Chap_3_ODE

UECM1653 Mathematics for Engineering I

UECM1683 Mathematics for Physics I Jan 2015

1

Topic 3: Ordinary Differential Equation

3.1 Ordinary differential equations An ordinary differential equation (ODE) is an equation involving an unknown function, y

with independent variable x say, ands its derivatives.

For example:

)(2)( xxyxy -- (1)

And

xxydx

dy

dx

yd2cos72sin432

2

2

-- (2)

are ordinary differential equations. The solution of an ODE is a function that satisfies the

equation at every point of its domain. For example, )exp()( 2xxy is a solution of (1), while

xxxy 2cos)exp()( is a solution of (2).

Other notations for OED are possible, such as an ODE for x as a function of t:

07532

2

xdt

dx

dt

xd -- (3)

Often derivatives with respect to t are denoted by dots, eg 2

2

,dt

xdx

dt

dxx , etc , so that (3)

may be also be written as

0753 xxx

Hence, the following ODE are equivalent:

034 yyy

0342

2

xdt

dx

dt

xd

034 xxx

ODEs arise quite naturally in a wide variety of physical situations, as described below.

Example:

Newtons law of cooling It states that the rate of decrease of the temperature of a body is proportional to the difference

between the temperature T of the body and the temperature T0 of the surrounding air, i.e.

0TTdt

dT

Or equivalently,

)( 0TTkdt

dT

for some constant k.



2

Example:

Motion of a mass on a spring in a resistive medium

x = 0 x = X (t)

Let X(t) denote the displacement of the mass at time t. The resistive force is given by

dt

dXkk 11 )speed( -- (4)

and the spring restoring force by

)()extension( 22 lXkk -- (5)

where k1 and k2 are constants and l is the natural length of the spring. By Newton's Second

Law, the total force F acting on the body must equal the acceleration a of the body times its

mass m, i.e. F = ma. Hence

2

2

21 )(dt

XdmlXk

dt

dXk -- (6)

lkXkdt

dXk

dt

Xdm 2212

2

-- (7)

Which is an ODE for the displacement X(t)

3.2 Classification of ODEs

ODEs may be separated into different categories, according to their general characteristics.

Independent and dependent variables

In equations (1) and (2) y is called the dependent variable and x is the independent variable.

In equation (3) x is called the dependent variable and t is the independent variable.

Order

The order of an ODE is the highest derivative occurring in the equation.

For example,

01062

2

ydx

dy

dx

yd

Is second-order

043 3 xydx

dy

Is first order



3

Degree The degree of an ODE is the power to which the highest-order derivative is raised, after the

equation has been rationalised to contain only integer powers of derivatives. Hence

0

72

4

4

y

dx

dyx

dx

yd

Is fourth-order and second degree

04

7

2

2

y

dx

dy

dx

yd

Is second order and degree 4, since in order to remove fractional powers, we rearrange as

follows 4

2

27

2

247

dx

ydy

dx

dy

dx

ydy

dx

dy

Hence the highest derivative is raised to the power of 4.

Linearity

An ODE is linear, if

(a) the y-dependent terms are y itself and derivatives of y and

(b) these terms do not appear multiplied together. ODEs containing products of y

dependent terms, or functions of y, are said to be nonlinear.

xeydx

dy

dx

yd

45

3

2

2

Is second order and nonlinear, whilst

06102

2

ydx

dy

dx

yd

Is second order and linear

03 xdx

dyy 0 ye

dx

dy

Are first order and nonlinear, whilst

xeyxdx

dy 434

Is first order and linear.

Notice that the coefficients in the ODE might depend upon the independent variable (x in this

case).



4

3.3 Homogeneous and inhomogeneous ODEs

Suppose we are given a linear differential equation. Let us assume that terms containing the

dependent variable are on the left-hand side of the equality sign and terms involving the

independent variable and constants on the right-hand side.

Then homogeneous equations have the right-hand side zero and inhomogeneous equations

have the right-hand side non-zero.

For example,

02

2

ydx

dyx

dx

yd

Is a homogeneous, linear second order ODE and

xydx

dyx

dx

ydcos2

2

2

is an inhomogeneous, linear, second-order ODE.

3.4 Solving ODEs

The general solution of an ODE is the most general function y that satisfies the equation; it

will contain constants of integration that may be determined by the application of suitable

initial or boundary conditions. The general solution of an nth-order ODE will contain n

arbitrary constants of integration and we therefore need n initial conditions in order to

determine these constants. When the initial conditions have been applied and the constants

determined then we are left with a particular solution to the ODE which obeys the given

initial conditions. When a possible solution to the ODE has been found it is always possible

to check its validity by substituting it into the original ODE and checking that it satisfies the

initial conditions.

