lecture 1

CHAPTER 2

Differential Equation

Introduction

Consider x as an independent variable and y as dependent variable. An equation that involves at least one derivative of y with respect to x, e.g. n

n

dx

yd

dx

yd

dx

dy,.....,,

2

2

Is known as a differential equation or common differential equation.

dxxdy 32

b) 52 3 xdx

dy

c) xydx

dysin2

2

d) 07432

2

dx

dy

dx

yd

a)a)

Example of Differential Equation

Order is the highest derivative

Degree is the highest power of the highest

derivative

Examples: a)

8233

2

2

yx

dx

dy

dx

yd

This DE has order 2 (the highest derivative appearing is the second derivative) and degree 1 (the power of the highest derivative is 1.)

Order & Degree

In this chapter we only deal with first order,

first degree differential equations.

A solution for a differential equation is a

function whose elements and derivatives may be

substituted into the differential equation. There

are two types of solution for differential

equations

General solution – The general solution of a

differential equation contains an arbitrary

constant c.

Particular solution - The particular solution of

a differential equation contains a specified

initial value and containing no constant.

Solutions

Examples Of General Solution

This is already in the required form, so we

simply integrate:

, c is constant

032 dxxdyy

cdxxdyy 32

cxy

43

43

Example

011 dx

dyyxxy

First we must separate the variables:

011 dyyxdxxy Multiply throughout by dx

0

11

y

dyyxdxx Divide throughout by y

0

11

y

dyy

x

dxxDivide throughout by x

This gives us: 01

11

1

dy

ydx

x

We now integrate,

constant is

11

11

c,cx

ylnyx

cxlnylnyxcylnyxlnx

cdyy

dxx

Example

21 ydx

dye x


dxydye x 21 Multiply throughout by dx

dx

e

ydy

x

21 Divide throughout by xe

dx

edy

y x

1

1

12

Divide throughout by 21 y

This gives us: dxedyy x 21

We now integrate:

constant is ,1

1

11

1

1

1

2

ccey

cey

dxedyy

x

x

x

Separable Variables and Integrating Factors

Example

Solve the differential equation

012 dx

dyyexx y


cdyyedxxx

dyyedxxxy

y

1

012

2

Consider dxxx 12 :

cx

ct

ct

dtt

x

dttxdxxx

2

32

2

3

2

3

2

1

2

13

13

1

23

21

2

1

21 Thus,

By using substitution,

12 xt

x

dtdx

xdx

dt

2

2

Consider

Let

dyye y

yu dydu

dy

du

1

andye

dy

dv

y

y

y

ev

dyedv

dyedv

1

Thus,

yeeye

dyeye

vduuvdyye

y

yy

yy

y

By using integration by

part

2

32 1

3

1x cye y 1

The General Solution :

Example

Solve the differential equation

when x =0, y=5


2 yxdx

dy

xdxy

dy

2We now integrate:

xdx

y

dy

2

cx

y 2

2ln2

We now use the information which means

at , to find c.

0x 5y

c2

025ln

2

3lnc

So the particular solution is:

3ln2

2ln2

x

y

gives

Exercise :

Solve the initial value problem. Express the solution implicitly.

0 when 0;)2( 22

xyyxdx

dye xa.

0 when 0;)cos1()1(sin xydx

dyxex yb.

lecture 1

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