1Lecture Student Notes ODE 2010

8/9/2019 1Lecture Student Notes ODE 2010

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Ordinary Differential Equations (ODE)

SK/EUM111/ODE/09 1


LEARNING OBJECTIVES

The objectives of this chapter are to:

1. Impart to students the knowledge of ordinary differential equations and their classifications.2. Impart to students the concept of solutions of differential equations.3. Equip students with the knowledge of solving first order ordinary differential equations using

the method of separating the variables.

4. Equip students with the knowledge of solving linear first order ordinary differential equationsusing integrating factor.

5. Equip students with the knowledge of solving homogenous first order ordinary differentialequations.

6. Equip students with the knowledge of solving linear second order ordinary differentialequations with constant coefficients.

Differential equations form the language in which the basic laws of science are expressed. The science

tells us how the system at hand changes "from one instant to the next." The challenge addressed by thetheory of differential equations is to take this short-term information and obtain information about

long-term overall behavior. So the art and practice of differential equations involves the followingsequence of steps: one "models" a system (physical, chemical, biological, economic, or even

mathematical) by means of a differential equation; one then attempts to gain information aboutsolutions of this equation; and one then translates this mathematical information back into the

scientific context.

Differential Equation:

Short terminformation

Solving Behaviour over time

Physical World

InterpretationModel


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A basic example is given by Newton's law, F = ma. a = acceleration, the second derivative of x =position. Forces don't effect x directly, but only through its derivatives. This is a second order ODE,

and we will study second order ODEs extensively later in the course.

Definition: A differential equation is an equation containing derivatives of an unknown function. A

differential equation (DE) contains one or more terms involving derivatives of one variable (thedependent variable,y) with respect to another single independent variable (the independent variable,x) is said to be anordinary differential equation.

In the DE 3+4x+7x=dx

dy 2 y=f(x) is the unknown function.dx

dyis the derivative of the

unknown function.

For example, xdx

dy wherey is the dependentvariable and

x is the independentvariable

txdt

dx wherex is the dependentvariable and

tis the independentvariable

The DE is an ordinary DE (ODE) if the unknown function depends on only one independent variable.

The following are ODE

5x=y+)dx

dy3(+)

dx

yd((d)

1=)dx

dy2(+

dx

dyec)(

t

1+x=

dt

dx(b)

4-2x=dx

dya)

263

2

2

3y

2

An equation involving thepartial derivatives of one or more terms involving derivatives of onevariable oftwo or more independent variables is called apartial differential equation.

(Partial differential equations)2

2

u u

t x

(heat equation)


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2 2

2 20

u u

x y

(Laplaces equation )

Order: The order of a differential equation is the order of the highest derivative that appears in the DE

equation.

First-Order : dy/dx = 3x2+10x+3 , y = 3x

3+4x+2

Second-Order: d2y/dx2 = 4x + 5, y = 5x4+3

Third-Order:y = 6x +2

Forth-Order:y(4) = 4x2

+ 4x+ 7

A general nth-order, ordinary differential equation is often represented by the

symbolism

, , ,...,n

n

dy d yF x y

dx dx

Thus first-order differential equations, contain onlyy' and may containy and functions ofx. Hence it is

written asF(x,y,y) =0

second-order differential equations, contain onlyyand may containy,y and functions ofx. Hence itis written as

F(x,y,y,y) =0

Degree: The degree of a DE is the highest power of the highest derivative (when the derivatives arecleared of radicals and fractions).

Linear: A DE. is said to be linear if the dependent variable and its derivative are of degree one andthere are no product involving the dependent variable ,the derivative and in transcendental functions

such as sine, exponential.

A DE is linear if it can be written in the form :1

1 1 0n 1

d y d y dy(x) +a (x) +...+a (x) +a (x) y = f(x)

dx dx dx

n n

n n n

a

. (1)

Two properties:

i) The dependent variables y and all its derivative are of first degree (power of each term isinvolving only 1)

ii) Each coefficient depends on only the independent variable x. iii) Not involving the dependent variable in transcendental functions such as sine, exponential

.For example


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Differential Equation Linearity

1.tx

dt

dx

Linear, ordinary and order is 1

2.t

dt

dx

dt

xd 3

2

2

Linear, ordinary and order is 2

3.y

dx

dy2)(

non-linear (y)2

2.tx

dt

dxx non-linear (

dxx

dt)

4. y x y y x4 2 cos is linear, ordinary and order is 2

5. y y y y x4 2 cos is nonlinear y y

6.

