ODE Lecture 2

7/25/2019 ODE Lecture 2

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11



2



3



4



5

Solve the differential equation

211

ydt dy

Example



6

The differential equation is

Just separate the variables and integrate

2

11

ydt

dy

dt y

dy2/11

Solution



7

C t y y )(tan 1

)(tan/1

12 y y

y1

dy

since

1

11

1/11

1

22

2

2 y y

y

yThus,



8


Solution:

Since

Therefore we can write as

2 2 2 21dy t y t y

dt

2 2 2 2 2 2

1 (1 )(1 )t y t y t y

)1)(1( 22 yt dt dy

Example



9

dt t

y

dy)1(

1

2

2

dt t

y

dy)1(

1

2

2

C t

t y3

)(tan3

1

3

tan3

t y t C

This can be written as



10

Solve

1 0 x dy ydx

Example



11

The equation

can be written as

Separate the variables and integrate

)(1

1

1 y

x x

y

dx

dy

1 0 x dy ydx

x

dx

y

dy

1

Solution



12

(1 ) y A x

1C

y e x

ln ln 1 y x C or

or



13

Solve

02 324

dye ydx xy x

Example



14

The equation

can be written as

02 324

dye ydx xy x

4

2

3

2

dy y x xe

dx y

Solution



15

The equation

can be written as

02 324

dye ydx xy x

4

2

3

2

dy y x xe

dx y

Solution

23

4

2 x ydy xe dx

y



16

1 13 3 3 3 9

22 4 1 323

x x x xe dx xe e

y y dy y y

Taking the integrals



17

1

3

3

213

9

13

3

1c y y

xe

x xe

223

4

y x

xe dx dy y

gives us

3

9 63(3 1)

xe x c

y y



18

Definition

A function f defined on some interval I ,which when substituted into a differential

equation reduces the equation to an

identity, is said to be a solut ion of the

equation on the interval.

A solution does not contain derivatives.



19

Remarks

i) A solution of differential equationthat is identically zero on an

interval I is referred to as a trivial

solution.

ii) Not every differential equation that

we write necessarily has a solution.



20

Definition

A solution of ODE in which the dependent

variable is expressed solely in terms ofindependent variables and constant is said to

be an expl ic i t so lut ion .

A relation is said to be an impl ic i t so lut ion

of ODE on an interval I provided it defines

one or more explicit solutions on I .



21

Initial Value Problem

We are often interested in solving a DE

y x f dx

dy

,

subject to a (side) condition 00 y x y

where

0 x is a number in an interval I and

0 y is an arbitrary real number.



22

00:Subject to

,Solve

y x y

y x f dxdy

is called an in i t ial-value p rob lem (IVP). The

side condition is known as initial condition.

The problem



23

00 , y x

I

Solutions of

ODE

Geometrically,

00

, y x

we are seeking at least one solution of theODE defined on some interval I such that

the graph of the solution passes the given

point .



24

on , y y

0 3 at 0, 3 y x y

Example

x

ce y As is its solution

0

3 3ce cSubstituting in solution

Now

Consider IVP

subject to 0 3 y

3 x

y e is the solution of IVP



25

y y

13 3ce c e

So would be the solution.1

3 xe y

Alternatively, If we say that a solution of

passes through the point (1,3)

rather than (0,3),

then y(1) = 3 would yield



26

Theorem

Let R be a rectangular region in the xy-plane

defined by a ≤ x ≤ b, c ≤ y ≤ d that contains

the point ( x0, y0) in its interior. If f ( x,y) and

∂ f/∂y are continuous on R, then there exist

some interval I 0:(x0-h, x0+h), h>0, contained

in [a,b], and a unique function y( x) defined on I 0, that is the solution of the IVP

Existence of a Unique Solution



27

Example

, 0 3 y y y

xe y 3

1,, y

f y y x f

,

For

is the only solution.

As are continuous

throughout the entire xy -plane .



28

Example

22 y xdx

dy

y

y

f y x y x f 2,, 22

For

are continuous throughout the xy - plane.

