Mathematics

Session

Differential Equations - 1

Session Objectives

Differential Equation

Order and Degree

Solution of a Differential Equation, General and Particular Solution

Initial Value Problems

Formation of Differential Equations

Class Exercise

Differential Equation

An equation containing an independent variable x,dependent variable y and the differential coefficients of the dependent variable y with respect to independent variable x, i.e.

2

2

dy d y, , …

dx dx

Examples

dy1 = 3xy

dx

2

2

d y2 + 4y = 0

dx

33 2

3 2

d y d y dy3 + + +4y = sinx

dxdx dx

2 24 x dx +y dy = 0

Order of the Differential Equation

The order of a differential equation is the order of the highest order derivative occurring in the differential equation.

2 32 23

2 2

d y dy d y dyExample: = =

dx dxdx dx

The order of the highest order derivative

2

2

d yis 2.

dx

Therefore, order is 2

Degree of the Differential Equation

The degree of a differential equation is the degree of the highest order derivative, when differential coefficients are made free from fractions and radicals.

3322 22 22

2 2

d y dy d y dyExample : + 1+ = 0 = 1+

dx dxdx dx

The degree of the highest order derivative is 2.2

2

d y

dx

Therefore, degree is 2.

Example - 1

Determine the order and degree of the differential

equation:2dy dy

y = x +a 1+dx dx

.

2dy dySolution: We have y = x +a 1+

dx dx

2dy dyy - x = a 1+

dx dx

2 22dy dy

y - x = a 1+dx dx

Solution Cont.

2 22 2 2 2dy dy dy

y - 2xy +x = a +adx dx dx

The order of the highest derivative is 1 and its degree is 2.

dydx

Example - 2

Determine the order and degree of the differential

equation:

324 2

4

d y dyc

dxdx

Solution: We have

3322 24 42

4 4

d y dy d y dy= c + = c +

dx dxdx dx

Here, the order of the highest order is 4 4

4

d y

dx

and, the degree of the highest order is 24

4

d y

dx

Linear and Non-Linear Differential Equation

A differential equation in which the dependent variable y and

its differential coefficients i.e. occur only in the

first degree and are not multiplied together is called a linear differential equation. Otherwise, it is a non-linear differential equation.

2

2

dy d y, , …

dx dx

Example - 3

is a linear differential equation of order 2 and degree 1.

is a non-linear differential equation because the dependent

variable y and its derivative are multiplied together. dydx

2

2

d y dyi - 3 + 7y = 4x

dxdx

dyii y × - 4 = x

dx

Solution of a Differential Equation

The solution of a differential equation is the relation between the variables, not taking the differential coefficients, satisfying the given differential equation and containing as many arbitrary constants as its order is.

For example: y = Acosx - Bsinx

is a solution of the differential equation2

2

d y+4y = 0

dx

General Solution

If the solution of a differential equation of nth order contains n arbitrary constants, the solution is called the general solution.

is the general solution of the differential equation 2

2

d y+y = 0

dx

y Bsin x

is not the general solution as it contains one arbitrary constant.

y = Acosx - Bsinx

Particular Solution

A solution obtained by giving particular values to the arbitrary constants in general solution is called particular solution.

y 3cos x 2sin x

is a particular solution of the differential equation 2

2

d y+y = 0.

dx

Example - 4

3 2Solution: We have y = x +ax +bx +c …(i)

2dy= 3x +2ax +b …(ii) Differentiating i w.r.t. x

dx

2

2

d y= 6x +2a …(iii) Differentiation ii w.r.t. x

dx

3 2

3

3

Verify that y = x +ax +bx+c is a solution of the

d ydifferential equation =6.

dx

Solution Cont.

3

3

d y= 6 Differentiating iii w.r.t x

dx

3

3

d y= 6 is a differential equation ofi .

dx

Initial Value Problems

The problem in which we find the solution of the differential equation that satisfies some prescribed initial conditions, is called initial value problem.

Example - 5

2x x

2

dy d y= e , = e

dx dx

xy = e +1 satisfies the differential equation2

2

d y dy- = 0

dxdx

Show that is the solution of the initial value

problem

xy = e +1

2

2

d y dy- = 0, y 0 = 2, y' 0 = 1

dxdx

xSolution : We have y = e +1

Solution Cont.

0 0

x=0

dyy 0 = e +1 and = e

dx

y 0 = 2 and 'y 0 1

xy = e +1 is the solution of the initial value problem.

x dyy = e +1 and

dxxe

Formation of Differential Equations

y = mx

Assume the family of straightlines represented by

dy= m

dxdy y

dx xdy

x ydx

is a differential equation of the first order.

X

Y

O

ymx m = tan

Formation of Differential Equations

Assume the family of curves represented by

where A and B are arbitrary constants.

y = Acos x +B …(i)

dyA sin x B ... ii

dx [Differentiating (i) w.r.t. x]

2

2

d yand Acos x B

dx [Differentiating (ii) w.r.t. x]

Formation of Differential Equations

2

2

d yy

dx [Using (i)]

2

2

d y+y = 0

dx

is a differential equation of second order

Similarly, by eliminating three arbitrary constants, a differential equation of third order is obtained.

Hence, by eliminating n arbitrary constants, a differential equation of nth order is obtained.

Example - 6

Form the differential equation of the family of curves

a and c being parameters. y = a sin bx + c ,

Solution: We have y = a sin bx + c

is the required differential equation.

2 22 2

2 2

d y d y= -b y + b y = 0

dx dx

[Differentiating w.r.t. x] dy= ab cos bx + c

dx

[Differentiating w.r.t. x] 2

22

d y= -ab sin bx + c

dx

Example - 7

Find the differential equation of the family of all the circles, which passes through the origin and whose centre lies on the y-axis.

If it passes through (0, 0), we get c = 0

2 2x +y +2gx +2ƒ y = 0

This is an equation of a circle with centre (- g, - f) and passing through (0, 0).

Solution: The general equation of a circle is2 2x +y +2gx +2ƒ y +c = 0.

Solution Cont.

Now if centre lies on y-axis, then g = 0.

2 2x +y +2ƒ y = 0 …(i)

This represents the required family of circles.

dyx y

dxƒdy

dx

dy dy2x +2y +2ƒ = 0 Differentiating i w.r.t. x

dx dx

Solution Cont.

2 2

dyx +y

dxx +y - 2y = 0 Substituting the value offdydx

2 2 2dy dyx +y - 2xy - 2y = 0

dx dx

2 2 dyx - y - 2xy = 0

dx

Thank you