Differential Equations Class 12th Notes - Free NCERT Class 12 Maths Chapter 9 Notes

Differential Equations Class 12th Notes - Free NCERT Class 12 Maths Chapter 9 Notes - Download PDF

Edited By Ramraj Saini | Updated on Apr 23, 2022 12:59 PM IST

Introduction: Differential equation belongs to the 9 chapter of NCERT. The NCERT Class 12 Maths chapter 9 notes is based on the differential equation. Class 12 Maths chapter 9 notes have important formulas and their derivations. A Class 12 Maths chapter 9 notes help a student to get detailed information about differential equations. Notes for Class 12 Maths chapter 9 is made in the sequence which is starting from the order of equation, degree of the equation to the types of the differential equation like a method of variable separation, homogeneous differential equation, and linear differential equation. NCERT Notes for Class 12 Maths chapter 9 not only covers the NCERT portion but also cover the CBSE class 12 maths chapter 9 notes portion.

Latest: JEE Main high scoring chapters | JEE Main 10 year's papers

After going through Class 12 differential equation notes students can also refer to,

Also read :

NCERT Class 9 Maths Chapter 9 Notes

The basic definition of a differential equation is it is such a kind of equation that involves a derivation of a function or functions.

For a given function g, find the function of f such that

$\frac{dy}{dx}=g(x) \ \ \ where \ \ y \ \ is \ \ f(x)$

The above-given equation is called a differential equation.

An example of a general differential equation is

$2 \frac{d^2y}{dx^2}+(\frac{dy}{dx})^3=0$

is an ordinary Differential equation

General notation

$\frac{dy}{dx}=y',\ \frac{d^2y}{dx^2}=y'', \ \frac{d^3y}{dx^3}=y''', \ \frac{d^ny}{dx^n}=y^n$

Order of the Differential equation

The order one differential equation is defined by the equation dy/dx = x. Here, the given function cannot be differentiated more than one time. So, it is called the first-order derivative. The same goes for equation (2).

$\frac{dy}{dx}-x=0 \ . \ . \ . \(1)$

Its order is 1

$\frac{d^2y}{dx^2}+y'=0 \ . \ . \ . \(2)$

Its order is 2

Degree of a Differential equation

It is defined as the highest order derivative it contains. In other words, it is the power to which the highest order derivative is raised.

Let us see an example

$\frac{d^3y}{dx^3}+2(\frac{d^2y}{dx^2})^2+\frac{dy}{dx}+y=0$

It has a degree of 1 because the power of the highest order is 1

Example:

Find the order and degree of the given equation $xy\frac{d^2y}{dx^2}+x(\frac{dy}{dx})^2-y\frac{dy}{dx}=0$

Solution:

The order of the equation is 2 and the degree of the equation is 1

General and Particular Solution of a Differential equation

Let us consider a curve y=f(x)=asin (x+b) where a, b R and

$\text{And let a=2 and}\ b=\frac{\pi}{4}\ \\ \text{Thus the function is} \ y=f(x)=2sin (x+\frac{\pi}{4})$

Here the function consists of 2 arbitrary constants a and b and they are known as general solutions of a differential equation. And is the function does not contain any arbitrary constant is called a particular solution.

Formation of a Differential equation whose general solution is given

We know that

x^2+y^2+2x-4y+a=0

Represents a circle with centre at (-1,2) and a radius of 1

On differentiating it we get

$\frac{dy}{dx}=\frac{x+1}{2-y}\ where\ (y \neq 2)$

This is a differential equation that represents a family of circles.

Now in the equation y=mx+c by giving different values to parameters m and c, we get different members of the family of a straight line.

1645091382489

Procedure to form a differential equation

a) If the family of curves F depends on one parameter then it is represented by

F(x, y, a) =0

On differentiating it we get

F(x, y, y’)=0

b) If the family of curves F depends on a two-parameter then it is represented by

F(x, y, a, b)

On differentiating it we get

F(x, y, y’, y’’)=0

Methods of solving first order, first-degree Differential equation

Here we will find 3 ways of solving the differential equation

Method of variable separation
Method of homogeneous differential equation
Method of Linear differential equation

Differential equation with variable separation

Firstly first order, first-degree equation is

$\frac{dy}{dx}=F(x,y)$

Now if F(x, y) is expressed as g(x) h(y) then

$\frac{dy}{dx}=h(y)g(x)$

Now separating variables we can write as

$\frac{dy}{h(x)}=g(x)dx$

Now integrating

$\int \frac{dy}{h(x)}= \int g(x)dx$

This gives the result as

H(y)=G(x)+c

Now let us see some examples

$\frac{dy}{dx} = xy^2$

$\frac{dy}{y^2} = xdx$

$-\frac{1}{y} = \frac{x^2}{2} + c_1$

$y = -\frac{2}{x^2+c_1}$

Homogeneous Differential equation

Let us consider a function

F(x, y)= y^2+2xy

Now if we replace x and y with αx and αy, for any non-zero constant we get

$\\ F( \alpha x, \alpha y)= \alpha^2( y^2+2xy)= \alpha^2F(x, y) \\ \\ Generalizing \\ \\ F (\alpha x, \alpha y)= \alpha^nF(x, y)\ where \neq 0$

Now in the case of differential equation

A differential equation of the form dy /dx = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of zero degrees. Solving a homogeneous differential equation of the type

dy/dx = F(x, y) = g(y/x) ----------(1)

