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

Have you ever considered what it would take for doctors to measure infection rates and implement proper safety measures quickly if a deadly disease like COVID-19 were to reappear? Have you ever wondered how a hot cup of tea or coffee cools down in your room at a normal temperature? These are the situations in which knowledge of Differential equations becomes relevant, and in Maths chapter 9 class 12, we will study this in detail. After going through the NCERT textbook solutions, students need a study material from which they can frequently revise the essential formulas and concepts. The purpose of these notes is to make learning easier for students.

Differential equations class 12 NCERT notes cover topics like general and particular solutions of a differential equation, formation of differential equations, some methods to solve a first-order - first-degree differential equation, and some applications of differential equations in different areas. These differential equations class 12 notes are prepared by Careers360 teachers who have multiple years of experience in this field. Students can also practice NCERT Exemplar Class 12 Maths Chapter 9 Solutions Differential Equations for a deeper understanding of this topic.

NCERT Class 12 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
$\frac{dy}{dx}=g(x),\text{ where }y\text{ is }f(x)$
The above-given equation is called a differential equation.

An example of a general differential equation is:

$2\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^3=0$
This is an ordinary Differential equation.
General notation
$\frac{dy}{dx}=y^{\prime },\frac{d^{2}y}{d{x}^2}=y^{\prime \prime },\frac{d^{3}y}{d{x}^3}=y^{\prime \prime \prime },\frac{d^{n}y}{d{x}^n}=y^n$

Order of the Differential equation

The order one differential equation is defined by the equation $\frac{dy}{dx}=x$. 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.

$\frac{d^{2}y}{d{x}^2}+y^{\prime }=0-----(2)$
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.

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

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 $xy\frac{d^{2}y}{d{x}^2}+x{\left(\frac{dy}{dx}\right)}^2-y\frac{dy}{dx}=0$
Solution:
The order of the equation is 2, and the degree of the equation is 1.

General and Particular Solutions of a Differential Equation
Let us consider a curve $\mathrm{y}=\mathrm{f}(\mathrm{x})=\mathrm{a}\sin (\mathrm{x}+\mathrm{b})$ where $\mathrm{a},\mathrm{b}\in R$ and
And let $\mathrm{a}=2$ and $b=\frac{\pi }{4}$
Thus the function is $y=f(x)=2\sin (x+\frac{\pi }{4})$

Here, the function consists of 2 arbitrary constants, a and b, and they are known as general solutions of a differential equation. If 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

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

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

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

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 two parameters, 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.

1. Method of variable separation

2. Method of homogeneous differential equation

3. Method of Linear differential equation

Differential equation with variable separation

The differential of the form $\frac{dy}{dx}=f(x)g(y)$,
Where $f(x)$ is a function of $x$ and $g(y)$ is a function of $y$,
are said to be in variable separable form.

Rewrite the equation as
$\frac{dy}{g(y)}=f(x)dx[\text{ where }g(y)\ne 0]$

This process is separating the variables.
Now, integrating both sides, we get
$\int \frac{\mathrm{dy}}{\mathrm{g}(\mathrm{y})}=\int \mathrm{f}(\mathrm{x})\mathrm{dx}+\mathrm{c}$

By this, we get the solution of the differential equation.

Let’s look at one illustration for a better understanding.

Solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=({\mathrm{e}}^{\mathrm{x}}+1)({\mathrm{y}}^2+1)$
Rewrite the differential equation as
$\frac{dy}{1+y^2}=(e^x+1)dx$

Integrating both sides, we get
$\begin{array}{rl}& \int \frac{\mathrm{dy}}{1+{\mathrm{y}}^2}=\int ({\mathrm{e}}^{\mathrm{x}}+1)\mathrm{dx}\\ & \Rightarrow {\tan }^{-1}y=e^x+x+c\\ & \Rightarrow y=\tan (e^x+x+c)\end{array}$

Homogeneous Differential equation

Let us consider a function.
$F(x,y)=y^2+2xy$
Now if we replace $x$ and $y$ with $ax$ and $ay$, for any non-zero constant we get
$F(\alpha x,\alpha y)={\alpha }^2(y^2+2xy)={\alpha }^{2}F(x,y)$
Generalizing
$F(\alpha x,\alpha y)={\alpha }^{n}F(x,y)\text{ where }\alpha \ne 0$

Now, in the case of differential equation,

A differential equation of the form $\frac{dy}{dx}=F(x,y)$ is said to be homogenous if F(x, y) is a homogenous function of zero degrees. Solving a homogeneous differential equation of the type.
$\frac{dy}{dx}=F(x,y)=g(\frac{y}{x})-------(1)$

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

Differentiating (2) with respect to x, we get
$\frac{dy}{dx}=v+x\frac{dv}{dx}-------(3)$
Substituting (3) in (1), we get,
$\begin{array}{rl}& v+x\frac{dv}{dx}=g(v)\\ & \Rightarrow x\frac{dv}{dx}=g(v)-v\end{array}$
Using variable separation in (4)

