Differential Equations

Differential Equations is the easiest and the most scoring topic in the Mathematics syllabus of the IIT JEE. This topic forms the base for many other topics. It is important to master this area to remain competitive in the JEE. It plays a vital role not only in applied mathematics but also in various other branches of engineering and physics.

Various heads covered under this topic include:

• Basic Definition

• Formation of Differential Equations

• Methods of Finding P.I.

A differential equation is an equation which involves a function and its derivatives. The function involved may be of one or several variables and the derivatives may also be of various orders. We will discuss the various heads in brief here as they have been discussed in detail in the coming sections.

Basically, the differential equations can be divided into two categories:

Ordinary Differential equations are those equations which involve an unknown function and its ordinary derivatives. The unknown function involved is the dependent variable and it is a function of single independent variable.

Mathematically, an ODE of order n is an equation of the form

F(x, y, y',…..,y ⁽ⁿ⁾ ) = 0, where y is a function of x and y' denotes the first derivative of y with respect to x.

Partial Differential equations are those equations which involve rate of change of continuous variables. In partial differential equations, the unknown function is a function of multiple independent variables and instead of simple derivatives, partial derivatives are involved.

The ordinary and partial differential equations can be further divided into linear and non-linear differential equations.

An ODE of order n is said to be linear if it is of the form

a_n (x)y⁽ⁿ⁾ + a _n-1(x) y ^(n-1) + …. + a₁ (x) y' + a₀ (x) y = Q(x)

These have a further classification into homogeneous but here we won’t be going into the intricacies as they have been discussed in detail in the coming sections.

Non-linear differential equations are those which involve higher order terms and the product of terms. There are a few methods of solving partial differential equations and they also depend on the type of equation. 

Formation of a Differential Equation

An ordinary differential equation is obtained by eliminating arbitrary constants (also called parameters) from a given relation involving the variables where the order of the equation is equal to the number of the constants to be eliminated.

A partial differential equation is obtained when we eliminate arbitrary constants or functions from a given relation involving the variables/functions. Whenever we consider differential equations we generally mean ordinary differential equations.

Let φ(x,y,c₁,c₂,……, c_n) =0 ……. (1)

be a relation involving variables x and y and n arbitrary constants c₁,c₂,...,c_n.

If we differentiate equation (1) n times with respect to ‘x’ and eliminate the n arbitrary constants between the (n + 1) relations, we get an ordinary differential equation of the form

F(x, y, y',…..,y ⁽ⁿ⁾ ) = 0 where

y' =dy/dx, y^” =d²y/dx²  ……… y⁽ⁿ⁾ = dⁿy/dxⁿ 

Example: Eliminate ‘a’ and ‘b’ from y = ax² + bx and form the differential equation.

Solution: The given equation is y= ax² +bx …… (1)

Differentiating the given equation with respect to x, we get

y' = 2ax +b ……. (2)

y” = 2a  …… (3)

Equation (3) gives, a= y” / 2

Equation (2) gives, b= y' - 2ax = y' -xy”

Equation (1) gives, y= y”/2 .x² + (y' -xy”) x

                                 = xy' - 1/2. x²y”

Hence, the required differential equation is

x²y” – 2xy' +2y =0.  

Example: Determine the differential equation by eliminating the constants A and B from y= Asin x +B cos 2x

Solution: The given equation is

y= Asin x +B cos 2x                                               …….(1) 

By differentiating twice with respect to x, we get,

y' = A cos x – 2B sin 2x                                          …….(2)             

y” = -A sin x – 4B cos 2x                                        …….(3)

Form equations (1) and (3) we get,

y” + 4y = 3A sin x                                                   ……..(4)

y” + y = -3B cos 2x                                                 .….....(5)

Substituting the values of A and B in equation (2), we obtain

y' = cos x {(y” +4y)/ 3 sinx} + y (4 cot x +2 tan 2x)

Hence, the required differential equation is given by

3y' = y” (cot x + 2 tan 2x) + y (4cot x + 2 tan 2x). 

There are various methods of solving differential equations like Variable Separable, Homogeneous, Exact etc. but they differ according to the type of equation. These methods have been discussed in the later sections. 