We make substitution y = v.x ------(2)

Differentiating (2) with respect to x we get

$\\ \frac{dy}{dx}=v+x\frac{dv}{dx} -------(3)\\ \\ \text{Substituting (3) in (1) we get}\\ v+x \frac{dv}{dx}=g(v) \\ \\ x \frac{dv}{dx}=g(v)-v --------(4) \\ \\ \text{Using variable separation in (4)} \\ \\ \frac{dv}{g(v)-v}=\frac{dx}{x} --------(5)\\ \\ \text{Now integrating (5) we get} \\ \\ \int \frac{dv}{g(v)-v}= \int \frac{dx}{x} \\$

Therefore (6) gives the solution f (1) when we replace v by y/x

Let us see an example

$\\ \frac{dy}{dx}=\frac{x+y}{x-y}\\ \\ \text{Substituting y =vx}\\ \\ \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \\ v+x \frac{dv}{dx}=\frac{x+vx}{x-vx} \\ \\ v+x \frac{dv}{dx}=\frac{1+v}{1-v} \\ \\ x \frac{dv}{dx}=\frac{1+v^2}{1-v} \\ \\ (\frac{1-v}{1+v^2}) dv = \frac{dx}{x} \\ \\ \text{Now integrating we get} \\ \\ \int (\frac{1-v}{1+v^2}) dv = \int \frac{dx}{x} \\ \\$

$\\ ln x =tan^{-1}v-\frac{1}{2}ln|1+v^2| +ln c\\ ln x =tan^{-1}(\frac{y}{x})-\frac{1}{2}ln |1+(\frac{y}x)^2| +ln c$

Linear Differential Equation

If an equation is

$\frac{dy}{dx}+Py=Q$

Where P and Q are constants of x and the equation is called linear differential Equation

Steps for solving linear differential equation

First, give the equation to a general form, i.e. $\frac{dy}{dx}+Py=Q$

Find the integrating factor(I.F.) which is equal to $e^{\int Pdx}$
Then write the equation in the following format $y(I.F)= \int Q^{*}(I.F)dx+c$

Let us see an example

$\\ \frac{dy}{dx}+\frac{1}{x}y=x^3\\ \\ here \ P=\frac{1}{x} and Q=x^3 \\ \\ I.F=e^{\frac{1}{x}dx}=e^{ln x} =x \\ \\ x.y=\int x^3.x dx+c \\ \\ xy=\frac{x^5}{5}+c$

From here we come to the end of the chapter.

Significance of NCERT Class 12 Math Chapter 9 Notes

Class 12 differential equation notes are helpful for the revision of the detailed chapter and for getting a glimpse of the important topics covered in the notes. Class 12 Math chapter 9 Notes is useful for getting a glance at the C lass 12 CBSE Syllabus and also for national competitive exams. Class 12 Maths chapter 9 notes pdf download can be used for getting a hard copy and to prepare for the exam.

differential equation Class 12 notes pdf download: link

NCERT Class 12 Notes Chapter Wise.

NCERT Class 12 Maths Chapter 1 Notes

NCERT Class 12 Maths Chapter 2 Notes

NCERT Class 12 MathsChapter 3 Notes

NCERT Class 12 Maths Chapter 4 Notes

NCERT Class 12 Maths Chapter 5 Notes

NCERT Class 12 Maths Chapter 6 Notes

NCERT Class 12 Maths Chapter 7 Notes

NCERT Class 12 Maths Chapter 8 Notes

NCERT Class 12 Maths Chapter 9 Notes

NCERT Class 12 Maths Chapter 10 Notes

NCERT Class 12 Maths Chapter 11 Notes

NCERT Class 12 Maths Chapter 12 Notes

NCERT Class 12 Maths Chapter 13 Notes

Subject Wise NCERT Exemplar Solutions

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook

Subject Wise NCERT Solutions

NCERT Books and Syllabus

NCERT Book for Class 12

NCERT Syllabus for Class 12

Also Read

Biomolecules class 12th notes- Free ncert class 12 Chemistry Chapter 14 Notes- Download PDF

Amines Class 12th Notes- Free NCERT Class 12 Chemistry Chapter 13 Notes- Download PDF

Quadratic Equations Class 10th Notes - Free NCERT Class 10 Maths Chapter 4 Notes - Download PDF

Aldehydes, Ketones and Carboxylic Acids Class 12th Notes- Free NCERT Class 12 Chemistry Chapter 12 Notes - Download PDF

NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Books for Class 12 Chemistry 2025-26: Download Free PDF

NCERT Book for Class 12 Physics 2024-25: Download PDF Here

NCERT Syllabus for Class 6-12 (2025-26) - Download All Subjects PDF Free

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

Articles

UPSC NDA Admit Card 2025 (Out) - Steps to Download NDA 1 Hall Ticket @ upsc.gov.in

Apr 03, 2025

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Apr 03, 2025

CDS Admit Card 2025 (Out) - Steps to Download UPSC CDS 1 Hall Ticket, Link

Apr 03, 2025

JKBOSE 12th Topper 2025- Check Toppers Name, Marks & Rank Here

Apr 03, 2025

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

Apr 03, 2025

HBSE 12th Toppers 2025: Check Toppers Name, Marks, Ranks & Schools

Application Date:24 March,2025 - 23 April,2025

Exam Pattern

Admit Card

Result

View All School Exams

Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is