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 the NCERT Class 12 Maths, we will study this in detail. The main purpose of these NCERT Notes of the Integrals class 12 PDF is to provide students with an efficient study material from which they can revise the entire chapter.

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Differential equations NCERT Class 12 Maths 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 subject matter experts who have multiple years of experience in this field. For the syllabus, solutions, and chapter-wise PDFs, head over to this link: NCERT.

NCERT Notes for Class 12 Chapter 9 Differential Equations: Free PDF Download

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NCERT Notes for Class 12 Chapter 9 Differential Equations

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

For a given function g, find the function f such that
$\frac{dy}{dx}=g(x),\text{ where }y\text{ is }f(x)$
The above 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 first order differential equation is defined by the equation $\frac{dy}{dx}=x$. Here, the given function cannot be differentiated more than once. 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 equations

Here, we will find 3 ways of solving the differential equation.

1. Method of variable separation

2. Method of homogeneous differential equations

3. Method of Linear differential equations

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 $y$,
It is said to be in a 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 Equations

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.

Differential Equations: Previous Year Questions and Answers

Given below are some previous year question answers of various examinations from the NCERT class 12 chapter 9, Differential Equations:

Question 1: $y=a{e}^{mx}+b{e}^{-mx}$ satisfies which differential equations?

Solution:
Given $y=a{e}^{mx}+b{e}^{-mx}$
upon differentiation, we get $\frac{dy}{dx}=a\cdot m{e}^{mx}-b\cdot m{e}^{-mx}$
atter differentiation again we get

$\begin{array}{rl}& \frac{d^{2}y}{d{x}^2}=a{m}^2{e}^{mx}-b{m}^2{e}^{-mx}\\ \Rightarrow & \frac{d^{2}y}{d{x}^2}=m^2(a{e}^{mx}-b{e}^{-mx})\\ \Rightarrow & \frac{d^{2}y}{d{x}^2}=m^{2}y\\ \Rightarrow & \frac{d^{2}y}{d{x}^2}-m^{2}y=0\end{array}$

Hence, the correct answer is $\frac{d^{2}y}{d{x}^2}-m^{2}y=0$.

Question 2: The solution of the differential equation $\cos x\sin ydx+\sin x\cos ydy=0$ is:

Solution:
$\begin{array}{rl}& \cos x\sin ydx+\sin x\cos ydy=0\\ & \Rightarrow \sin x\cos ydy=-\cos x\sin ydx\\ & \Rightarrow \frac{\cos y}{\sin y}dy=-\frac{\cos x}{\sin x}dx\\ & \Rightarrow \cot ydy=-\cot xdx\end{array}$

Upon integration of both sides,
$\begin{array}{rl}& \int \cot ydy=-\int \cot xdx\\ \Rightarrow & \log |\sin y|=-\log |\sin x|+\log c\\ \Rightarrow & \log |\sin y|+\log |\sin x|=\log c\\ \Rightarrow & \log |\sin y\cdot \sin x|=\log c\\ \Rightarrow & \sin y\sin x=c\end{array}$

Hence, the correct answer is $\sin y\sin x=c$.

Question 3: The solution of the differential equation $\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{{(1+x^2)}^2}$ is:

Solution:
$\begin{array}{rl}& \frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{{(1+x^2)}^2}\text{ is }\\ & P=\frac{2x}{1+x^2}\text{ and }Q=\frac{1}{{(1+x^2)}^2}\\ & \text{ Here, }\\ & \therefore \text{ I.F. }=e^{\int \frac{2x}{1+x^2}dx}=e^{\log (1+x^2)}=1+x^2\end{array}$

$\therefore$ The general solution is

$\begin{array}{rl}& y(1+x^2)=\int (1+x^2)\frac{1}{{(1+x^2)}^2}+C\\ & \Rightarrow y(1+x^2)=\int \frac{1}{1+x^2}dx+C\\ & \Rightarrow y(1+x^2)={\tan }^{-1}x+C\end{array}$

Hence, the correct answer is $y(1+x^2)={\tan }^{-1}x+C$.

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