2

2

u

x

v

tu v u sin

is linear in v but nonlinear in u sinu . Theequation is nonlinear

7. d x

dt

dy

d tx y t

2

2 sin

is linear in each of the dependent variables x and

y. But it is nonlinear in the set of {x, y}. The

equation is nonlinear

Homogeneous or NonHomogeneous Equation

From equation (1), it can be written as

1

1 1 0n 1

d y d y dy(x) +a (x) +...+a (x) +a (x) y = f(x)

dx dx dx

n n

n n na

.

1

1 01

n 1

a (x) a (x)a (x)d y d y dy f(x)+ +...+ + y =

dx (x) dx (x) dx (x) (x)

n n

n

n

n n n na a a a

1

1 01

n 1

a (x) a (x)a (x)d y d y dy f(x)+ +...+ + y =

dx (x) dx (x) dx (x) (x)

n n

n

n

n n n na a a a

( ) ( 1) '

1 1 0( ) ... ( ) ( ) ( )n n

ny p x y p x y p x y r x

(2)


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where 1 01a (x) a (x)a (x) f(x)

,..., , and(x) (x) (x) (x)

n

n n n na a a a

are represented with1 1 0( ),..., ( ), ( ) ( )np x p x p x and r x

respectively.

When1 1 0( ),..., ( ) ( )np x p x and p x are constants, we say that equation (2) has constant

coefficients; otherwise it has variable coefficients. Ifr(x) = 0, equation (2) becomes

( ) ( 1) '

1 1 0( ) ... ( ) ( ) 0n n

ny p x y p x y p x y

(3)

and is calledhomogeneous.

Ifr(x) is not identically zero, the equation is callednonhomogeneous.

Constant-coefficient linear equations

A linear differential equation has constant coefficients if the coefficients of the dependent variable

and its derivatives are constants.

For example, 1) tdt

dx

dt

xd 3

2

2

constant-coeff linear ODE

2) 32

2

745 xydx

dy

dx

yd constant-coeff linear ODE

3) teydt

dyt 5 linear ODE but not constant-coeff

Exercise :State the linear or nonlinear, and homogeneous or nonhomgeneous of the following

differential equations.

i. x5 y''' + x2 y' + 8y = 0 - the homogeneous linear equationii. y'' + 2y = 6e-xsinx - nonhomogeneous linear differential equation

iii. (2 x2)y'' 5xy' + 6y = 0 - homogeneous linear differential equationiv. x(y''y + y'2) + 2y'y = 0 - homogeneous nonlinear differential equation

Solution of an ODE


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A solution is an expression which satisfies the differential equation. A solution of a DE which

involves the

independent variable x and dependent variable y is the finding of the equation f(x,y)=0 which satisfies

the DE.

General Solution: A solution is a set of all possible solutions of the DE.A general solution of a

differential equation represents an infinite number of solutions, where the arbitrary constant C may be

any real or complex number. For example, the general solution ofy = 2x yieldsy = x2 + C.

Exercise. 1 : Show thaty = e2x is a solution of y

dx

dy2

Exercise 2: Show thaty =5e2x is a solution of y

dx

dy2

There are many different expressions which can satisfy a differential equation, that is, there are

many solutions. A solution from which many solutions can be found is called the general solution. It

contains the same number of arbitrary constants as the order of the equation.

The general solution of ydx

dy2 isy = Ce2x where C is any constant. C is called an arbitrary

constant.

Particular Solution: Particular Solution is any one solution of the DE.A particular solution is created

when the constant C has a specific value, which occurs when initial conditions are given and the

correspondence of a particular value of the independent variablex gives a specific valuey.

Example 1 - the particular solution ofy = 2x with the initial condition ofy(0) = 2, yieldsy = x2 + 2.

Initial Conditions and Boundary Conditions: These two conditions determine the value(s) of the

constant(s) in the general solution and form the particular solution.

Example 2 - the general solution of ydx

dy

2 isy = Ce2x

If y(0) = 4

then y = 4e2x which is a particular solution


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If the initial conditions (e.g. at t=0 or at initial displacement) are given, the problem is called an

initial value problem. If the conditions are given at different times, then the problem is called a

boundary value problem.

Explicit and Implicit Solutions

A solution of an ordinary differential equation that can be expressed in the formy =f(x) is said to be an explicit solution.

A relation G(x,y) =0 is said to be an implicit solution of an ordinary differentialequation on an intervalIprovided it defines one or more explicit solutions onI.