Therefore through any point ( x0, y0) therepasses one and only one solution of the IVP.



29

),( y x f dx

dy

)()( y g xhdx

dy

The differential equation of the form

is called separable if it can be written in the

form

Variable Separable



30

Step 1: We solve the equation tofind the constant solutions of the equation.

Step 2: For non-constant solutions we write

the equation in the form.

dx xh y g

dy

)()(

0)( y g

Method Of Solution



31

C x H yG )()(

to obtain a solution of the form

Step3: We list the entire constant and the

non-constant solutions of the equation.

Step 4: For an IVP, we use the initial

condition to find the particular solution.

1

( )( ) dy h x dx g y

then integrate

E l



32


Solution:

1. No constant solutions as the equation

has no real roots.

2. For non-constant solutions, we separatethe variables and integrate

211

ydt dy

01

12 y

dt

y

dy2

/11

Example



33

3. Since there exist no constant solution, allsolutions are given by the implicit equation

found in step 2.

C t y y )(tan 1

)(tan/1

1

2 y y

y1

dy

since

1

11

1/11

122

2

2 y y

y

yThus,

E l



34

Solve the initial value problem

Solution:

Since

Therefore

1. No constant solution because there exist noreal roots of the equation

10 ,1 2222 ) y( yt yt dt dy

2 2 2 2 2 21 (1 )(1 )t y t y t y

)1)(1( 22 yt dt

dy

21 0. y

Example



35

dt t y

dy )1(1

2

2

dt t y

dy

)1(1

2

2

2. For non-constant solutions we separate the

variables and integrate.

C t

t y

3

)(tan3

1

3

tan

3

t y t C

This can be written as



36

3. Since there exist no constant solutions, all

solutions are given by the implicit or explicit

equation.4. Using initial condition y(0) = 1, we obtain

The solution to the initial value problem is4

)1(tan 1

C

43)(tan

31 t

t y

43

tan3

t t y

E l



37

Solve

Solution: The equation can be written as

1. The only constant solution is y = 0.

2. For non-constant solution we separate thevariables and integrate

)(1

1

1 y x x

y

dx

dy

01 ydxdy x

x

dx

y

dy

1

Example



38

0

1

y

xc y

1

1 ,

c

y c x C e

3. All the solutions to the given equation are

1ln ln 1 y x cor

or

Example



39

Solve

Solution:The given DE can be written as

Since . Therefore, the only

constant solution is .

02324

dye ydx xy x

2 3

2

4

y

y x xedx

dy

0

22

4

y

y

y

0 y

Example



40

1)0( ,12

y ydx

dy

01.1)0( ,12

y ydx

dy

Example

Solve the initial value problems

(a)

(b)

and compare the solutions to observe: how a

very small change in the data changes the

solution.

Solution



41

12

dxdy y c x y 1

1

2. For non-constant solutions, we separate the

variables and integrate

or

3. All the solutions of the equation are

Solution

1, 1

1 x c y

y

1.Since .

Therefore, the only constant solution is y = 1

2( 1) 0 1 y y

4 A l i diti t fi d ti l



42

4. Applying conditions to find particularsolutions of both the problems

(a) For

Thus

So that the solution of problem (a) is y1 = 1,

which is same as constant solution

0 1 1 when 0. y y x

1 10

1 1 0c c c

11

1 y y

and



43

(b) For0 when01.101.10 x y y

1000101.1

1cc

x y x

y 100

11100

1

1

2

11

100 y

x

Thus

and

So the solution of the problem (b) is



44

5. Comparison:We can see

as

Therefore, there can be a radical change in thesolution corresponding to a very small change

in the initial conditions.

