njc differential equation lecture notes teachers edition

National Junior College Mathematics Department 2010

2010 / SH1 / H2 Maths / Differential Equations (Teacher’s Edition) 1

National Junior College

2010 H2 Mathematics (Senior High 1)

Differential Equations (Lecture Notes)

Differential Equations

Objectives:

At the end of this chapter, students will be able to:

� formulate a simple statement involving a rate of change as a differential equation.

� use direct integration to solve differential equations of the form

• d

f ( )d

yx

x= ,

•

2

2

df ( )

d

yx

x= .

� use the method of separable variables to solve differential equations of the form d

f ( )d

yy

x= .

� use a given substitution to reduce a first-order differential equation to the form d

f ( )d

yx

x= or

df ( )

d

yy

x= .

� formulate differential equations based on physical interpretation and modelling.

� understand that the general solution of a differential equation can be represented graphically

by a family of curves, and sketch typical members of the family using a graphing calculator.

� use an initial condition to find a particular solution to a differential equation, and interpret

the solution in the context of a problem modelled by the equation.

� comment on the appropriateness of the models used and the assumptions made.



§1 Introduction

1.1 Definition

A differential equation is an equation that states a relationship between an

independent variable x, a dependent variable y and at least one of the derivatives x

y

d

d,

2

2

d

d

x

y,

3

3

d

d

x

y etc. The order of a differential equation is the order of the highest

derivative that occurs in it. The degree of a differential equation is the degree of the

highest derivative that occurs in it.

Differential Equation (DE) Order of DE Degree of DE

1d

d 2+=

y

x

yx 1 1

0d

d

d

d3

2

2

=+

+ y

x

yx

x

y 2 1

03d

d2

2

2=+

+ y

x

yx 1 2

Note: We will only be dealing with first order and second order differential equations.

§2 Solutions of Differential Equations

A solution of a differential equation involving x and y is an equation expressing the

relationship between x and y but containing NO derivatives.

Note that the solution of an nth

order differential equation involves n arbitrary

constants. Therefore, the solution of a 1st order differential equation involves 1

arbitrary constant, and the solution of a 2nd

order differential equation involves 2

arbitrary constants.

Solutions involving arbitrary constants are called general solutions.

If some conditions are given so that definite values can be found for the arbitrary

constants, then particular solutions are obtained.



( )

2

2

2

1

d 1 d(a) 1 1

d d1

1d 1 d

1

tan

tan .

y yy

x xy

y xy

y x C

y x C

−

= + ⇒ =+

⇒ =+

⇒ = +

⇒ = +

∫ ∫

( )

d(b) e

d

e d 1 d

e

e

ln

y

y

y

y

y

x

y x

x C

B x

y B x

−

−

−

=

⇒ =

⇒ − = +

⇒ = −

⇒ = − −

∫ ∫

In the following sections from 2.1 to 2.5, we will be dealing with various types of

differential equations and the methods in solving them.

2.1 Solving differential equations of the form d

f ( )d

yx

x=

Differential equations of the form d

f ( )d

yx

x= can be solved by direct integration i.e.

integrating both sides of the equation with respect to x.

Example 1

Find the general solution of the following differential equations:

(a) d

3d

yx

x= − (b)

dsin

d

yx x

x=

Solution

2

d(a) 3

d

1 d ( 3) d

3 .2

yx

x

y x x

xy x C

= −

⇒ = −

⇒ = − +

∫ ∫

d(b) sin

d

1 d cos cos d

cos sin .

yx x

x

y x x x x

y x x x C

=

⇒ = − +

⇒ = − + +

∫ ∫

2.2 Solving differential equations (d

g( )d

yy

x= ) using the method of separable variables

Differential equations of the form d

g( )d

yy

x= can be solved by first bringing g( )y to

the other side and then integrating both sides of the equation with respect to x.