Watch this video for more on differential equations

 

Methods of finding the Particular Integral:

While solving a differential equation, it is generally very easy to find the C.F, but the problem arises in the particular integral. There is one general method which can be used for all types of functions but the drawback is that it is quite lengthy. There are short methods depending on the types of functions involved. We give an outline of these methods:

 General method:

If both m₁ and m₂ are constants, the expressions (D – m₁) (D – m₂) y and (D– m₂) (D – m₁) y are equivalent i.e. the expression is independent of the order of operational factors.

 We discuss some of the short methods of finding the particular integral

(1) When X = e^ax in f(D) y = X, where  a is a constant

     Then 1/f(D) e^ax = 1/f(a) e^ax , if f(a) ≠ 0 and

     1/f(D) e^ax = x^r/f ^r(a) e^ax , if f(a) = 0, where f(D) = (D-a)^rf(D)

(2) To find P.I. when X = cos ax or sin ax

      f (D) y = X

     If f (– a²) ≠ 0   then 1/f(D²) sin ax = 1/f(-a²) sin ax 

     If f (– a²) = 0 then (D² + a²) is at least one factor of f (D²)

(3) To find the P.I.when X = x^m  where m ∈ N

     f (D) y = x^m

     y = 1/ f(D) x^m

(4) To find the value of 1/f(D) e^ax V where ‘a’ is a constant and V is a function of x

      1/f (D) .e^ax V = e^ax.1/f (D+a). V

      To find 1/f (D). xV where  V is a function of x

      1/f (D). xV = [x- 1/f(D). f'(D)] 1/f(D) V

To get an idea about the types of questions asked you may refer the Past year Papers. 

Illustration: Find a general solution of the equation

dy/dx + 3y/ (x+2) = x+2, where x ≠2

Solution: Given equation is dy/dx + 3y/ (x+2) = x+2

Here P(x) = 3/ (x+2)

Integrating this we get, 3 ln|x+2| = ln |x+2|³

Integrating factor is e ∫ P(x) =(x+2)³.

Hence, d/dx(y(x+2)³) = (x+2)³ (x+2)

(y (x+2)³) = ∫ (x+2)⁴ dx

y (x+2)³ =  (x+2)⁵/ 5 dx +c

y= (x+2)² /5 +c /(x+2)³  is the general solution.

Illustration : If y = y(x) and [{2 + sinx)/ (y+1)] dy/dx = -cos x, y (0) =1, then find the value of y(π/2).

Solution: The given differential equation is

cos x /( 2 + sinx ) dx + dy/(y+1) = 0.

Hence, ln (2 + sin x) + ln (y+1) = ln c

So, (2 + sin x) (y+1) = c

So, 2 x 2 = c

So, c =4.

Thus, y+1 = 4/ (2 + sin x)

So, y = 2 - sin x/ 2 + sinx

Hence, y (π/2) = 1/3.

Illustration : For the primitive integral equation
ydx + y²dy = xdy, x ∈ R, y > 0,

y = y(x), y(1) = 1, then y (-3) is ?

Solution: The given equation can be written as

(ydx – xdy) / y² = - dy

So, d(x/y) = - dy

Hence, x/y = -y + c

Also, we know y(1) = 1

This gives the value of c as 2.

y² -2y + x = 0

For y (-3) we have y² - 2y – 3 = 0

This means (y-3) (y+1) = 0

So, y = 3, -1.

Illustration : The differential equation dy/dx = √(1-y²)/ y determines a family of circle with:

1. variable radii and  affixed centre at (0, 1)

2. variable radii and  affixed centre at (0, -1)

3. fixed radius 1 and variable centres along the x-axis

4. fixed radius 1 and variable centres along the y-axis

Solution: The given equation is

dy/dx = √(1-y²)/ y

Taking the terms of y and x on separate sides

y/√(1-y²) dy = dx

Integrating both sides, we get

∫ y/√ (1-y²) dy = ∫ dx

-√(1-y²) = x + c

x² +y² + 2cx + c² - 1 =0

or 1- y² = (x + c)²

This clearly shows that this differential equation represents a circle of fixed radius 1 and variable centres along x –axis. 

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