Example : For -1


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Exercise 2: Show that xx BeAxey is the general solution of 022

2

ydx

dy

dx

yd

Hence, determine the particular solution satisfying 1)0(,0)0( dx

dyy

Exercise 3: Obtain the solution of

a) xdx

dycos

b) tdt

xd

2

2

with conditions 7)2(and3)0( xx

Exercise 4: Determine the values of the constants C1 and C2 in y=C1sin x + C2cos x given

(a) the initial conditions y(0)=1, y(0)=2

(b) the boundary conditions 1=)(y'1,=y(0)


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First-Order Ordinary Differential Equations

First-order differential equation: ( , , ) 0F x y y

-- involves a first but no higher derivatives

Example: ( ) 2cos 4y x x

yfunction of x and xindependent variable

( , ( ), ( )) 0F x x x ( )x : solution

Whenfis independent of the variabley, that isf (x,y) =g(x) ,

The differential equation

( )dy g xdx

can be solved by direct integration.

Direction Field as a geometrical representation of first order DE

-- A set of line segments tangent to a curve

-- Give a rough outline of the shape of the curve

( , , ) 0F x y y

Solve forslope

( , )direction field

y f x y

SLOPE FIELDS


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Slope fields (also called vector fields or direction fields) are a tool to graphically obtain the solutions

to a first order differential equation.

Example 1: Draw a slop field for dy/dx = y

This is the FIELD in which we plot the SLOPES.

So we first tried it with the expression ( dy/dx = y ). This

means that the derivative of any point in the function is the

value of the function at that point (y-coordinate).

Let's say we want to plot the point (1,1). The derivative

(SLOPE) of the function at that point is 1, so we draw a

small line with slope 1 at that point.

Next, take the point (3,2). The derivative at that point is 2,

so there's another small line, but with a slope 2 at the point

(3,2).).

Now, you understand how they are constructed, this is how

the SLOPE FIELD of the function(s) that have ( dy/dx = y)


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From here, you can solve for the point (1,1).

To do this, draw a curve that follows the slope of the line in

either direction until the next whole number is reached. At

that point, change the curve's slope to that of the new point

and continue until the next whole number and so on...

WHAT FUNCTION DOES THE CURVE LOOK LIKE?It

is easy to see that the function that has a derivative equal tothe value of the function and passes through the point (1,1)

is y=ex.

However, you can also see that there are many other functions with a similar shape that could have

been drawn, depending on the starting point.

This is because the slope fields shows the entire family of functions [ (x)dx + C ].So, by solving for a point, you can find the exact one function from that family.

Let say the slope fields for dy/dx = 2x and dy/dx = -x/y are

Can you guess by observation what functions they are?


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Example 2:

y= -2xy

The slope,y'(x), of the solutionsy(x), is determined once we know the values forx andy , e.g., ifx=1

andy=-1, then the slope of the solutiony(x) passing through the point (1,-1) will be (-2).1.(-1)=2. Ifwe graphy(x) in thex-y plane, it will have slope 2, givenx=1 andy=-1. We indicate this graphicallyby inserting a small line segment at the point (1,-1) of slope 2.

Thus, the solution of the differential equation with the initial conditiony(1)=-1 will look similar to thisline segment as long as we stay close tox=-1.

Of course, doing this at just one point does not give much information about the solutions. We want to

do this simultaneously at many points in thex-y plane.

We can get an idea as to the form of the differential equation's solutions by " connecting the dots." So

far, we have graphed little pieces of the tangent lines of our solutions. The " true" solutions should notdiffer very much from those tangent line pieces!


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Example 3:2

y y The slope:2

y

General Solution:1

yx k

Direction field fory'=y and integral curves through (0,1), (0,2), (0,3), - 1, (0, - 2), and (0, - 3).

Example 4: Let's consider the following differential equation:

y = e-y


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Here, the right-hand side of the differential equation depends only on the dependent variabley, not onthe independent variablex. Such a differential equation is called autonomous. Autonomous

differential equations are always separable.

Autonomous differential equations have a very special property; their slope fields are horizontal-shift-invariant, i.e. along a horizontal line the slope does not vary.


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Directly Integrable Equations

Equation type:y = g(x)

Integrate on both sides of the equation will directly yield the solution.

( )y g x Solve by integrating both sides directly.

Example 1:y = 2x + 5 yieldsy = x2 + 5x + C

Example 2:y = 6e3x + 2/x yields y = 2e3x + 2ln x + C

Separation of Variables

( ) ( ) ( ) ( ) 0F x G y dx f x g y dy

This type of DE is called separable because it can be written in the form (variables can be separated)

0dy)y(Ndx)x(M

The first equation is usually not exact but multiplying it by the appropriate integrating factor will make

it exact, but use of an integrating factor may eliminate solutions or may lead to extraneous solutions.

After multiplying by the integrating factor)y(G)x(f

1the equation becomes:

0dy)y(G

)y(gdx

)x(f

)x(F

where)x(f

)x(F)x(M and

( )( )

( )

g yN y

G y .