The ODE may possess an analytical solution where the dependent variable can be expressed

as a function of the independent variable. For example,

xdx

dy4

has an analytical solution

cxxy 22)(

where c is a constant of integration. Here the dependent variable y has been expressed as a

function of the independent variable x.



5

Alternatively, it might be possible to use graphical methods. For example, suppose we

consider

xdx

dy

2

1

We can compute dx

dy at any point in the xy plane, as follows. Suppose we consider a grid of

points in the xy plane. At each point, draw a short line segment with gradientdx

dy. This is the

direction field which consists of line-segments that are tangent to the solution curves. Thus

the solution curves may be inferred

Direction fields of the ODEs 2/xy with solution curve

3.5 First order ODEs

These are usually written in the form

),( yxfdx

dy

Where f is a given function. Sometimes they are written in the equivalent form

),()( yxfxy



6

Separable first order ODEs

In this case, the function f separates into the product of a function of x and a function of y.

Then we can write

)()(),( yhxgyxf

So that the equation becomes

)()(),( yhxgyxfdx

dy

We can then rearrange this into the form

)()(

1xg

dx

dy

yh

Integrating with respect to x we get

dxxgdyyh

)()(

1+ c

remember to include the arbitrary constant of integration c

Example:

Solve the equation

kydx

dy

We have

kxcckx eeey

ckxy

cdxkdyy

kdx

dy

y

||

||ln

1

1

Hence, kxkxc Aeeey

Where ceA is an arbitrary constant.

Example:

Solve the differential equation

y

x

dx

dy

Solution:

Axy

Cxy

Cxdxydy

22

22

22



7

Example:


xydx

dy 2/1

Solution:

Cx

y

Cxy

Cx

dyy

Cxdxdyy

22

22/1

2

1

2

22

1

2

2

1

2/1

Example:


yxedx

dyx 2)1( 2

Solution:

Cxe

Cdxx

xdye

Cdxx

xdy

e

y

y

y

|1|ln

)1(

2

)1(

21

2

2

2

Example:


yey

x

dx

dy

1

2

Solution:

Cxeye

Cxdyey

Cxdxdye

y

ey

x

dx

dy

yy

y

y

y

2

2

)1(

)1(

21

1

2



8

Linear first order ODEs

Recalling our definition of a linear differential equation, we see that for

),( yxfdx

dy

To be linear we must have

)()(),( xqyxpyxf

for given functions p and q. Then the equation becomes

)()( xqyxpdx

dy -- (8)

No `y'-terms are multiplied together and there are no functions of y. If a linear first-order

equation is written as in (8) it is said to be in standard form. Notice that if q(x) = 0 the

equation is also separable. For then

0)( yxpdx

dy yxp

dx

dy)(

Hence,

cdxxpdyy)(

1

Where c is a constant of integration, so that

cdxxpy )(||ln

Taking exponentials, dxxpAey )(

Where ceA is an arbitrary constant.

Integrating factors

dxxpdx

dye

dx

dyee

dx

dy

dx

dyeye

dx

d dxxpdxxpdxxpdxxpdxxp)(

)()()()()(

And also

)())(()()( dxxpdxxp

yedx

dyxp

dx

dye -- (9)

and integrating the RHS then just gives dxxp

ye)(



9

Theory Hence, to solve the linear ODE in standard form

)()( xqyxpdx

dy

We multiply by the integrating factors dxxp

e)(

dxxp

e)(

yxp

dx

dy)(

dxxp

e)(

)(xq

From (9), we then have

dxxpdxxp exqyedx

d )()()(

Now integrate both sides with respect to x:

cexqyedxxpdxxp

)()(

)(

Final result:

dxxpdxxpdxxp

cedxexqey)()()(

)(

If, q(x) = 0, then we get the same result as before.

Example:

Solve 332 xydx

dyx

Solution:

232

xyxdx

dy )()( xqyxp

dx

dy

Integrating factor 2ln22

xee xdx

x

Cxyx

dxxyxdx

d

xxxyxdx

d

52

42

4222

5

3

3

33



10

Example: (Bernoulli differential equation)

Solve )cos(432 txdt

dxtxt

Solution:

3

4 )cos(1

t

txx

tdt

dx ayxQyxP

dt

dy)()(

Let 31 )()()( txtxty a

Therefore, dt

dxx

dt

dy 43 and dt

dyx

dt

dx

3

4

Then 3

34

3

4 cos33cos1

3 t

tx

tdt

dyx

t

tx

tdt

dyx

3

cos33

t

ty

tdt

dy

The integrating factor 3ln33

tee tdt

t

Ctyt

Cdttytdt

d

tt

ttyt

dt

d

sin3

cos3

cos3cos3

3

3

3

33

But 31 )()()( txtxty a or 3

1

xy

So

3/13

3

3

3

sin3

sin31

sin3

Ct

tx

Cttx

Ctyt



11

Example:

Solve 222

yxyx

y .