Example 2


(a) 2d1

d

yy

x= + (b)

de

d

yy

x=

Solution



2.3 Solving differential equations (d

f ( )g( )d

yx y

x= ) using the method of separable

variables

Steps to solve:

1. Bring g( )y to the other side, i.e. 1 d

f ( )g( ) d

yx

y x= ,

i.e. manipulate the equation to the form: d

G( ) F( )d

yy x

x= ,

where G(y) and F(x) are functions in y and x respectively.

2. Integrate both sides with respect to x to get:

1

d f ( )d or G( ) d F( ) dg( )

y x x y y x xy

= =∫ ∫ ∫ ∫ .

3. Integrate with respect to y and x respectively.

Example 3


(a) d

( 3)d

yx y

x− = (b)

2d 12sec

d

y yx

x y

+ =

(c) 2

d0

d

y x

x y+ = (d)

2

ed

d yx

x

yy

−=

For (d), find the particular solution given that 0=y when 0=x .

Solution

( )

d(a) ( 3)

d

d d

3

ln ln 3

e3

3 ,

C

yx y

x

y x

y x

y x C

y

x

y B x

− =

⇒ =−

⇒ = − +

⇒ =−

⇒ = −

∫ ∫

where eCB = ± is an arbitrary

constant. ( )

(b)

d 1 2 sec

d

2 d cos d

1

22 d cos d

1

2 2 ln 1 sin

1 ln 1 sin .

2

y yx

x y

yy x x

y

y x xy

y y x C

y y x C

+ =

⇒ =+

⇒ − =+

⇒ − + = +

⇒ − + = +

∫ ∫

∫ ∫



2

2

2

3 2

3 2

d(c) 0

d

d

d

d d

3 2

3, 3 .

2

y x

x y

y x

x y

y y x x

y xC

y x A A C

+ =

⇒ = −

⇒ = −

⇒ = − +

⇒ = − + =

∫ ∫

2

2

2

2

2

2d(d) e

d

e d e d

1 e e

2

e 2e 2

ln(2e )

For 0 and 0, we have 0 ln(2 ) 1.

Hence the particular solution is ln(2e 1).

x y

y x

y x

y x

x

x

yy

x

y y x

C

C

y A

x y A A

y

− =

⇒ =

⇒ = +

⇒ = +

⇒ = +

= = = + ⇒ = −

= −

∫ ∫

2.4 Solving first order differential equations using a given substitution

Steps to solve:

1. Differentiate the given substitution with respect to x. Use of implicit

differentiation may be necessary.

2. Replace the d

d

y

x term in the differential equation by the expression obtained in

Step 1, and the terms in y are also replaced accordingly by the new variable.

3. A separable variables differential equation will be obtained upon

simplification, after which we solve it as in 2.1, 2.2 or 2.3.

4. Replace the new variable by y in the general solution.

Example 4

Using the substitution y vx= , solve the differential equation

d

2d

yx y x

x+ = .

Solution

d dSince , we have . ----- Step 1

d d

dThen 2 becomes

d

d ( ) 2 ----- Step 2

d

d 2

d

d

y vy vx v x

x x

yx y x

x

vx v x vx x

x

vv x v

x

vx

= = +

+ =

+ + =

+ + =

2 2 ----- Step 3d

vx

= −



2

2

2

2

1 1 1 d d

2 1

1 ln 1 ln

2

ln 1 ln( )

ln( ) ln 1

ln 1 ----- Step 4

1

v xv x

v x C

v x A

x v A

yx A

x

yx

x

⇒ =−

⇒ − − = +

⇒ − − = +

⇒ + − = −

⇒ − = −

⇒ −

∫ ∫

22

e

or .

A B

x B Bx xy B y x

x x

−= ± =

−⇒ − = = = −

Example 5

Show that the substitution yxu += reduces the differential equation yx

yx

x

y

−−

++=

1

1

d

d to

d 2

d 1

u

x u=

−. Hence solve the differential equation.