Solutions are of the form cdy)y(Ndx)x(M where 0)y(G0)x(f & .

Solve by separation of variables intody dy f(x)

= f(x)/g(y) ( or y' = = )dx dx g(y)

or Equation type:f(y)

dy/dx = g(x)

Solve by separation of variables intof(y) dy = g(x) dx, and integrate each side.


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( ) ( )dy

g y dx f x dx cdx

=> ( ) ( )g y dy f x dx c

Example 1:

1. ( 2 )' x yy xe is separable2. y = 3x-y is not separable

Example 2:y = xy2

(1/y2) dy = x dx

1/y = x2/2 + C

Example 3:y = y sin x with initial conditionsy(0) = 1

General Solution: (1/y) dy = sin x dx

ln y = - cos x + C

Particular Solution: Solve to find C = 1 (ln 1 = -cos 0 + C => 0 = -1 + C)

ln y = - cos x + 1

Example 4:y = 4x (y-2)

1/(y-2) dy = 4x dx

ln (y-2) = 2x2

+ C which is reducible toy = 2 + eC => y = 2 + C1

First Order DE. Linear

A first order differential equation is said to be linear if:

(1) the dependent variable and its derivatives occur to the first order only,

(2) there are no products involving the dependent variable with its derivatives, and

(3) there are no non-linear functions of the dependent variable such as sine, exponential, etc.


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Afirst order linear DE may be written in the standard formdy

+ P(x) y =Q(x).dx

Exercise : What are P(x) and Q(x) referred to the standard form?

2

2 3

dy(a) - 3x y - 5x = 0dx

dy(b) x = 5y

dx

dy(c) x + 3x y + 2x = 5y.

dx

Linear First Order Homogeneous Differential Equation

If the DE given is a linear 1st

order DE., i.e.,dy

+ P(x) y =0dx

, the general solution of the DE can be

solved as follows.

The general solution of the DE can be found if the integral ( )P x dx exists.Example : y'+3y=0 with initial conditiony(1)=1

Solution : Since3( ) 3 3 3 e xP x dx x y C

and3 3[1 ](1) 1 e ( ) e xy C y x

First Order Linear Non-homogeneous DE

By definition, a linear first-order differential equation in y cannot contain products, powers or

other nonlinear combinations of y or y'. Have its most general form is

( )

( ) 0

( )

ln ( )

eP x dx

dyP x y

dx

dyP x dx

y

y P x dx C

y C


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( ) ( )dy

P x y Q xdx

withf(x)0, we can obtain the general solution of above equation as follows:

Multiply both sides by ])(exp[ dxxa :

[ ' ( ) ]exp[ ( ) ] ( )exp[ ( ) ]

exp[ ( ) ] ( )exp[ ( ) ]

exp ( ) ] ( )exp[ ( ) ]

y P x y P x dx Q x P x dx

dy P x dx Q x P x dx

dx

y P x dx Q x P x dx C

or exp ( ) ( )exp ( )y P x dx Q x P x dx dx C Here ])(exp[ dxxa is called the integration factor. Using integration factor to find the solution of DE.is a standard way in other form of DE. as well.

Integrating Factor

The linear DE in standard form has the integrating factor

P(x)dxe=x)(

Exercise : Find the integrating factor of the DE

dy(a) = 5y

dx

dy(b) x + 2y = 2x

dx

dy(c) + y tan x = sin x

dx

General Solution

The general solution of the DEdy

+ P(x)y = Q(x)dx

is

(x) y = (x) Q(x) dx + C

Example 1 Solve the differential equation

1x

y

dx

dy


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Solution

Here we have

xP

1

and 1Q

The integrating factor,

xeex

dxx ln1

We integrate with respect tox:

Cxxdxxy2

2

1

x

Cxy

2

1

We muptiply through the given equation with x :

dy

x y xdx

This is an exact equation because it can be written as

xxydx

d)(

Exercise :Find the general solution of the DE.

4 x

dy(a) = 5y

dx

dy(b) x + 2y = 2x

dx

dy(c) + y tan x = sin x

dx

dy(d) x -3y = x e

dx

EXACT EQUATIONS AND INTEGRATING FACTORS

First-order Differential Equations for which we can find exact solutions


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Study the patterns carefully. The first step of any solution is correct identification of the type of

differential equation.

A Linear First Order DE with Variable Coefficients can be written in the following form.

)x(Qy)x(Pdxdy

This equation is exact only when P(x) = 0.

Proof:

Write DE in the form 0dy)y,x(Ndx)y,x(M

)x(Qy)x(Pdx

dy

dx)x(Qydx)x(Pdy

0dydx)x(Qydx)x(P

0dy1dx)x(Qy)x(P

But ( , ) ( ) ( ) ( )M x y P x y Q x P xy y

0)1(x

)y,x(Nx

P(x) must equal 0 for the DE to be exact.

