ADVANCED HIGHER MATHEMATICS
DIFFERENTIAL EQUATIONS
A differential equation is an equation involving one or more derivatives.
Examples
(1)
(2)
( x 2  1)
dy
 2 xy  x is a first-order differential equation, as the equation
dx
contains a first derivative only.
d2y
dy
5
 4 y  0 is a second-order differential equation, as the
2
dx
dx
equation contains a second derivative.
VERIFICATION OF SOLUTIONS OF DIFFERENTIAL EQUATIONS
Given a function y  f (x) , you can verify that the function satisfies a particular
differential equation.
Worked Example
Verify that the function y  Ae 5 x  Be 2 x , where A and B are constants, satisfies the
differential equation
d2y
dy
 3  10 y  0 .
2
dx
dx
Solution
y  Ae 5 x  Be 2 x

dy
 5 Ae 5 x  2 Be  2 x
dx

d2y
 25 Ae 5 x  4 Be  2 x
2
dx
d2y
dy
 3  10 y
2
dx
dx
5x
 25 Ae  4 Be 2 x  3(5 Ae 5 x  2Be 2 x )  10( Ae 5 x  Be 2 x )
 25 Ae 5 x  4 Be 2 x  15 Ae 5 x  6 Be 2 x  10 Ae 5 x  10 Be 2 x
0
Hence y  Ae
5x
 Be
2 x
d2y
dy
 3  10 y  0 .
satisfies the differential equation
2
dx
dx
YOU CAN NOW ATTEMPT THE WORKSHEET
"VERIFICATION OF SOLUTIONS OF DIFFERENTIAL EQUATIONS."
GENERAL SOLUTIONS OF DIFFERENTIAL EQUATIONS
Given a function y  f (x) , it is usually straightforward to verify that y satisfies a
particular differential equation.
For example, we have already verified that the function y  Ae 5 x  Be 2 x , where A
and B are constants, satisfies the differential equation
d2y
dy
 3  10 y  0 .
2
dx
dx
Given a differential equation on its own, however, it is not so straightforward to find
the general form of the function y which satisfies the differential equation. For
d2y
dy
 3  10 y  0 , how do we solve this
example, given the differential equation
2
dx
dx
5x
2 x
differential equation to find that y  Ae  Be
is the general solution of the
differential equation.
FIRST-ORDER DIFFERENTIAL EQUATIONS WITH VARIABLES
SEPARABLE
One approach which can sometimes be used to solve first-order differential equations
is to separate the variables in the differential equation.
If a first-order differential equation can be written in the form
f ( y )dy  g ( x)dx ,
where f ( y ) is a function of y only and g (x ) is a function of x only, we say that we
have separated the variables x and y.
The general solution of the differential equation is then found be integrating each
sides with respect to the appropriate variable,
i.e.
 f ( y)dy   g ( x)dx .
The following examples illustrate this.
Worked Example 1
Find the general solution of the differential equation
dy x 2
.

dx
y
Solution
dy x 2

dx
y

ydy  x 2 dx
The variables have been separated.
 ydy   x


2
dx
1 2 1 3
y  x C
2
3
2
3y  2x 3  C
[ 6 ]
NOTES
(1)
When solving differential equations, it is convenient for C to always denote
the "total" constant so far in each line of working, although strictly speaking a
different letter should be used for each new constant.
(2)
It is not always necessary, or convenient, to express the general solution for y
explicitly in terms of x; that is, in the form y  f (x) . Often it is more
convenient, as in the above example, to express y implicitly in terms of x, for
example in the form y 2  f ( x) or sin y  f ( x) . Of course, the explicit
solution must be given if requested.
Worked Example 2
Find the general solution of the differential equation
e4y
dy
 x  2.
dx

e4y
Solution
e4y
dy
x2
dx
dy
x2
dx
e 4 y dy  ( x  2)dx

The variables have been separated.
e


4y
dy   ( x  2)dx
1 4y 1 2
e  x  2x  C
4
2
e 4 y  2 x 2  8x  C
[ 4 ]
[If it is required to express y explicitly in terms of x, this can be done as follows:


e 4 y  2 x 2  8 x  C [ln]
4 y  ln( 2 x 2  8x  C )
1
y  ln( 2 x 2  8 x  C ) ]
4
Worked Example 3
Find the general solution of the differential equation
dy
2