Solution

d dSince , we have 1 ----- Step 1

d d

d 1Then becomes

d 1

d 1 1 ----- Step 2

d 1

u yu x y

x x

y x y

x x y

u u

x u

= + = +

+ +=

− −

+− =

−

2

2

d 1 2 1 (shown) ----- Step 3

d 1 1

(1 ) d 2 d

22

( ) ( ) 2 ---

2

u u

x u u

u u x

uu x C

x yx y x C

+= + =

− −

⇒ − =

⇒ − = +

+⇒ + − = +

∫ ∫

2

2

-- Step 4

2( ) ( ) 4 2

( ) 2( )

x y x y x C

x y x y A

⇒ + − + = +

⇒ + + − =

(Note that e A−± is an arbitrary constant.)



Example 6

Solve the differential equation xxyx

yy =−

2

d

d2 using the substitution 2yu = , given

that 0 when 0y x= = .

Solution

2

2

d dSince , we have 2 . ----- Step 1

d d

dThen 2 becomes

d

d ----- Step 2

d

d (1 )

d

u yu y y

x x

yy xy x

x

uux x

x

ux u

x

= =

− =

− =

= + ----- Step 3

2

2

2

22 2

1 d d

1

ln 12

1 e .e

e 1 ----- Step 4

Now when 0 and 0, 1. Ther

xC

x

u x xu

xu C

u

y A

y x A

⇒ =+

⇒ + = +

⇒ + =

⇒ = −

= = =

∫ ∫

22 2efore the particular solution is e 1.

x

y = −



2.5 Solving differential equations of the form 2

2

df ( )

d

yx

x=

Differential equations of the form 2

2

df ( )

d

yx

x= can be solved by direct integration as

in 2.1, except that we now have to integrate the equation with respect to x twice to

reduce the 2

2

d

d

y

x term to the y term, to obtain the general solution.

Example 7

Solve the following second order differential equations:

(a) 2

2

dcos

d

yx

x= (b) ( )

2

2

d2 e e , 0 when 0

d

and 2 when 1

x xyy x

x

y x

−= + = =

= =

Solution

2

2

2

2

d(a) cos

d

dd cos d

d

dsin

d

dd sin d

d

cos .

yx

x

yx x x

x

yx A

x

yx x A x

x

y x Ax B

=

⇒ =

⇒ = +

⇒ = +

⇒ = − + +

∫ ∫

∫ ∫

( )

( )

( )

( )

( )

( )

2

2

2

2

1 1

1

d(b) 2 e e

d

dd 2 e e d

d

d2 e e

d

dd 2 e e d

d

2 e +e .

0, 0 : 0 2(1 1) 4

1, 2 : 2 2(e e ) 4 6 2(e e )

2 e +e 6 2(e e ) 4.

x x

x x

x x

x x

x x

x x

y

x

yx x

x

yA

x

yx A x

x

y Ax B

x y B B

x y A A

y x x

−

−

−

−

−

− −

− −

= +

⇒ = +

⇒ = − +

⇒ = − +

⇒ = + +

= = = + + ⇒ = −

= = = + + − ⇒ = − +

∴ = + − + −

∫ ∫

∫ ∫



§3 Formulation of Differential Equations

3.1 Mathematical Interpretation of Physical Properties

Example

A body falls from rest in a medium that causes the velocity to decrease at a rate

proportional to the velocity.

Let v be the velocity of the falling body in the medium at time t. Then

d

d

d,

d

vv

t

vkv k

t

+

∝ −

∴ = − ∈R

3.2 Mathematical Interpretation of Laws of Growth & Decay

Example

The rate of decay of a radioactive material is proportional to the amount of material

present.

Let m be the amount of radioactive material at time t. Then

d

d

d,

d

mm

t

mkm k

t

+

∝ −

∴ = − ∈R

Note that the half–life of a substance is the time taken for half of the amount of

substance to decay.

3.3 Mathematical Interpretation of Laws of Cooling

Example

The rate of cooling of a body is proportional to the difference between the

temperature of the body and that of the surrounding.