The equation 0dy1dx)x(Qy)x(P can be solved though by finding a proper integrating factor.The factor depends only on x, so call it )x( .

The new equation 0dy)x(dx)x(Q)x(y)x(P)x( is exact so

since )x(P)x()x(Q)x(y)x(P)x(y

)x(dxd)x(

x

( ) ( ) ( )d

x P x xdx

ordx

d)x(P

d1

dx)x(P


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d1

dx)x(P

lndx)x(P

dx)x(P

e)x( assuming 0)x(

The solution of the DE is therefore of the form

cdx)x(Qeey

dx)x(Pdx)x(P

Total Differential of a Function F(x,y)

F is a function of two variables which has continuous partial derivatives over a domain D. The total

differential of F is defined as :

dF(x,y) = dyy

)y,x(Fdx

x

)y,x(F

Dyx ).(

Exact Differential

M(x,y)dx + N(x,y)dy is an exact differential over D if )y,x(F such that

)y,x(dFdy)y,x(Ndx)y,x(M

dyy

)y,x(Fdx

x

)y,x(F

= dF(x,y)

In other words, M(x,y) =x

)y,x(F

N(x,y) =y

)y,x(F


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M(x,y)dx + N(x,y)dy = 0 is an exact differential equation if

M(x,y)dx + N(x,y)dy is an exact differential.

Then

i.e. dF x, y 0.

Integrating, we have F x, y c arbitrary constant

Given a DE that can be written in the form M(x,y)dx + N(x,y)dy = 0,

find F(x,y) such that .dy)y,x(Ndx)y,x(M)y,x(dF

The following theorem supplies a method for determining whether or not a DE is exact.

Theorem

M(x,y) and N(x,y) are functions with continuous first partial derivatives in the domain D.

M(x,y) and N(x,y) form the DE

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

The above DE is exact in D if and only if

)y,x(Nx

)y,x(My

Dyx ),(


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Solution Method for Exact Equations

1. Showy

dx)y,x(M =

x

dy)y,x(N

Set )y,x(Mx

)y,x(F

. Integrate with respect to x to get F ),( yx

F ),( yx = dx)y,x(M + )y( 2. Differentiate with respect to y to get )y,x(N

y

[ dx)y,x(M + )y( ] = )y,x(N 3. Solve for )y( 4. The answer will be of the form : F ),( yx = g(x,y) + )(y , if no boundary value

is given.

Example : Determine whether the following equations are exact and solve the ones that are exact.

0)2( dyyxedxe yy

Solution:

(a) ,yeM yxeN y 2

so thatx

Ne

y

M y

Hence the given equation is exact and

yeyxMx

f ),( (1)

and yxeyxNy

f y2),(

(2)

We integrate (1) with respect to x:


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dxeyxf y),(

)(ygxey

Differentiating with respect toy:

yxeyxNdy

dgxe

y

f yy 2),(

ydy

dg2

Integrating, we have

1

2)( cyyg

The desired solution

21

2),( ccyxeyxf y

cyxey 2

Exercise 1 : Which of these equations are exact ?

a. dxy )3(2 + ( dyxy )42 = 0

b. dxyx )23(2 dyyx )( 3 = 0

c. 0sin2cos)1(2 rdrdr

d. 0dy)yy

x(dx)x

y

x(

3

2

2

e. dxx

y)

12(

2/1

2/3 + (3x

2/1y dy)1

2/1 =0


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Exercise 2: Solve

a. 0dy)1xy3yx2(dx)x2yyx3(23322 given 12y

b. 0dyxy2edxye2ye x2xx

c. Find A so that this equation is exact. 0dyxy4xdxy2yAx 322

Modified Exact First Order DE

Sometimes, a non-exact differential equation M x y dx N x y dy( , ) ( , ) 0 can be turned into an

exact differential equation by multiplying the whole equation by an appropriate factor, called an

integrating factor.

What ifx

)y,x(N

y

)y,x(M

We may be able to rewrite the equation so that it is exact.

If 0dy)y,x(Ndx)y,x(M is not exact, but 0)y,x(N)y,x(dx)y,x(M)y,x( is exact, then

)y,x( is called an integrating factor.

Example : Show that 0dyxdxxy2y 22 is not exact, then find n such that ynis an integrating factor.

i. x2y2xy2yy

2

x2)x(x

2

therefore the DE is not exact.

ii. Multiply the DE by yn, then solve.

