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.

22

dy d y, , …dx dx

Examples

dy1 =3xydx

2

2d y2 + 4y = 0dx

33 2

3 2d y d y dy3 + + +4y = sinxdxdx 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 2d y dy d y dyExample: = =dx dxdx dx

The order of the highest order derivative

22

d y is 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.

3 322 22 222 2

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

The degree of the highest order derivative is 2.2

2d ydx

Therefore, degree is 2.

Example - 1Determine the order and degree of the differential

equation:2dy dyy = x +a 1+dx dx .

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

2dy dyy - x = a 1+dx dx

2 22dy dyy - x = a 1+dx dx

Solution Cont.

2 22 2 2 2dy dy dyy - 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

4d y dy

cdxdx

Solution: We have

3 322 24 424 4

d y dy d y dy= c + = c +dx dxdx dx

Here, the order of the highest order is 4 4

4d ydx

and, the degree of the highest order is 24

4d ydx

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.

22

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

2d y dyi - 3 + 7y = 4xdxdx

dyii y × - 4 = xdx

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

2d y +4y = 0dx

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

2d y +y = 0dx

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

2d 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. xdx

2

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

3 2

33

Verify that y = x +ax +bx+c is a solution of thed ydifferential equation =6.dx

Solution Cont.

3

3d y = 6 Differentiating iii w.r.t xdx

3

3d 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

2dy d y= e , = edx dx

xy = e +1 satisfies the differential equation2

2d y dy- = 0dxdx

Show that is the solution of the initial value

problem

xy = e +1

2

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

xSolution : We have y = e +1

Solution Cont.

0 0

x=0

dyy 0 = e +1 and = edx

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 = mdxdy ydx x

dyx ydx

is a differential equation of the first order.

X

Y

O

ymx m = tan

Formation of Differential EquationsAssume 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

2d yand Acos x Bdx

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

Formation of Differential Equations

2

2d y

ydx

[Using (i)]

22

d y +y = 0dx

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 2d y d y= -b y + b y = 0dx dx

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

[Differentiating w.r.t. x] 2

22

d y = -ab sin bx + cdx

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

dxƒdydx

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

Solution Cont.

2 2dyx+y dxx +y - 2y = 0 Substituting the value offdy

dx

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

2 2 dyx - y - 2xy = 0dx

Thank you