Let T be the temperature of the body at time t and 0T be the temperature of the

surroundings (constant). Then

0

0

dT ( )

d

dT( ),

d

T Tt

k T T kt

+

∝ − −

∴ = − − ∈R



Example 9

A colony of bacteria is being grown in a shallow dish of area 100 cm2. After t

minutes, the area of the colony is x cm2. The growth of the colony is modeled by the

differential equation )100(25

1

d

dxt

t

x−= .

Given that initially, the colony occupies 1 cm2, solve the differential equation giving

x in terms of t, and calculate the time, correct to 2 decimal places, that will elapse

before the colony has covered half the remaining area of the dish (i.e. x = 50.5 cm2).

Solution

( )2

d 1Integrating (100 ) on both sides w.r.t. , we get

d 25

1 1 d d

100 25

Since 100,

1 ln 100

25 2

xt x x

t

x t tx

x

tx C

= −

=−

<

⇒ − − = +

⇒

∫ ∫

2

2

2

50

50

50

1 e . e

100

100 e

When 0 and 1, we have 1 100 99.

Therefore the particular solution is 100 99e .

t

C

t

t

x

x A

t x A A

x

−

−

=−

⇒ = −

= = = − ⇒ =

= −

2 2

2

Remaining area = 99 cm . Half of remaining area = 49.5 cm .

Required area = Initial area + Covered area = 49.5 + 1 = 50.5 cm .

Now when 50.5, we need to solve for .

50.5 1

x t=

=

2

2

2

50

50

00 99e

1 e

2

1 ln

50 2

1 50ln 5.89 minutes

2

t

t

t

t

−

−

−

⇒ =

⇒ − =

⇒ = − =

Hence the time that will elapse before the colony has covered half the remaining area

of the dish is 5.89 minutes.

Something to ponder on:

What can you gather about the size of the colony (i.e. x) after a prolong period of time

(i.e. as time t tends to infinitely large value)?



Example 10

In a small town, there was an outbreak of the SARS (Severe Acute Respiratory

Syndrome) virus and the rate of infection can be modeled by the differential equation

( )d

1d

yy y

t= − , where t is in days and y is the proportion of the population in the town

who are infected.

Given that by the 9th

day of the outbreak, 10% of the population were infected,

estimate the time taken before 60% of the population becomes infected with the virus,

assuming that no preventive measures are taken to curb the spread of the virus.

Solution

( )

( )

d1

d

1d d

1

1 1+ d d

1

ln ln 1

1

t C

yy y

t

y ty y

y ty y

y y t C

ye

y

+

= −

=−

=−

− − = +

=−

∫ ∫

∫ ∫

When 0.1y = at 9t = , 90.1

1 0.1

Ce

+=

−11.2C⇒∴ ≈ −

When 0.5y = , 11.20.611.6

1 0.6

te t

−= ⇒∴ ≈

−days

Approximately by the 12th

day of the outbreak, 60% of the town population will be

infected by the virus.

Note that it is not necessary to make y the subject in order to solve for the arbitrary

constant C and finding out the required t value in this question.

Do you know?

Singapore was among the first nations hit by SARS in 2003. Over two hundred cases

of SARS were reported in Singapore and 27 people died from the outbreak.



Example 11

The rate at which a substance evaporates is K times the amount of substance that has

not yet evaporated where K is a positive constant. If the initial amount of substance

was A and the amount which has evaporated at time t is x, write down a differential

equation involving x, and solve it to give x in terms of A, K and t.

Sketch the graph of x against t. Show that the time taken for half the substance to

evaporate is 2ln1

K.

Solution

( )

d( ) d( ) ( )

d d

d ( )

d

1 d d

ln

A x xK A x K A x

t t

xK A x

t

x K tA x

A x Kt

−= − − ⇒ − = − −

⇒ = −

⇒ =−

⇒ − − =

∫ ∫

e

Now when 0 and 0, we have 0 .