0dyxydx)xy2y(y2n2n

n1n1n2n xy1n2y2nxy2yy

must equal


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xy2xyx

n2n

which means 1ny2n must equal 0 and

nxy1n2 must equal nxy2 for this to be so n must equal -2

iii. Now, solve the equation.

0dy)x(ydxxy2yy 2222

Solution:

1 2 21 2 0xy dx x y dy

2 1

( , )F x y x x y c

Theorem (a) If1

N

M

y

N

x

is a function ofx only, sayf(x), then exp ( )f x dx is an integrating

factor of the equation M x y dx N x y dy( , ) ( , ) 0

(b) if

y

M

x

N

M

1is a function ofy only, say g(y), then exp ( )g y dy is an integrating factor

of the equation M x y dx N x y dy( , ) ( , ) 0

Proof

(a) By the hypothesis, if ( )x is the integrating factor which depends on variablex only. We see

that

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

is the exact differential. From theorem, we have the necessary condition

yM

xN

M

y

N

xN

d

d x

( ) ( )

Eventually, we obtain


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SK/EUM111/ODE/09 27

d

d x

M

y

N

x

Nf

which has a general solution ( ) exp ( )x f x dx

The proof for (b) is similar and it is left for Exercise.

Example

Find the integration factor of the following ODE and solve the corresponding equation

3 2 02 3 2 2x y x y y dx x y dy

Solution

For this ODE,

M x y x y x y y

M

yx x y

M

xx y y

( , )

3 2

3 2 3 6 2

2 3

2 2

and

N x y x yN

xx

N

yy( , ) 2 2 2 2

and

We can see that1

3N

M

y

N

x

. Therefore the integration factor is given by

exp 3 3d x e x and the ODE is reduced into

e x y x y y dx e x y dyx x3 2 3 3 2 23 2 0

The differential function is f x y( , )

fx

e x y x y yx 3 2 33 2


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SK/EUM111/ODE/09 28

f

ye x y f x y e x y

yC x

f

xe xy e x y

yC x

f

xe xy x y y C x

x x

x x

x

3 2 2 3 23

3 3 23

3 2 3

3

2 33

2 3

( , ) ( )

' ( )

' ( )

By comparing the above equations, we have

constantthenand0 )()(' xCxC

Therefore the general solution of the ODE is

33 2( , )

3

x yf x y e x y k

Homogeneous DE

If 0dy)y,x(Ndx)y,x(M can be written in the form )y,x(fdx

dy where

x

yg)y,x(f then

the DE is homogeneous or is homogeneous if the functionf(x,y) is homogeneous, that is-

( , ) ( , )f tx ty f x y . If ( , ) ( , )nf tx ty t f x y then f is homogeneous of degree n.

Check that the functions

Example 1:2 2

( , )xy

f x yx y

2 2

2 2 2 2 2 2( , ) ( , )

txty xyf tx ty t t f x y

t x t y x y

is homogeneous degree 2.

Example 2:

2

3 3 2( , ) ( ) 4

x

yf x y x y e xy

2

3 3 2

2

3 3 3 3 2

3

( , ) (( ) ( ) ) 4 ( )

(( ) ( ) ) (4 )

( , )

tx

ty

x

y

f tx ty tx ty e tx ty

t x y e t xy

t f x y

is homogeneous degree 3.


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SK/EUM111/ODE/09 29

A homogeneous equation can be transformed to a separable equation by a change of variable vxy .

The new equation is separable in v and x, so it can be solved.

A homogeneous equation 0dy)y,x(Ndx)y,x(M , is always transformed into a separable one by

letting vxy .

Thendy dv

v xdx dx

.

Since the given DE is homogeneous, we know it can be written in the form

x

yg

dx

dy. Since vxy

,

vgx

vxg

x

yg

.

And since

x

yg

dx

dy

( )dv

v x g vdx

.

Separating variables dx)v(gxdvvdx : 0xdvdx)]v(gv[ : 0dv)v(gv

1dx

x

1

To solve, integrate: cdv)x(g(v 1dxx1

c)v(Fxln or cx

yFxln

Let us summarize the steps to follow:

i) Recognize that your equation is an homogeneous equation; that is, you need to check thatf(tx,ty)=f(x,y), meaning thatf(tx,ty) is independent of the variable t;

ii) Write out the substitution v=y/x;iii)Through easy differentiation, find the new equation satisfied by the new functionz.

You may want to remember the form of the new equation:


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SK/EUM111/ODE/09 30

dx

dvxv

dx

dy

iv)Solve the new equation (which is always separable) to find v;v) Go back to the old functiony through the substitutiony =x v;vi) If you have an IVP, use the initial condition to find the particular solution.vii)Since you have to solve a separable equation, you must be particularly careful about the

constant solutions.

