EUT 102EUT 102 11

CHAPTER 2 :CHAPTER 2 : First Order Differential First Order Differential

EquationsEquations

EUT 102EUT 102 22

Learning Outcomes Learning Outcomes

a)a) At the end of this chapter, it is expected that all At the end of this chapter, it is expected that all students will be able to know:students will be able to know:

What is ODE?What is ODE? What is order of ODE?What is order of ODE? How to solve ODE?How to solve ODE? How to use IVP and BVP?How to use IVP and BVP?

b) And also at the end of the chapter students will have b) And also at the end of the chapter students will have to possess the skills of solving first-order ODE.to possess the skills of solving first-order ODE.

EUT 102EUT 102 33

An equation that contains an An equation that contains an unknown function and one or unknown function and one or

more of its derivativesmore of its derivatives

WHAT IS A DIFFERENTIAL WHAT IS A DIFFERENTIAL EQUATION?EQUATION?

EUT 1024

ODE BASICS

“An ordinary differential equation (ODE) is an equation that contains one or several derivatives of an unknown function.” (Kreyzig). For example:

22''''''2

''

'

)2(2

04

cos

yxyeyyx

yy

xy

x

EUT 1025

Order of a Differential Equation

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

First order differential equations can be written as:

),('

0),,( '

yxfy

or

yyxF

EUT 1026

Example 2.1

22''''''2

''

'

)2(2

04

cos

yxyeyyx

yy

xy

x

EUT 1027

Solution of an ODE

A function f is called a solution of a differential equation if the equation is satisfied when y=f(x) and its derivatives are substituted into the equation.

EUT 1028

Example

Solve the following differential equations :

1 b)

cos a)

2

x

x

dx

dy

xdx

dy

EUT 1029

FIRST ORDER DIFFERENTIAL EQUATIONS

Separable Equations

Homogeneous Equations

Linear Equations

Exact Equations

Applications

EUT 10210

2.1 : SEPARABLE EQUATIONS

The differential equation ;

y’ = f(x,y) is said to be separable if the equation can be written

as the product of a function of x, u(x) and a function of y,v(y) i.e.

y’ = u(x)v(y)

EUT 10211

Example 2.2

y’ - xy = x can be written as y’ = x(1+y) i.e u(x) = x & v(y) = 1+y

y’ siny cosx – cosy sinx = 0 or can be written as y’ = tan x cot y

EUT 10212

Exercise

Determine each of the following ODEs is separable equations or not.

a) y’ = xey-x

b) y’x = x – 2y

EUT 10213

Solution of Separable Equations

The equation y’ = u(x)v(y) can be written in the form ;

Then, we integrate on both sides of the equation:

where A is constant

dx)x(u)y(v

dy

Adx)x(u)y(v

dy

EUT 10214

Example 2.3

Solve the following differential equations :

a) (x + 2)y’ = y

b) y’ex + xy2 = 0

c) x2y’ = 1 + y

EUT 10215

Solutions

a) y-1dy = (x+2)-1dx

By integrating both sides :

ln |y| = ln |x+2| + A

ln |y| - ln |x+2| = A

and becomes y = B(x+2).

ABexBy

b x ,)1(1

)

Cx

yc 1

1ln)

EUT 10216

Example 2.4

Let θ be the temperature (in oC) for a mass in a room with a constant room temperature at 18oC. The mass cools from 70oC to 57oC in 5 minutes, how much longer will be needed by the mass to cool down to 40oC.

EUT 10217

Solution

Hint…

7040)18(kdt

d

EUT 10218

= 18 + Ae-kt

– From the initial condition, we obtain A = 52 and k = 0.05753.

– Solve…

The mass will need approximately 9.952 minutes more to cool.

EUT 10219

2.2 : HOMOGENEOUS EQUATIONS

Definition 2.2 :

A differential equation y’ = f(x,y) is said to be homogeneous if

f(λx , λy) = f(x , y)

for any real number λ.

EUT 10220

Example 2.5

Determine whether y’y = x(lny – lnx) is a homogeneous equation.

EUT 10221

Solution

Then, show that f(λx , λy) = f(x,y).

x

yln

y

x)y,x(f

EUT 10222

Example 2.6

equation shomogeneou a is that Show22

'

yx

xyy

EUT 10223

Solution of Homogeneous Equations

A homogeneous equation can be transformed into separable equations by substituting

y = xv or v = y/x.

Hence,

y’ = v + x (dv/dx) This will result in separable equations in variables x

and v. Solve and resubstitute the value v=y/x.

EUT 10224

Example 2.7

Solve the initial value problem.

y(0) = 2.

,'22 yx

xyy

EUT 10225

Solution

Test the homogenity Substitute y = xv

Integrate

x

dxdv

v

1

v

13

2v2

1k|xv|ln

EUT 10226

Substitute v = y/x and determine the general solution.

The initial condition, y(0) = 2, so that A = 2.

22 y2/xAey

EUT 10227

Sekalipun mereka dapat memahami hakikat ‘bagaimana hendak bermula tetapi tidak pernah bersiap’…. Sampai bila pun mereka tidak akan bermula

“Peluang sentiasa ada kepada setiap orang. Hanya kesanggupan yang membezakan”

EUT 10228

Example 2.8

xy

AxeyxAnswer

xxy

yx

dx

dy

2

2

22

)(:

a)

equations aldifferenti following theSolve 1.

