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.

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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.

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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

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