Example 3: Find all the solutions of2 5

2

dy x y

dx x y

Follow these steps:

i) It is easy to check that 2 5( , )2

x yf x y

x y

is homogeneous;

ii) Consider v=y/x ;iii) We have

2 5 2 5'

2 2

x xv zxv v

x yv z

which can be rewritten as

1 2 5' .

2

vv v

x v

This is a separable equation. If you don't get a separable equation at this point, then yourequation is not homogeneous, or something went wrong along the way.

iv) All solutions are given implicitly by 4 ln | 2 | 3ln(| 1|) ln(| |)

1

2

v v x C

v

v

v) Back to the functiony, we get


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SK/EUM111/ODE/09 31

4 ln | 2 | 3ln(| |)

2

y x y x C

y x

y x

Note that the implicit equation can be rewritten as

3 4

1( ) ( )y x C y x

Example 4:2

yxy y

x

22

2 ( ) (A)

y y y y

y x x xx

Lety

vx

, y vx , 'y v x v

(A) 2'v x v v v 2'v x v

2

1 1dv dx

v x 1 ln | |x c

v

1ln | | yv x c x ln | |xy x c

Non-Homogeneous DE Reducible to Homogeneous Form

Consider the non-homogeneous equation

1 1 1

2 2 2

a x b y cdy

dx a x b y c

(1)

where a1, b1,c1,... c2 are all constants.

Case1:

If

1 11 1

2 22 2

, . ., 0a ba b

i ea ba b

then the transformation (shift of origin)


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SK/EUM111/ODE/09 32

reduces the non-homogeneous equation (1) to the homogeneous equation of the form

1 1 1 1

2 1 2 1

a x b ydy

dx a x b y

(2)

here the unknown constants h,k are determined by solving the pair of equations

1 1 1

2 2 2

0

0

a h b k c

a h b k c

Now the equation (3) is homogenous in the new variables x1 and y1, it can be solve homogenous

equation method

.

Case 2:

If

1 11 1

2 22 2

, . ., 0a ba b

i ea ba b

then the transformation

reduces the non-homogeneous equation (1) in the variables x and z which can be solved using

homogenous equation method

Example 1: Solve the non-homogeneous equation10 2 2

3 9

dy x y

dx x y

1 1,x x h y y h

1 1 1z a x b y


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SK/EUM111/ODE/09 33

Example 2: Solve the non-homogeneous equation7 3 7

7 3 3

dy x y

dx y x


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SK/EUM111/ODE/09 34

Bernoulli Equations as First Order Linear Non-homogeneous differential equation.

The Bernoulli differential equation has the form

d y

d xP x y Q x y

n ( ) ( )

n=0 ( ) ( )y P x y Q x linear

n=1 ( ) ( )y P x y Q x y separable

( ( ) ( )) ( )

1 1( ) ( )

y Q x P x y R x y

dyR x dy R x dx

y dx y

If n 1 then Bernoulli equations can be transformed to linear equations by changing the dependent

variablen1

yv

.

Proof:

ny)x(Qy)x(Pdx

dy


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SK/EUM111/ODE/09 35

(y-n

) multiply => )x(Qy)x(Pdx

dyy n1n

letn1

yv

dxdyyn1

dxdv n => )x(Qv)x(P

dxdy

dxdv

dydx

)n1(1

dxdv

dy

dx

n1

1y n

(1-n) multiply => )x(Q)n1(v)x(Pn1dx

dv

let )x(P)n1()x(P1 => )x(Qv)x(Pdx

dv11

which is linear in v!

)x(Q)n1()x(Q1

Example : y3xy4dx

dy1x 2

change to Bernoulli y1x

3y

1x

x4

dx

dy22

where1x

x4)x(P

2 &

1x

3)x(Q

2

since yn

= y1

y-1

multiply =>1x

3

1x

x4

dx

dyy

22

1

Let2)1(1n1

yyyv solving for 2/1vy

dx

dyy2

dx

dv solving for

dx

dv

y2

1

dx

dy

1x

3

1x

x4

dx

dvv

2

1v

22

2/12/1

1x

x43

dx

dv

v2

12

separate and integrate

dx1x

x43dv

v2

12


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SK/EUM111/ODE/09 36

ylnvlndvv1

2

1 2/1=

dx

1x

x432

= dx1x1

31x

x22

22

yln = xarctan31xln2 2

solving for y

3arctan

22 1

xe

yx

Solving Second Order Differential Equation by Reduction of Order

Let us consider a second-order differential equation in the form

d y

d xF y y

2

2 ,

where F y y, is a continuous function. To solve the differential equation of this type, we can at leastreduce its order by one

Let pd y

d x , then

d y

d x

d p

d x

d p

d y

d y

d xpd p

d y

2

2

Therefore, the original differential equation Eq.(6.6) is reduced to a first-order differential equation

pyFdy

dpp ,

The above method is known as the reduction of order.