211

100 y

x

100 x

R k



45

Remark An equation of the form

,0, bcbyax f dx

dy

can be reduced to a separable equation by

means of the substitution .

u ax by c



46

Example

Solve2

2 y xdx

dy

Solution: substitute

2

1

u x y

du dy

dx dx

Therefore differential equation becomes



47

2

2

1

1

1

tan

tan( )

2 tan( )

duu

dx

dudx

u

u x c

u x c

x y x c

which is the required solution

Integrating



4848

Differential Equations with

Boundary Value Problems

by

Dennis G. Zill

7th Edition



4949

A function f is said to be

homogenous if for some real

number n it satisfies the following

property

y x f t tytx f n,,



5050

Example

y x f t

y xy xt

yt xyt xt tytx f

y xy x y x f

,

53

53,

53,

2

222

22222

22

This function is homogeneous

of degree 2



5151

Example

y x f t

yt xt tytx f

y x y x f

,

,

,

3/2

3 2222

3 22

Homogeneous of degree 2/3



5252

y x f t

ty

txtytx f

y

x

y x f

,

42

,

42,

0

Example

Homogeneous of degree 0



5353

Remarks:

i) Homogeneous functions can berecognized by examining the total

degree of each term.

e.g.

Not homogeneous

,, 2

y x y x f



5454

ii) If f is homogeneous function of degree

n then we can write

1,,and

,1,

y

x f y y x f

x

y f x y x f

n

n

1,,1 y x f and

x y f Where are

homogeneous of degree 0.



5555

Example

x

y f x

x

y

x

y x

y xy x y x f

,1

31

3,

2

2

22

22

f(1,y/x) is homogeneous of degree 0.



5656

A homogeneous differential equation

can be solved by means of an algebraic

manipulation specifically, either substitution

, where u and v

are dependent variables, will reduce the

equation to a separable first order ODE.

0,, dy y x N dx y x M

vy xux y or

Example:



5757

Solve:

0222 dy xy xdx y x

Solution:

xduudxdyux y

Example:

Both M and N are homogeneous of

degree two. If we let



5858

2 2 2 2 20 x u x dx x ux udx xdu



5959

2 2 2 2 2

2 2 2 2 2 2 3 3

0

0

x u x dx x ux udx xdu

x u x dx x u u x dx x ux du

2 2 2 2 2



6060

2 2 2 2 2

2 2 2 2 2 2 3 3

2 2 3 3

0

0

0



x x u dx x ux du

2 2 2 2 2



6161

2 2 2 2 2

2 2 2 2 2 2 3 3

2 2 3 3

2 3

0

0

0

1 1 0

1 1 0



x x u dx x ux du

x u dx x u du

u dx x u du

Which is a separable equation

d1



6262

C x x y

x y

x yuux y

C xuu

x

dxdu

u

x

dxdu

u

u

lnln1ln2

,/

lnln1ln2

01

21

01

1

so

as



6363

Using the logarithmic properties

x

y

Cx

y x 2

ln

On simplification we get

x

y

Cxe y x2

is the solution

E l



6464

Coefficients are homogeneous of degreetwo, we could choose

but the relative simplicity of the term

suggests that we put

022 dy y x xydx

vx y

ydvvdydxvy x

Example:

Solve

Solution:



6565

2 2 2 2

3 2 2 2 2 2

3 2 2 2

0

0

1 0

vy vdy ydv v y y dy

vy dv v y y v y dy

vy dv y v v dy



6666

2 2 2 2

3 2 2 2 2 2

3 2 2 2

2

2

2

0

0

1 0

2 1 0

1 40

4 2 1

1ln 2 1 ln ln

4

vy vdy ydv v y y dy

vy dv v y y v y dy

vy dv y v v dy

vydv v

vdv dy

v y

v y C



6767

C y x y

C v y

C yv

222

24

2

2

or

12

lnln412ln

Therefore ,

Which is the desired result

General Solution



6868

0,1,

,1

0,1,1,1

0,1,1

0,,

0,,

uuN ui M duu N

xdx

duu xN dxuuN u M

xduudxu xdxu M x

xduudxux x N dxux x M

xduudxdyux y

dy y x N dx y x M

nn

putso

Separating the variables

ODE Lecture 2

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