.
dx sin y
Solution
dy
2

dx sin y

sin ydy  2dx
The variables have been separated.



 sin ydy   2dx
 cos y  2 x  C
cos y  2 x  C
cos y  C  2 x
[  (1) ]
Worked Example 4
Find the general solution of the differential equation
dy
 2x 9  y 2 .
dx
Solution
dy
 2x 9  y 2
dx


dy  2 x 9  y 2 dx
1
9 y2
dy  2 xdx
The variables have been separated.




1
9  y2
dy   2 xdx
 y
sin 1    x 2  C
3
y
 sin( x 2  C )
3
y  3 sin( x 2  C )
[sin]
YOU CAN NOW ATTEMPT THE WORKSHEET
"DIFFERENTIAL EQUATIONS 1."
In some cases the properties listed below of the logarithmic and exponential functions
can be used to simplify general solutions of differential equations.
(1)
e a b  e a e b
(2)
ln a  ln b  ln( ab)
(3)
a
ln a  ln b  ln  
b
(4)
n ln a  ln( a n )
Worked Example 5
Find the general solution of the differential equation
dy
 4y .
dx
Solution
dy
 4y
dx

dy  4 ydx

1
dy  4dx
y
The variables have been separated.
1
 y dy   4dx




ln y  4 x  C
ye
[e^]
4 x C
y  e 4x  eC
y  Ae 4 x [where A  e C ]
Worked Example 6
Find the general solution of the differential equation
dy y  2

.
dx x  1
Solution
dy y  2

dx x  1

( x  1)dy  ( y  2)dx

1
1
dy 
dx
y2
x 1
The variables have been separated.
1
1
 y  2 dy   x  1 dx




ln( y  2)  ln( x  1)  C
y  2  ( x  1)  e
y  2  A( x  1)
y  A( x  1)  2
[e^]
C
[where A  e C ]
Worked Example 7
Find the general solution of the differential equation
dy
y

.
dx 2 x  1
Solution
dy
y

dx 2 x  1

(2 x  1)dy  ydx

1
1
dy 
dx
y
2x  1
The variables have been separated.
1
1
 y dy   2 x  1 dx

ln y 
1
ln( 2 x  1)  C
2
1
2

ln y  ln( 2 x  1)  C

ln y  ln 2 x  1  C

y  2x  1  e

y  A 2x  1
[e^]
C
[where A  e C ]
1
ln( 2 x  1) should be written as ln 2 x  1 before taking the exponential
2
of both sides of the equation.
Note that
YOU CAN NOW ATTEMPT QUESTION 1 OF THE WORKSHEET
"DIFFERENTIAL EQUATIONS 2."
Worked Example 8
2x  3
in partial fractions.
x( x  1)
(a)
Express
(b)
Hence find the general solution of the differential equation
x( x  1)
dy
 y (2 x  3) ,
dx
expressing y explicitly in terms of x.
Solution
(a)
2x  3 3
1
 
.
x( x  1) x x  1
It can be shown that
dy
 y (2 x  3)
dx
x( x  1)dy  y (2 x  3)dx
1
2x  3
dy 
dx
y
x( x  1)
x( x  1)
(b)


The variables have been separated.
1
2x  3
1
3
 y dy  x( x  1) dx
1 

 y dy    x  x  1 dx


ln y  3 ln x  ln( x  1)  C



ln y  ln( x 3 )  ln( x  1)  C
 x3 
  C
ln y  ln 
x

1


3
 x  C
  e
y  
 x  1
y
Ax 3
x 1
[e^]
[where A  e C ]
Worked Example 9
x 1
in partial fractions.
x(2 x  1)
(a)
Express
(b)
Hence find the general solution of the differential equation
dy
y ( x  1)

