Therefore, Order=1, Degree=3, Non- linear

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RD Sharma Class 12th Exercise 21.1 Chapter 21 is the best material to learn about the chapter and practice well. It contains step-by-step solutions which make it easier to understand even for students who are weak in maths.

RD Sharma class 12 chapter 21 exercise 21.1 material is available for free through Career360'sCareer360's website. Students can take advantage of RD Sharma Class 12th Exercise 21.1 by studying from it and scoring good exams.

## RD Sharma Chapter-wise Solutions

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1. Where can students get the RD Sharma class 12th exercise 21.1 Solutions?

RD Sharma Class 12th Exercise 21.1 Chapter 21 material can be accessed through Career360’s website by searching the book name and exercise number.

2. Who can use RD Sharma class 12 chapter 21 exercise 21.1?

Class 12 Maths RD Sharma Chapter 21 material is explicitly made for CBSE students who want to gain more knowledge and score better marks in exams.

3. Does the RD Sharma Class 12 Maths Solutions Chapter 21 material help score good marks?

The Class 12 RD Sharma chapter 21 exercise 21.1 solution follows the CBSE syllabus and covers all topics. It can help students score good marks in exams.

4. What is the cost of class 12 RD Sharma chapter 21 material?

The Class 12 RD Sharma chapter 21 exercise 21.1.material is free for students through Career360’s website.

5. Can I solve NCERT questions after preparing this material?

As RD Sharma and NCERT books have the same concepts, students can solve them after referring to this material.

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