Solution:

2 1 2

2

2

12

2ln 2

2 2 2

2 4

52

2 5

2

2

1Let

2

solve this differential equation by First order linear equation

Integrating factor is

2

5

5

5

dxxx

y y y xx

wy

yw

y

w w xx

e e x

x w w x xx

dx w dx x dx C

dx

xx w C

x xC

y

xy

x

5 A



12

Example:

Solve 234

yxyx

y .

Solution:

||ln1

||ln1

||ln1

11

141

1 isfactor gIntegratin

equationlinear order First by equation aldifferenti thissolve

4

1Let

4

4

4

4

4

3

44

4

ln4

14

3

2

312

xCxy

xCxy

xCwx

Cdxx

dxwxdx

d

xx

wx

wx

xee

xwx

w

y

yw

yw

xyx

yy

xdx

x



13

Example:

Solve 4

3

1yey

dx

dy x .

Solution:

x

x

x

eww

eww

yy

w

yw

eydx

dy

y

3

3

1

3

1

3

1Let

3

11

4

3

3

4

xdx ee

1

isfactor gIntegratin


)3(1

31

3

3

3

3

3

3

xCey

Cxy

e

Cxwe

Cdxdxwedx

d

wedx

d

eewwe

x

x

x

x

x

xxx



14

Example:

Solve .3xyydx

dyx

Solution:

22

11

2

1

2

1Let

1111

3

2

23

3

wx

w

wx

w

yy

w

yw

xydx

dy

y

yx

y

dx

dy

2

12 1

isfactor gIntegratin


xe

dxx

2

2

2

2

2

2

22

22

22

2

1

21

2

21

21

21

2121

Cxxy

Cxxy

Cxxw

Cx

wx

Cdxx

wx

xw

xdx

d

xw

xw

x



15

Example:

Solve 2ByAydx

dy , where A and B is constant.

Solution: 2ByAyy

a = 2, therefore let y

yyu a1211

and yy

u 2

1 yuy 2

then 2ByAyy 122 AyBuByAyuy

so BAuuAuBu

so the general solution is A

BCeu At

A

BCe

uy

At

11

Exercise:

Solve xeydx

dy 2

Solution:

xeydx

dy 2 )()( xqyxp

dx

dy

Integrating factor xdx

ee 22

xx

xx

xx

xxxx

Ceey

Ceye

dxeyedx

d

eeeyedx

d

2

2

2

22



16

Euler homogeneous equation

Some functions have the property that, for any

),(),( yxfyxf n

Such functions are called homogeneous of degree n. For example,

i. ),()(),(),( yxfxyxyyxfxyyxf

ii. 2 2 2 2 2 2 2 2 2 2( , ) ( , ) ( ) ( , )g x y y x g x y y x y x g x y

In the above, f is homogeneous of degree one, g is homogeneous of degree two.

A homogeneous differential equation is one that may be written in the form

),(

),(

yxB

yxA

dx

dy

Where A(x,y) and B(x,y) are homogeneous functions of the same degree. The RHS of a

homogeneous ODE can be written as a function of y/x, because

)/()/,1(

)/,1(

),(

),(xyF

xyB

xyA

yxB

yxA

dx

dy

The homogeneous equation, rewritten in the form

x

yF

dx

dy

May then be solved by making the substitution y = v x so that,

vFdx

dvxv

dx

dy or vvF

dx

dvx

This equation is now separable and can be solved as follows

xdx

vvF

dv

)( -- (11)

Hence, the solution method is first to check whether the equation is homogeneous. If so,

make the substitution y = vx, then separate variables as in (11) and solve directly. Finally

replace v by y/x to obtain the solution.



17

Example:

Solve

xy

xy

dx

dy

Homogeneous equation with degree one,

Let ),(),(),( yxfyxfxyyxf

),(),(),( yxgyxgxyyxg

Since f and g is homogeneous degree one, so substitute y = vx

dx

dvxv

dx

dy

Then we have

xy

xy

dx

dy

1

1

1

1 2

v

vv

v

v

dx

dvx

This equation is separable and we get

Cx

yyx

Cxvv

dxx

dvv

v

)(tan||ln2

1

||ln)(tan|1|ln2

1

1

1

1

122

12

2

Example:

Solve

x

y

x

y

dx

dytan

This equation is homogeneous of degree zero, let y = vx to get

vvdx

dvxv tan

Hence,

Axxy

Cxv

Cdxx

dvv

1sin

||ln|sin|ln

1cot



18

3.6 Initial conditions

We have seen that the general solution of a first-order ODE involves one unknown constant.