,
dx x (2 x  1)
expressing y explicitly in terms of x.
Solution
(a)
x 1
1
1
 
.
x(2 x  1) x 2 x  1
It can be shown that
(b)


dy
y ( x  1)

dx x (2 x  1)
x(2 x  1)dy  y ( x  1)dx
1
x 1
dy 
dx
y
x(2 x  1)
The variables have been separated.
x 1
1
 y dy  x(2 x  1) dx
1
1
1


 y dy    x  2x  1 dx

ln y  ln x 

ln y  ln x  ln( 2 x  1) 2  C

ln y  ln x  ln 2 x  1  C
1
ln( 2 x  1)  C
2
1




x 
  C [e^]
ln y  ln 
 2x  1 

x  C
  e
y  
 2x  1 
Ax
y
[where A  e C ]
2x  1
YOU CAN NOW ATTEMPT QUESTIONS 2 TO 18 OF THE WORKSHEET
"DIFFERENTIAL EQUATION 2."
PARTICULAR SOLUTIONS OF DIFFERENTIAL EQUATIONS
Worked Example 1
Find the particular solution of the differential equation
y2
dy
 4x 2  1,
dx
given that y  2 when x  1 , expressing y explicitly in terms of x.
Solution
dy
 4x 2  1
dx
2
y dy  (4 x 2  1)dx
y2

The variables have been separated.
y


2
dy   (4 x 2  1)dx
1 3 4 3
y  x  xC
3
3
3
y  4 x 3  3x  C
y  2 when x  1



Hence y  4 x  3x  1
3
3
[ 3]
2 3  4  13  3  1  C
7 C 8
C 1

y  (4 x  3x  1)
3
1
3
Worked Example 2
Find the particular solution of the differential equation
dy
 2xy 2 ,
dx
given that y  
3
when x  2 , expressing y explicitly in terms of x.
10
Solution


dy
 2xy 2
dx
dy  2 xy 2 dx
1
dy  2 xdx
y2
The variables have been separated.
y




y
2
dy   2 xdx
 y 1  x 2  C
1
  x 2  C [  (1) ]
y
1
 x 2  C
y
1
y
C  x2
3
when x  2
10





Hence y 
1
2
 x2
3

3
.
2  3x 2
3
1

10 C  (2) 2
3
1


10 C  4
 3(C  4)  10
 3C  12  10
2
C
3

Worked Example 3
Find the particular solution of the differential equation
( x  4)
dy
1 y ,
dx
given that y  7 when x  1, expressing y explicitly in terms of x.
Solution
dy
1 y
dx
dy
( x  4)
 y 1
dx
( x  4)dy  ( y  1)dx
1
1
dy 
dx
y 1
x4
( x  4)



The variables have been separated.
1
1
 y  1 dy   x  4 dx

ln( y  1)  ln( x  4)  C



y  1  ( x  4)  e C
y  1  A( x  4) [where A  e C ]
y  A( x  4)  1
y  7 when x  1



Hence y  2( x  4)  1 
[e^]
7  A(3)  1
3A  1  7
A2
y  2x  9
Worked Example 4
2
in partial fractions.
( y  1)( y  3)
(a)
Express
(b)
Hence find the particular solution of the differential equation
2x
given that y  
dy
 ( y  1)( y  3) ,
dx
5
when x  1, expressing y explicitly in terms of x.
3
Solution
(a)
It can be shown that
2
1
1


.
( y  1)( y  3) y  1 y  3
dy
 ( y  1)( y  3)
dx
2 xdy  ( y  1)( y  3)dy
2
1
dy  dx
( y  1)( y  3)
x
2x
(b)


The variables have been separated.
2
1
 ( y  1)( y  3) dy   x dx
 1
1 
1

  y  1  y  3 dy   x dx

ln( y  1)  ln( y  3)  ln x  C








 y 1
  ln x  C [e^]
ln 
 y  3
y 1
 x  eC
y3
y 1
 Ax [ A  e C ]
y3
y  1  Ax( y  3)
y  1  Axy  3 Ax
y  Axy  3 Ax  1
y (1  Ax)  3 Ax  1
3 Ax  1
y
1  Ax
y
5
when x  1 
3





5 3 A(1)  1

3 1  A(1)
5  3A  1
 
3
1 A
 5(1  A)  3(3 A  1)
 5  5 A  9 A  3
4A  2
1
A
2

3
x 1
3x  2
Hence y  2
.