$\frac{dv}{g(v)-v}=\frac{dx}{x}--------(5)$
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 $\frac{y}{x}.$

Let us see an example.
$\frac{dy}{dx}=\frac{x+y}{x-y}$
Substituting $\mathrm{y}=\mathrm{vx}$
$\begin{array}{rl}& \frac{dy}{dx}=v+x\frac{dv}{dx}\\ & \Rightarrow v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}\\ & \Rightarrow v+x\frac{dv}{dx}=\frac{1+v}{1-v}\\ & \Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1-v}\\ & \Rightarrow \left(\frac{1-v}{1+v^2}\right)dv=\frac{dx}{x}\end{array}$

Now, integrating, we get,

$\int \left(\frac{1-v}{1+v^2}\right)dv=\int \frac{dx}{x}$

$\begin{array}{rl}& \ln x={\tan }^{-1}v-\frac{1}{2}\ln |1+v^2|+\ln c\\ & \Rightarrow \ln x={\tan }^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\ln |1+{\left(\frac{y}{x}\right)}^2|+\ln c\end{array}$

Linear Differential Equation

The linear differential equations are those in which the variable and its derivative occur only in the first degree.

An equation of the form
$\frac{dy}{dx}+P(x)\cdot y=Q(x)----(i)$

Where P(x) and Q(x) are functions of x only or constant is called a linear equation of the first order.

To solve the differential equation (i),
multiply both sides of Eq (i) by $\int e^{P(x)dx}$, we get,
$\begin{array}{ll}& e^{\int P(x)dx}(\frac{dy}{dx}+P(x)y)=e^{\int P(x)dx}\cdot Q(x)\\ \text{ i.e. }& e^{\int P(x)dx}\cdot \frac{dy}{dx}+y\cdot P(x)\frac{d}{dx}\left(e^{\int P(x)dx}\right)=Q{e}^{\int P(x)dx}\\ \Rightarrow & \frac{d}{dx}\left(y{e}^{\int P(x)dx}\right)=e^{\int P(x)dx}\cdot Q(x)\end{array}$
Integrating both sides, we get,
$\Rightarrow \int \mathrm{d}\left(y{\mathrm{e}}^{\int \mathrm{P}(\mathrm{x})\mathrm{dx}}\right)=\int ({\mathrm{e}}^{\int \mathrm{P}(\mathrm{x})\mathrm{dx}}\cdot \mathrm{Q}(\mathrm{x}))\mathrm{dx}$
$\Rightarrow {\mathrm{ye}}^{\int P(x)dx}=\int Q(x)e^{\int P(x)dx}dx+C$

Which is the required solution of the given differential equation.

The term ${\mathrm{e}}^{\int \mathrm{P}(\mathrm{x})\mathrm{dx}}$, which converts the left-hand expression of the equation into a perfect differential, is called an Integrating factor (IF).

Thus, we remember the solution of the above equation as
$y(\mathrm{IF})=\int Q(\mathrm{IF})dx+C$

Note :
Sometimes, a given differential equation becomes linear if we take x as the dependent variable and y as the independent variable.

Application of Differential Equation

Differential equations are used in a variety of disciplines, such as biology, economics, physics, chemistry, and engineering.

Growth and Decay Problem:

Let the amount of substance (or population) that is either growing or decaying be denoted by N(t). If we assume the time rate of change of this amount of substance, $\frac{dN}{dt}$, is proportional to the amount of substance present, then

$\frac{dN}{dt}=kN$
$\Rightarrow \frac{dN}{dt}-kN=0$

Where k is the constant of proportionality. We are assuming that N(t) is a differentiable, hence continuous, function of time.

Importance of NCERT Class 12 Math Chapter 9 Notes

NCERT Class 12 Maths chapter 9 notes often prove to be a convenient tool for revision and recalling the important concepts and formulas. Here are some more points why these notes are important.

• These notes are prepared by following the latest Class 12 CBSE Mathematics Syllabus to help students learn the topics that are required to get good marks on the exam.
• A considerable number of questions come from this chapter on the class 12 board exam as well in some of the competitive exams. So, frequent revision of these topics to get a better conceptual clarity is important. That is what these notes are for.
• Completing these notes will boost the confidence of the students and create a positive vibe, which is a necessity before an exam.
• These notes come with many graphs and necessary pictures, which will make the learning easier and interesting for students.

Subject Wise NCERT Exemplar Solutions

After completing the NCERT textbook solutions, students can practice these NCERT Exemplar questions. These w

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

Subject Wise NCERT Solutions

The following links will lead students to the solutions of other subject exercises. These well-structured solutions will be very useful for them.

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

NCERT Books and Syllabus

Students can use the following links to check the latest CBSE syllabus and some reference books.