NCERT Solutions for Exercise 9.1 Class 12 Maths Chapter 9- Differential Equations

NCERT solutions for exercise 9.1 Class 12 Maths chapter 9 introduces the questions related to differential equations. In the NCERT Class 11 Mathematics Book and also chapter 5 of Class 12 Maths, the concepts of derivatives are discussed. Exercise 9.1 Class 12 Maths gives an idea about equations involving derivatives. NCERT solutions for Class 12 Maths chapter 9 exercise 9.1 give clarity about the concept of degree and order of a differential equation. A few examples are also given in the NCERT Book to understand the same. Here are solutions to Class 12 Maths chapter 9 exercise 9.1 prepared by expert Mathematics faculties. In continuation with the Class 12th Maths chapter 6 exercise 9.1, the NCERT Class 12 chapter differential equations have the following 6 exercises.

Also check -

Differential Equations exercise 9.2
Differential Equations exercise 9.3
Differential Equations exercise 9.4
Differential Equations exercise 9.5
Differential Equations exercise 9.6
Differential Equations miscellaneous exercise

Differential Equations Class 12 Chapter 9 Exercise: 9.1

Question:1 Determine order and degree (if defined) of differential equation $\frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0$

Answer:

Given function is
$\frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0$
We can rewrite it as
$y^{''''}+\sin(y''') =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''''}$

Therefore, the order of the given differential equation $\frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0$ is 4
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, it's a degree is not defined

Question:2 Determine order and degree (if defined) of differential equation $y' + 5y = 0$

Answer:

Given function is
$y' + 5y = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'}$
Therefore, the order of the given differential equation $y' + 5y = 0$ is 1
Now, the given differential equation is a polynomial equation in its derivatives and its highest power raised to y ' is 1
Therefore, it's a degree is 1.

Question:3 Determine order and degree (if defined) of differential equation $\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$

Answer:

Given function is
$\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$
We can rewrite it as
$(s^{'})^4+3s.s^{''} =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $s^{''}$

Therefore, the order of the given differential equation $\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$ is 2
Now, the given differential equation is a polynomial equation in its derivatives and power raised to s '' is 1
Therefore, it's a degree is 1

Question:4 Determine order and degree (if defined) of differential equation.

$\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$

Answer:

Given function is
$\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$
We can rewrite it as
$(y^{''})^2+\cos y^{''} =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, the order of the given differential equation $\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$ is 2
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, it's a degree is not defined

Question:5 Determine order and degree (if defined) of differential equation.

$\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$

Answer:

Given function is
$\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$
$\Rightarrow \frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}\left ( \frac{d^2y}{dx^2} \right )$

Therefore, order of given differential equation $\frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0$ is 2
Now, the given differential equation is a polynomial equation in it's dervatives $\frac{d^2y}{dx^2}$ and power raised to $\frac{d^2y}{dx^2}$ is 1
Therefore, it's degree is 1

Question:6 Determine order and degree (if defined) of differential equation $(y''')^2 + (y'')^3 + (y')^4 + y^5= 0$

Answer:

Given function is
$(y''')^2 + (y'')^3 + (y')^4 + y^5= 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'''}$

Therefore, order of given differential equation $(y''')^2 + (y'')^3 + (y')^4 + y^5= 0$ is 3
Now, the given differential equation is a polynomial equation in it's dervatives $y^{'''} , y^{''} \ and \ y^{'}$ and power raised to $y^{'''}$ is 2
Therefore, it's degree is 2

Question:7 Determine order and degree (if defined) of differential equation

$y''' + 2y'' + y' =0$

Answer:

Given function is
$y''' + 2y'' + y' =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'''}$

Therefore, order of given differential equation $y''' + 2y'' + y' =0$ is 3
Now, the given differential equation is a polynomial equation in it's dervatives $y^{'''} , y^{''} \ and \ y^{'}$ and power raised to $y^{'''}$ is 1
Therefore, it's degree is 1

Question:8 Determine order and degree (if defined) of differential equation

$y' + y = e^x$

Answer:

Given function is
$y' + y = e^x$
$\Rightarrow$ $y^{'}+y-e^x=0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'}$

Therefore, order of given differential equation $y^{'}+y-e^x=0$ is 1
Now, the given differential equation is a polynomial equation in it's dervatives $y^{'}$ and power raised to $y^{'}$ is 1
Therefore, it's degree is 1

Question:9 Determine order and degree (if defined) of differential equation

$y'' + (y')^2 + 2y = 0$

Answer:

Given function is
$y'' + (y')^2 + 2y = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $y'' + (y')^2 + 2y = 0$ is 2
Now, the given differential equation is a polynomial equation in it's dervatives $y^{''} \ and \ y^{'}$ and power raised to $y^{''}$ is 1
Therefore, it's degree is 1

Question:10 Determine order and degree (if defined) of differential equation

$y'' + 2y' + \sin y = 0$

Answer:

Given function is
$y'' + 2y' + \sin y = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $y'' + 2y' + \sin y = 0$ is 2
Now, the given differential equation is a polynomial equation in it's dervatives $y^{''} \ and \ y^{'}$ and power raised to $y^{''}$ is 1
Therefore, it's degree is 1

Question:11 The degree of the differential equation $\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$ is

(A) 3

(B) 2

(D) not defined

Answer:

Given function is
$\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$
We can rewrite it as
$(y^{''})^3+(y^{'})^2+\sin y^{'}+1=0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$ is 2
Now, the given differential equation is a not polynomial equation in it's dervatives
Therefore, it's degree is not defined

Therefore, answer is (D)

Question:12 The order of the differential equation $2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ is

(A) 2

(B) 1

(D) Not Defined

Answer:

Given function is
$2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$
We can rewrite it as
$2x.y^{''}-3y^{'}+y=0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ is 2

Therefore, answer is (A)

Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.1

One fill in the blank or multiple choice type or very short answer can be expected from exercise 9.1 Class 12 Maths for CBSE Class 12 Maths Board Exams
Not only CBSE, but certain state boards also follow the NCERT Syllabus. Therefore the NCERT solutions for Class 12 Maths chapter 9 exercise 9.1 can be used to prepare for state boards that follow NCERT.

Also see-

NCERT Exemplar Solutions Class 12 Maths Chapter 9
NCERT Solutions for Class 12 Maths Chapter 9

NCERT Solutions Subject Wise

NCERT Solutions Class 12 Chemistry
NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Biology
NCERT Solutions for Class 12 Mathematics

Subject Wise NCERT Exemplar Solutions

NCERT Exemplar Class 12 Maths
NCERT Exemplar Class 12 Physics
NCERT Exemplar Class 12 Chemistry
NCERT Exemplar Class 12 Biology

NCERT Solutions for Exercise 9.1 Class 12 Maths Chapter 9- Differential Equations

Differential Equations Class 12 Chapter 9 Exercise: 9.1

More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.1

Also Read| Differential Equations Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.1

Also see-

NCERT Solutions Subject Wise

NCERT Solutions Class 12 Chemistry

Subject Wise NCERT Exemplar Solutions