1
2x
1 x
2
[Alternatively the value of A can be found by substituting y  
into the equation
y 1
 Ax .
y3
5
1
3
 A(1)
5
 3
3

Then



y 1
 Ax
y3







5
and x  1
3
53
 A
59
2
 A
4
1
A
2
y 1 1
 x
y3 2
y 1 x

y3 2
2( y  1)  x( y  3)
2 y  2  xy  3x
2 y  xy  3x  2
y ( 2  x)  3 x  2
3x  2
y
]
2x
YOU CAN NOW ATTEMPT THE WORKSHEET
"PARTICULAR SOLUTIONS OF DIFFERENTIAL EQUATIONS."
DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS
Worked Example 1
The number of strands of bacteria, x, present in a culture after t days of growth is
assumed to be increasing at a rate proportional to the number of strands present.
(a)
Write down a differential equation which represents this and find the general
solution for x in terms of t.
(b)
Given that there are 326 strands initially present and that the number of
strands observed after 4 days is 1833, estimate the number of strands likely to
be present after one week.
Solution
(a)
The rate of growth of the bacteria is
dx
and the number of strands present is
dt
x.
Hence
dx
is proportional to x 
dt
dx
 kx
dt

dx  kxdt

1
dx  kdt
x
dx
 kx ,
dt
The variables have been separated.
1
 x dx   kdt
(b)

ln x  kt  C



x  e kt  C
x  e kt  e C
x  Ae kt
[e^]
When t  0 , x  326 


326  Ae k 0
Ae 0  326
A  326
When t  4 , x  1833 

1833  Ae k 4
326e 4 k  1833
1833
e 4k 
[ln]
326
 1833 
4k  ln 

 326 


where k is a constant.

k
1  1833 
ln 
  0  4317
4  326 
Hence x  326e 04317t .
When t  7 :
x  326e 043177  326e 30219  6693
About 6693 strands are likely to be present after one week.
Worked Example 2
According to Newton's law of cooling, the rate at which an object cools is
proportional to the difference in temperature between the object and the surrounding
medium. When an object cools surrounded by air at a temperature of 75 oC, the
cooling of the object is governed by a differential equation of the form
dT
 k (T  75) ,
dt
where T oC is the temperature of the object after t hours of cooling and k is a constant.
(a)
Find the general solution of this differential equation, expressing T as a
function of t.
(b)
Given that a certain object cools from 125 oC to 100 oC in half an hour when
surrounded by air at a temperature of 75 oC, find:
(i)
(ii)
the temperature of this object at the end of another half-hour
the time taken for the temperature of this object to fall to 80 oC.
Solution
(a)
dT
 k (T  75)
dt


dT  k (T  75)dt
1
dT  kdt
T  75
The variables have been separated.
1
 T  75 dT    kdt
(b)

ln( T  75)  kt  C




T
T
T
T
When t  0 , T  125 



When t 
1
, T  100 
2


[e^]
 75  e  kt C
 75  e  kt  e C
 75  Ae  kt
[where A  e C ]
 75  Ae  kt
125  75  Ae  k 0
75  Ae 0  125
75  A  125
A  50
100  75  Ae
Ae
1
 k
2
50e
1
 k
2
 25
 25
k
1
2
1
 k
2

e




1
2
[ln]
1
1
k  ln  
2
2
1
k  2 ln    1  3863
2
Hence T  75  50e 13863t .
(i)
When t  1:
T  75  50e 138631
 75  50e 13863
 87  5
The temperature of the object if 875 oC at the end of another half hour.
(ii)
When T  80 :




80  75  50e 13863t
50e 13863t  5
1
e 13863t 
[ln]
10
1
 1  3863t  ln  
 10 
1
ln  
 10 
t
 1  66
 1  3863
It will take 166 hours for the temperature of the object to fall to 80 oC.
Worked Example 3
When a valve is opened, the rate at which the water drains from a pool is proportional
to the square root of the depth of the water. This can be represented by the differential
equation
dh
 k h ,
dt
where h metres is the depth of the water t minutes after the valve was opened and k is
a positive constant.
(a)
Find the general solution of the differential equation.
(You need not express h explicitly as a function of t.)
(b)
Given that the pool was initially 9 metres deep and that the depth of water was
4 metres after 20 minutes of draining, how long did it take to drain the pool
completely?
Solution
(a)
dh
 k h
dt
dh  k hdt

1

dh  kdt
h
The variables have been separated.