Generally, the number of unknown constants in the general solution equals the order of the

ODE. To determine a particular solution, we must find these constants. We therefore need as

many conditions on the solution as there are unknown constants. These conditions usually

involve the value of the unknown function and/or some of its derivatives at certain points. For

a first-order equation we only need one such condition, called initial condition. The

terminology stems from the fact that one may think of the independent variable in the ODE as

representing time t. Specifying the value of the solution at a given time, for example at t = 0,

then amounts to fixing the initial value of the dependent variable.

Example:

Solve

xyx

x

dx

dy

21

With initial condition y (1) = 0

Solution:

Integrating factor 2/12)( 1 xe dxxp

2

2

2/322

22

22

11

3

1

13

11

11

11

x

Cxy

Cxxy

dxxxxydx

d

xxxydx

d

Given y = 0 when x = 1 0)1( y

22

11 1 0

3 1 1

2 2

3

C

C

2

2

1

1

3

221

3

1

xxy



19

Example:

Suppose that a variable current I(t), where t is time in seconds, flows through a coil with

inductance L and a resistor of resistance R with an applied voltage V , where we assume that

L, R and V are constants. Suppose also that the current is zero at time t = 0. Then the current

satisfies the ODE

VRIdt

dIL

Subject to the initial condition

I(0) = 0

Solution:

The general solution is t

L

R

ceR

VtI

)( and given I(0) = 0, so we get C = R

V

Therefore the particular solution is

tL

R

eR

VtI 1)(

Exercise:

The rate of increase of the concentration of a chemical is proportional to the concentration at

that time. Experiment shows that the concentration doubles in four hours. If the initial

concentration is c0, what is the concentration after six hours?



20

3.7 Second order ODEs

Homogeneous We start by considering second-order ODEs of the general form

02

2

cydx

dyb

dx

yda

where a, b and c are given constants.

These are called constant coefficient, homogeneous, linear second-order ODEs. Recall that

homogeneous here refers to the right-hand side being zero; we will also consider the case

when there is a non-zero function there, i.e. inhomogeneous equations. Other notation is

possible; for example, the ODE might be in terms of x(t) and its derivatives with respect to t.

Suppose we start by considering the case where a = 0: Then

0 cydx

dyb

and we know how to solve this (by separating the variables). The solution is

xb

c

Aey

Alternatively, if we try y = Aemx

as a solution

0 cydx

dyb

0 ( ) 0mx mx mxbmAe cAe Ae bm c

Hence,

b

cm

And we have got the same answer as before.

Suppose we adopt the same approach for the second-order equation:

If y = Aemx

then mxmAedx

dy and mxAem

dx

yd 22

2

Substituting into the ODEs, we get

0)(

0

2

2

cbmamAe

cAebmAeAeam

mx

mxmxmx

00 2 cbmamAemx -- (12)

So m satisfies a quadratic equation called the auxiliary equation.

It turns out that if the auxiliary equation has two distinct roots, say m1 and m2, then the

general solution is

xmxm

BeAey 21

Where A and B are arbitrary constants.



21

Example:

We start with an example. Suppose we wish to solve

ydx

yd 22

2

We first check that xy sin is a solution:

yxyxy 22 sin,cos

Next, we check that xy cos is a solution:

yxyxy 22 cos,sin

We can show that xy cos2 is a solution:

yxyxy 22 cos2,sin2

More generally, it is possible to show that the general solution is

xBxAxy cossin)(

Where A and B are constants.

Example:

Solve

022

2

ydx

dy

dx

yd

Solution:

Try mxAey then we have 0)2( 2 mmAemx

Solve the equation m = 2 , m = -1

The general solution for y is then

xx BeAey 2



22

Example

Solve

0342

2

ydx

dy

dx

yd

Solution:


Solve the equation m = 3 , m = 1

The general solution for y is then xx BeAey 3

Since as we've seen in (12) m1 and m2 are the roots of a quadratic equation with real

coefficients, three different cases are possible:

- m1 and m2 are real and distinct (as in the previous two examples) - m1 and m2 are real and repeated (i.e. m1 = m2) - m1 and m2 are complex conjugates

We now consider the second and third cases, the first having already been dealt with.