Therefore the particular solution is (1 e ).

Kt

Kt

C

x A B

t x A B B A

x A

−

−

+

⇒ = −

= = = − ⇒ =

= −

1When , we need to find .

2

1 i.e. (1 e )

2

1 1 e

2

1 e

2

1 ln

2

1 ln 2 (shown)

Kt

Kt

Kt

x A t

A A

Kt

tK

−

−

−

=

= −

⇒ − =

⇒ =

⇒ − =

⇒ =

x

t

A

O



§4 Sketching of Family of Solution Curves

4.1 Family of Solution Curves

Recall that the general solution of a first order differential equation contains an

arbitrary constant.

Geometrically, the general solution of the differential equation represents a family of

solution curves. If the value of the arbitrary constant is known, we obtain a particular

solution which represents a typical member of the family of solution curves.

Refer to Example 1 (a).

The general solution of d

3d

yx

x= − is 21

32

y x x C= − + , where C is an arbitrary

constant.

Thus, the general solution represents a family of solution curves 213

2y x x C= − + ,

where C takes on any real value.

If we are given that a particular solution curve passes through the point (0, 1), then we

obtain the value of C = 1. Hence, the equation of the solution curve is 213 1

2y x x= − + ,

which is member of the family of solution curves 213

2y x x C= − + .

We may use the GC to sketch typical members of a family of solution curves.

The following table illustrates the use of GC to sketch 3 members of the family of

curves 213

2y x x C= − + , for C = − 1, 0, 1.

Steps Screen

1 Enter the equation of the curve.

Note that instead of entering 3 equations for the 3

different values of C, we may enter just one equation

using the notation { − 1, 0, 1} to graph the 3 members

of the family of curves.

2 The GC shows 3 solution curves, each belonging to a

different value of C.

Note: When you present your answers on paper, remember to label your graphs clearly,

i.e. label any axial intercepts, stationary points and asymptotes.



Note: C = − 1, 0, 1 is only a suggestion to the possible constants to be used. In general,

we should use a variety of constants that can illustrate all possible variations of the

family of curves. Example 12 illustrates an example that all possible variations could

be obtained with C = − 1, 0, 1 without the need for other constants. However, not all

family of solutions curves could be obtained by just using C = − 1, 0, 1.

Example 12

Consider the family of solution curves represented by 2y x Cx= + , where C is an

arbitrary constant..

Sketch, on the same diagram, the solution curves for C = − 1, 0, and 1.

Summary

• A first order variables separable differential equation is of the form ( ) ( )yxx

ygf

d

d=

where f is a function of x only and g is a function of y only.

Steps to solve:

1. Bring g( )y to the other side, i.e. 1 d

f ( )g( ) d

yx

y x= ,

i.e. manipulate the equation to the form: d

G( ) F( )d

yy x

x= ,

where G(y) and F(x) are functions in y and x respectively.

2. Integrate both sides with respect to x to get:

1

d f ( )d or G( ) d F( ) dg( )

y x x y y x xy

= =∫ ∫ ∫ ∫ .

3. Integrate with respect to y and x respectively.

C = 1 C = 0

C = -1



• In solving first order differential equations using a given substitution, adopt the

following steps:

1. Differentiate the given substitution with respect to x.

2. Replace the d

d

y

x term in the differential equation by the expression obtained in

Step 1, and the terms in y are also replaced accordingly by the new variable.

3. A separable variables differential equation will be obtained upon simplification,

after which we solve it.

4. Replace the new variable by y in the general solution.

• Differential equations of the form 2

2

df ( )

d

yx

x=

Steps to solve:

1. Integrate both sides with respect to x to getd

f ( ) dd

yx x

x= ∫ .

2. Integrate both sides with respect to x again to get ( )f ( ) d dy x x x= ∫ ∫ .

• The general solution represents a family of curves, and a particular solution represents

one of the curves.

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njc differential equation lecture notes teachers edition