APPLIED MATHEMATICS III
MATH 2042
APPLIED MATHEMATICS III
FEB 2025
Content
Chapter I: ORDINARILY DIFFERENTIAL EQUATIONS OF THE FIRST ORDER
1.Basic concepts and ideas
2.Equations with separated and separable variables.
3.Problems in setting up differential equations
4.Exact differential equations
5.Integrating factors
6.Homogeneous first-order equations
7.First-order linear equations
01
BASIC
CONCEPTS
AND IDEAS
BASIC CONCEPTS AND IDEAS
Let
y
be a function of x. Then we write it as
y  f (x)
Its first, second, third and so on, derivatives are written as
y, y, y, , y
(n)
or
dy d 2 y d 3 y
d2y
,
,
, , n
dx dx 2 dx 3
dx
Definition A differential equation is one which connects the independent variable x,
unknown function y = f(x), and its derivatives y, y,, y ( n )
(n)
Symbolically, F ( x, y, y, y,, y )  0
If the sought – for function y = f(x) is a function of one independent variable, then the
differential equation is called Ordinary. The order of a differential equation is the order
of the highest derivative which appears. The solution or integral of a differential
equation is any function y = f(x) which, when put in to the equation, converts it in to an
identity.
BASIC CONCEPTS AND IDEAS
Example 1 :
y   2 xy 2  5  0
y   ky   by  sin x  0
(First -order)
(Second -order)
d2y
y0
dx 2
The functions y = sinx, y = 2cosx, y = 3sinx – cosx and, in general, functions of the form
y = C1 sinx + C2 cos x
are solutions of the given equation for any choice of constants C1 and C2 , Indeed,
Example 2 : Suppose we have the equation
y   C1 sin x  C 2 cos x
Putting the expressions y and y  in to the initial equation, we get the identity
 C1 sin x  C2 cos x  C1 sin x  C2 cos x  0
BASIC CONCEPTS AND IDEAS
1. Show that the indicated function which depend on arbitrary constants, satisfy the
corresponding differential equations:
sin x
;
a) y  sin x  1  Ce
dy
1
 y cos x  sin 2 x
dx
2
2
b) y  Cx  C  C ;
dy
dy
 dy 


