EUT 102 1 CHAPTER 2 : First Order Differential Equations

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

EquationsEquations

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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.

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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?

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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

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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

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Example 2.1

22''''''2

''

'

)2(2

04

cos

yxyeyyx

yy

xy

x

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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.

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Example

Solve the following differential equations :

1 b)

cos a)

2

x

x

dx

dy

xdx

dy

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FIRST ORDER DIFFERENTIAL EQUATIONS

Separable Equations

Homogeneous Equations

Linear Equations

Exact Equations

Applications

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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)

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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

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Exercise

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

a) y’ = xey-x

b) y’x = x – 2y

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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

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Example 2.3

Solve the following differential equations :

a) (x + 2)y’ = y

b) y’ex + xy2 = 0

c) x2y’ = 1 + y

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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)

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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.

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Solution

Hint…

7040)18(kdt

d

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= 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.

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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.

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Example 2.5

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

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Solution

Then, show that f(λx , λy) = f(x,y).

x

yln

y

x)y,x(f

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Example 2.6

equation shomogeneou a is that Show22

'

yx

xyy

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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.

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Example 2.7

Solve the initial value problem.

y(0) = 2.

,'22 yx

xyy

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Solution

Test the homogenity Substitute y = xv

Integrate

x

dxdv

v

1

v

13

2v2

1k|xv|ln

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

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

22 y2/xAey

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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”

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Example 2.8

xy

AxeyxAnswer

xxy

yx

dx

dy

2

2

22

)(:

a)

equations aldifferenti following theSolve 1.

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Ax eBBxeyAnswer

yxydx

dyx

,:

2 b)

2

2

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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).

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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.

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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

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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

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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

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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.

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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

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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?

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Answer

13.45 liter

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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).

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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

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Example 2.13

Show that

(6xy+2y)y’ = –(2x + 3y2)

is an exact equation.

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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

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Example 2.14

Solve the following differential equation;

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

Solution :

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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

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Cont…

2) Write

(II)

and

(I)

Ny

u

Mx

u

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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

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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

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Solution

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

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

Show that

x

N

y

M

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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)

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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

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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

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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

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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

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Answer

x2y = A , A is a constant