Real and repeated roots

If m1 = m2, the general solution appears to be xmxmxmxm

CeBAeBeAey 1111 )(

The solution now only contains one constant of integration, but we know that it should

contain two. In this case, the solution is xm

eBxAy 1)(

Where A and B are arbitrary.

Example:

Solve

096 yyy


Solve the equation m = 3 , m = 3 (repeat)


xeBxAy 3



23

Complex conjugate roots

Suppose the roots are

im 1 im 2

The general solution for y is now xixi BeAey )()( xixxix eBeeAe

xixix BeAee xDxCe x sincos

Where C and D are arbitrary constants (they are, in fact, combinations of A and B)

This is the general solution in this case.

Example:

Solve

04 yy

Solution:

Try mxAey then we have 0)4( 2 mAemx

Solve the equation m = 2i , m = -2i (repeat)

0 and 2


xDxCy 2sin2cos

3.8 Inhomogeneous second order ODEs

Here we consider second-order ODEs of the general form

)(2

2

xfcydx

dyb

dx

yda

The general solution consists of the complementary function + the particular integral.

The complementary function is the solution of the homogeneous equation

02

2

cydx

dyb

dx

yda



24

And can be found using the methods of previous section. Here we will discuss how to find

the particular integral. There is no general rule for finding the particular integral. In certain

cases it is possible to find a particular integral by first guessing the general form of the

particular integral in terms of a function containing a number of as yet undetermined

constants, and then adjusting the constants so as to turn the function into a solution. This

technique is known as the method of trial functions and we will illustrate it by a number of

examples.

Example:

Find the general solution of

xydx

dy

dx

yd 2

2

2

Solution:

The complementary function is the solution of

022

2

ydx

dy

dx

yd

Which we know from above is xx BeAey 2

To find the particular solution, we consider the trial function

DCxy

Then direct substitute into the differential equation and get

xDCxC )(2

Equating the coefficients of the various power of x on the left and right sides give

02 DC 12 C Which we have

2

1C and

4

1D

So that the particular integral is

4

1

2

1)( xxy

Thus, the general solution is

pc yyxy )(

)(xy xx BeAe 2

4

1

2

1 x



25

Example:


xydx

dy

dx

yd 42

2

495

Solution:


xydx

dy

dx

yd 42

2

495

Which we know from above is xx

c BeAey615615

2

1

2

1


xCey 4


xxxx eCeCeCe 4444 92016


27

1C


x

p exy4

27

1)(


pc yyxy )(

)(xy xx

BeAe615615

2

1

2

1 xe4

27

1



26

Example:


xydx

dy

dx

yd2cos95

2

2

Solution:


xydx

dy

dx

yd2cos95

2

2

Which we know from above is xx

c BeAey615615

2

1

2

1


xDxCy 2sin2cos


4 cos2 4 sin 2 5( 2 sin 2 2 cos2 ) 9( cos2 sin 2 ) cos2C x D x C x D x C x D x x


269

13C and

269

10D


xxxy p 2cos132sin10269

1)(


pc yyxy )(

)(xy xx

BeAe615615

2

1

2

1 xx 2cos132sin10

269

1

In summary, trial function for particular integrals of equations of the form

)(2

2

xfcydx

dyb

dx

yda

Are as follows

If f (x) is a polynomial in x of degree p, the trial function should be a polynomial in x of

degree p.

e.g. if f (x) = 6x2 + 4x + 3 try Cx

2 + Dx + E.



27

If f (x) is an exponential function of x, the trial function should be an exponential function of

x.

e.g. if f (x) = 6e5x

try Ce5x

If f (x) consists of the sine or cosine of a multiple of x, the trial function should be a linear

combination of sine and cosine of the same multiple of x.

e.g. if f (x) = cos 5x try Csin 5x + Dcos 5x.

(In all the above, C, D an E are arbitrary constants)

3.9 Initial conditions second order ODEs

We have seen that the general solution for a second-order ODE contains two unknown

constants. To determine these constants we need two initial conditions. Note the initial

conditions must only be applied once the general solution (complementary function plus

particular integral) has been determined.

Example:

Solve

xydx

dy

dx

yd 2

2

2

Subject to the initial condition y(0) = 0 and 0)0( y

Solution:

The general solution is

4

1

2

12 xBeAey xx

To determine A and B, we now apply the initial conditions

12

1

3

1

02

12)0(

04

1)0(

B

A

BAy

BAy

Therefore we get 4

1

2

1

12

1

3

1 2 xeey xx

Documents

Chap_3_ODE