EUT 10229

Ax eBBxeyAnswer

yxydx

dyx

,:

2 b)

2

2

EUT 10230

2.3 : LINEAR EQUATIONS

Definition 2.3 : (First OLDEs)

An equation is said to be a first-order linear

equation if it has the form

a(x)y’ + b(x)y = c(x)

where a(x), b(x) & c(x) are continuous functions on a given interval.(or a constant).

EUT 10231

Example 2.9

a) xy’ – 2y = x + 1 is a linear equation with

a(x) = x, b(x) = -2 dan c(x) = x+1

b) 2x2y’ + xey = sin x is not a linear equation.

EUT 10232

Example 2.10

Determine whether the following ODEs are linear equations or not.

a) (1 - x2)y’ = x(y + sin-1x)

b) y’ + xy2 = ex

EUT 10233

Solution of Linear Equations

We rewrite the linear the equation as

y’ + p(x)y = q(x)• a(x) = 1

find ∫p(x)dx

An integrating factor is ρ = e∫p(x)dx

EUT 10234

Cont.....

multiplying both sides of the linear equation by ρ we get, ρy’ + ρp(x)y = ρq(x)

Or can be rewrite into the form;

The solution to the linear equation is by integrating both sides with respect to x.

)x(q)y(dx

d

EUT 10235

Example 2.11

Solve the following linear equations.a) y’ + y = x

b) y’ +y tan x = kos x,

given that y = 1 when x = 0.

EUT 10236

Solutions

a) y’ + y = x

we know that p(x) = 1 and q(x) = x.

Answer : y = (x – 1) + Ae-x

b) Answer : y = (x + 1) kos x

EUT 10237

Example 2.12: Linear Equations Application

A tank containing 50 liters of liquid with composition, 90% water and 10% alcohol. A second liquid with composition, 50% water and 50% alcohol is poured into the tank at the rate of 4 liters per minute. While the second liquid is poured into the tank, the liquid in the tank is drained out at the rate of 5 liters per minute. Assuming that the liquid in the tank mix uniformly, how much alcohol left in the tank 10 minutes later?

EUT 10238

Answer

13.45 liter

EUT 10239

2.4 : EXACT EQUATIONS

Definition 2.4

A first order differential equation of the form

M(x,y)dx + N(x,y)dy = 0

is called an exact diffrential equation if the differential form M(x,y)dx + N(x,y)dy is exact,that is,

du = M(x,y)dx + N(x,y)dy

of some function u(x,y).

EUT 10240

Theorem 2.1 : (Condition of an exact equation)

* The equation

M(x,y)dx + N(x,y)dy = 0

is an exact equation if and only if :

x

N

y

M

EUT 10241

Example 2.13

Show that

(6xy+2y)y’ = –(2x + 3y2)

is an exact equation.

EUT 10242

Solution

We rewrite in the form

(2x + 3y2)dx + (6xy + 2y)dy = 0

i.e. M = (2x + 3y2) and N = (6xy + 2y)

Then, we test for exactness

x

N

y

M

EUT 10243

Example 2.14

Solve the following differential equation;

(1- kos2x) dy + (y sin 2x) dx = 0.

Solution :

EUT 10244

Solution of An Exact Equations

1) Write in the form of exact equation :

M(x,y)dx + N(x,y)dy = 0.

Test for Exactness :

x

N

y

M

EUT 10245

Cont…

2) Write

(II)

and

(I)

Ny

u

Mx

u

EUT 10246

Cont....

3) By integrating Equation (I) with respect to x, we have u(x,y) = ∫ M dx + Φ(y).

4) Differentiate u with respect to y, and equating the result with equation (II) to determine the function Φ(y).

5) Write the solution in the form of u(x,y) = A, where A is a constant.

5) By substituting the initial condition, we will get the particular solution of the initial value problem

EUT 10247

Example 2.14

Show the the following ODE is an exact equation and solve the differential equation.

(6x2 - 10xy + 3y2)dx + (-5x2 + 6xy -3y2)dy = 0

EUT 10248

Solution

Test for Exactness(6x2 - 10xy + 3y2) = M and

(-5x2 + 6xy -3y2) = N

Show that

x

N

y

M

EUT 10249

Cont …

Let u(x,y) be the solution, then

u = ∫Mdx +Φ(y)

u = ∫(6x2 - 10xy + 3y2)dx + Φ(y)

= 2x3 – 5x2y +3xy2 + Φ(y)

differentiate u with respect to y, we obtain

∂u/∂y = -5x2 +6xy + Φ’(y)

EUT 10250

Cont .....

By equating with N we obtain,

-5x2 +6xy + Φ’(y) = (-5x2 + 6xy -3y2)

hence; Φ’(y) = -3y2 and we obtain

Φ(y) = -y3 + B

The general solution is u(x,y) = A, where

u(x,y) = 2x3 – 5x2y +3xy2 - y3 + B = A

u(x,y) = 2x3 – 5x2y +3xy2 - y3 = C, C = A - B

EUT 10251

Example 2.15

Solve the following differential equations

1) xy’ + y + 4 = 0

2) sin x dy + (y kos x – x sin x) dx = 0

EUT 10252

Answer

1) u(x,y) = (y + 4)x = C , C = A – B

2) u(x,y) = y sin x + x kos x – sin x

= C , C = A – B

EUT 10253

Exercise 2

Show that the following differential equation

2y dx + x dy = 0

is not an exact equation.

If the integrating factor,

solve the differential equation.

,1

),(xy

yx

EUT 10254

Answer

x2y = A , A is a constant