Example

Solve the differential equationsd y

dxy

dy

dx

2

2 by reduction of order:


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SK/EUM111/ODE/09 37

Solution

Let pd y

d x , we have

dy

dpp

dx

yd

2

2

and the ODE becomes

pd pd y

y p d p ydy d p y dy

p yC d y

d xy C

d y

C yd x

C

y

Cx C

y c c x c

1

2 2

1

2

1

2

1 1

2

2

22

2 2

2 2

1

1 1 2

tan '

tan

Riccati Equation

2( ) ( ) ( )y P x y Q x y R x

( ) 0P x linear

Let S(x) be a solution and let ( ) 1/ y S x z

The Riccati equation is transformed into linear

Example: 21 1 2 (A)y y yx x x

( ) 1 / P x x , ( ) 1 / Q x x , ( ) 2 / R x x

By inspection, ( ) 1y S x is a solution of (A)

Let1 1

( ) 1y S xz z

2

1y z

z

2

21 1 1 1 1 2(A) (1 ) (1 )

1 1 1 2(1 )(1 1)

1 1 1 2(1 )(2 )

zz x z x z x

x z z x

x z z x


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SK/EUM111/ODE/09 38

2 2

2 2

2

2 2

1 1 2(1 )(2 )

3 1 2(2 )

2 3 1 2 3 1

z zz

x z z x

z z

x z z x

z z z zx x x x x x

3 1z zx x

(linear)

Integrating factor

3( )

3dx

xe x

3 2 3 23 ( )x z x z x z x

Integrate 3 313

x z x c and3

1

3

cz

x

Solution:3

3 3

1 1 3 21 1

1/ 3 / 3

c xy

z c x c x


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SK/EUM111/ODE/09 39

Application Problems

Exponential decay or increaseFor radioactive decay, the decay rate is proportional to the number of the mother nuclei left. For

example, if the decay rate constant is and the initial no. of mother nuclei is N0,then we have,

dN

dtN and the initial condition is N(0)=N0.

( )t

dNN

dt

N t Ce

Initial condition implies C=N0 and thus N(t)=N0t

e

Note : Although the above two examples are very simple, it should be pointed this is not the usualcase. The remained sections in this chapter will try to illustrate some methods to solve some of the

standard form of DE.

Another example:


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SK/EUM111/ODE/09 40

Solution of this problem is:

Population Dynamics

The easiest mathematical model offered to govern the population dynamics of a certain species is

called the exponential model, that is, the rate of change of the population is proportional to the

existing population. In other words, ifP(t) measures the population, we have

dPkP

dt ,

where the rate kis constant. It is fairly easy to see that ifk> 0, we have growth, and ifk


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SK/EUM111/ODE/09 41

P(t)=P0kt

e

,

where P0 is the initial population, i.e. P(0)=P0. Therefore, we conclude the following:

ifk>0, then the population grows and continues to expand to infinity, that is, lim P t

x

ifk0, is not adequate and the model can be dropped. The main argument for this

has to do with environmental limitations. The complication is that population growth is eventually

limited by some factor, usually one from among many essential resources. When a population is far

from its limits of growth it can grow exponentially. However, when nearing its limits the populationsize can fluctuate, even chaotically. Another model was proposed to remedy this flaw in the

exponential model. It is called the logistic model (also called Verhulst-Pearl model). The differential

equation for this model is

(1 )dP P

kPdt M

,

whereMis a limiting size for the population (also called the carrying capacity). Clearly, when P issmall compared toM, the equation reduces to the exponential one. In order to solve this equation we

recognize a nonlinear equation which is separable. The constant solutions are P=0 and P=M. The non-constant solutions may obtained by separating the variables

(1 )

dPkdt

PP

M

and integration

(1 )

dPkdt

PP M

The partial fraction techniques gives

11

(1 ) (1 )

dP M dPP PP

PM M


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which gives 1P

In P In kt cM

|

Easy algebraic manipulations give

(1 )

ktP CeP

M

where Cis a constant. Solving for P, we get

If we consider the initial condition P(0)=P0 (assuming that P0 is not equal to both 0 orM), we get

,

which, once substituted into the expression for P(t) and simplified, we find

It is easy to see that

lim P tx

However, this is still not satisfactory because this model does not tell us when a population is facing

extinction since it never implies that. Even starting with a small population it will always tend to thecarrying capacityM.

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1Lecture Student Notes ODE 2010

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