1
1
 h 2 dh    kdt
(b)

2h 2  kt  C

2 h  C  kt
When t  0 , h  9


2 9 C k 0
C 6
When t  20 , h  4



2 4  C  k  20
6  20k  4
20k  2
1
k
10

Hence 2 h  6 
1
t
t , i.e. 2 h  6  .
10
10
2 0 6
When h  0 :

06

t
6
10
t  60

t
10
t
10
It takes 60 minutes (i.e. 1 hour) to drain the pool completely.
Worked Example 4
In a chemical reaction, two substances X and Y combine to form a third substance Z.
Let Q denote the number of grams of Z formed t minutes after the reaction begins.
The rate at which Q varies is represented by the differential equation
dQ (20  Q)(10  Q)

.
dt
400
400
in partial fractions.
(20  Q )(10  Q )
(a)
Express
(b)
Given that Q  0 when t  0 , find the general solution of the differential
equation, expressing Q explicitly as a function of t.
(c)
Find:
(i)
(ii)
(d)
the number of grams of Z formed one hour after the reaction begins
the time taken to form 5 grams of Z.
Find the limiting value of Q as t   .
Solution
400
40
40


,
(20  Q)(10  Q)
20  Q 10  Q
(a)
It can be shown that
(b)
dQ (20  Q)(10  Q)

dt
400

400dQ  (20  Q)(10  Q)dt

400
dQ  dt
(20  Q)(10  Q)
The variables have been separated.
400
 (20  Q)(10  Q) dQ   dt






40
40 



  20  Q 10  Q dQ   dt
 40 ln( 20  Q)  40 ln( 10  Q)  t  C
40 ln( 20  Q)  40 ln( 10  Q)  t  C
40ln( 20  Q)  ln( 10  Q)  t  C
 20  Q 
  t  C
40 ln 
 10  Q 


 20  Q  t
 
ln 
C
 10  Q  40
t
C
20  Q
 e 40
10  Q

20  Q
 e 40  e C
10  Q

20  Q
 Ae 40
10  Q
[e^]
t
t
[where A  e C ]
20  0
 Ae 0
10  0
A2

Q  0 when t  0

20  Q
 2e 40
Hence
10  Q
t





t
40
20  Q  2e (10  Q)
20  Q  20e
2Qe
t
40
t
40
 2Qe
t
40
t
 Q  20e 40  20
 t

 t

Q 2e 40  1  20 e 40  1




t


20 e 40  1


Q
t
2e 40  1
(c) (i) When t  60 : Q 
 60

20 e 40  1


60
2e 40  1
20(e15  1)

 8  74
2e15  1
874 grams of Z are formed on hour after the reaction begins.
5
(ii) When Q  5 :
 t

20 e 40  1


t
40
1

 t

 1  20 e 40  1



2e


5 2e


10e 40  5  20e 40  20
t
t
40
t

10e
t
40
 15
t
40
e  1  5 [ln]
t
 ln 1  5
40
t  40 ln 1  5  16  2



It takes 162 minutes to form 5 grams of Z.
(d)
Q
 40t

20 e  1


t
2e 40  1
As t   , both the numerator and denominator of the above expression will
t
tend to infinity (since e 40   ) and the limiting value of Q cannot be found
using this expression. An alternative expression for Q must therefore be found
before investigating the limiting value.
t
Dividing all terms in the above expression by e 40 :

1

201  t

 e 40
Q
1
2 t






t



40 

201  e 


2e

t
40
e 40
Now as t   , e

t
40
 0 and Q 
20(1  0)
 10 .
20
Hence the limiting value of Q as t   is 10.
YOU CAN NOW ATTEMPT THE WORKSHEET
"DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS."