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

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RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
Differential Equations Excercise: 21.1
RD Sharma Chapter-wise Solutions

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.

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise: 21.1

Differential Equation exercise 21.1 question 1
Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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:

Solution:

Here in this question, the dependent variable is_tand 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_tand its derivatives are multiplied together with. So this equation is non-linear differential equation.

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

Differential Equation exercise 21.1 question 11

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Class 12 RD Sharma chapter 21 exercise 21.1 is prepared by a team of experts to help students score good marks in their exams. This material contains questions and answers in the same place, which makes it convenient for students for preparation and revision. This material follows the CBSE syllabus and is updated to the latest version of the book. This means that it contains questions from the newest addition of the RD Sharma maths book.

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