x
y0
 
dx
dx
 dx 
y 2  2Cx  C 2 ;
y ( y ) 2  2 xy   y  0
2
c)
d) y 
C1
 C2;
x
y  
2
y  0
x
BASIC CONCEPTS AND IDEAS
A differential equation of the first-order is of the form
F ( x, y, y)  0
If this equation can be solved for y , it can be written in the form
y  f ( x, y)
Theorem: If in the equation y  f ( x, y) , the function f(x, y) and its partial derivative
with respect to y, f , are continuous in some domain D, in an xy-plane,
y
containing some point (x0, y0), then there is a unique solution to this equation,
y   (x ) which satisfies the condition y  y0 at x  x0 . The condition that
for x = x 0 the function y must be equal to the given number y0 is called the
initial condition
BASIC CONCEPTS AND IDEAS
Definition : The general solution of a first–order differential equation is a function
y   ( x, c )
which depends on a single arbitrary constant C and satisfies the following
conditions:
a) It satisfies the differential equation for any specific value of the constant C;
b) No matter what the initial condition y = y0 for x = x0, it is possible to find a value C  C 0
such that the function y   ( x, c0 ) satisfies the given initial condition.
In searching for the general solution of a differential equation, we often arrive at a relation
like ( x, y, c)  0 which is not solved for y. An equation of the form ( x, y, c)  0 which
gives an implicit general solution is called the complete integral of the differential
equation.
BASIC CONCEPTS AND IDEAS
Example For the first–order equation
dy
y
  . The general solution is
dx
x
The particular solution that will satisfy the initial condition y0 = 1 when
x0 = 2 is
have
. Indeed putting these values into the formula
or c = 2
,
we
BASIC CONCEPTS AND IDEAS
Let us now consider the following problem. Let there be given a family of functions that
depends on a single parameter C:
y   ( x, c )
(1)
and let only one curve of this family pass through each point of the plane ( or some region
in the plane). For what differential equation is this family of functions a complete integral?.
The equation is found as follows: From relation (1), differentiating with respect to x, we
find
y
dy
  x ( x, c )
dx
(2)
Since only one curve of the family passes through each point of the plane, for every number
pair x and y, a unique value of C is determined from equation (1). Putting this value of C
in to (2), we find f  as a function of x and y.
EQUATIONS WITH SEPARATED AND SEPARABLE VARIABLES
Definitions 2.2
1.An equation of the form M ( x ) dx  N ( y ) dy  0 is called an equation
Solution of an equation with separated variables
M ( x)dx  N ( y )dy  0
with separated variables
integrating with respect to x and with respect y respectively, we obtain
2. An equation of the form M 1 ( x) N1 ( y )dx  M 2 ( x) N 2 ( y )dy  0 is called
 M ( x)dx   N ( y)dy  c
which is the general integral
an equation with separable variables
Examples
1) xdx  ydy  0
2)
ydx  xdy  0
(Separated variables)
(Separable variables)
Solution to an equation with variables separable
M 1 ( x ) N 1 ( y ) dx  M 2 ( x ) N 2 ( y ) dy  0
Dividing by N 1 ( y ) M 2 ( x ) , we obtain
M 1 ( x)
N ( y)
dx  2
dy  0
M 2 ( x)
N1 ( y )
which is of separated variables. Then by integrating, we get the general integral
(1  u )vdu  (1  v)udv  0
M 1 ( x)
N 2 ( y)
2
1
 M ( x) dx   N ( y ) dy  C
EQUATIONS WITH SEPARATED AND SEPARABLE VARIABLES
Example 2.3.1 Prove that a curve having the slope of the tangent to any point
proportional to the abscissa of the point of tangency is a parabola
Proof: Let the equation of the curve be y = f(x). Then the differential equation of the
problem is
y  kx
dy  kxdx
 dy   kxdx  C
x2
yk
C
2
y  k1 x 2  C , where k1  k
2
First Order Linear Equations
HOMOGENEOUS FIRST–ORDER EQUATIONS
Definition2.4
The differential equation Mdx  Ndy  0 is called homogeneous if
The ratio
M
y
can be represented as a function of the ratio . We denote this ratio
N
x
by t: t  y
x
Example 2.4.1
y  x  y  dx  xdy  0 is homogeneous, because
2
2
2
M y  x2  y 2
y
 y

   1     t  1  t 2
N
x
x
 x
EQUATIONS WITH SEPARATED AND SEPARABLE VARIABLES
Solution of a homogeneous equation


Example 2.4.2 : Integrate the equation y  x 2  y 2 dx  xdy  0
Solution: Let y = tx. Then dy = tdx + xdt. By substituting these in to the given
equation, it can be reduced to an equation with variables separable.
 tx  x  x t  dx  x(tdx  xdt)  0
2
2 2
x 1  t 2 dx  x 2 dt  0
dx
dt

0
x
1 t2
dx
dt
 x   1 t  C
2


ln x  ln t  1  t 2  ln c1
x
t  1 t2

 c1
x  c1 t  1  t 2

2
y
 y 
x  c1   1    
x
 k  

(General integral)
First Order Linear Equations
Integrate the following homogeneous differential equations:
Ans. y  2 xy  x  C
2
1. ( y  x)dx  ( y  x)dy  0 ;
2.
2.4
( x  y )dx  xdy  0 ;
2
Ans. x2 + 2xy = C
FIRST–ORDER LINEAR EQUATIONS
Definition 2.5:
An equation of the form dy  P( x) y  Q( x) , where P(x) and Q(x) are
dx
given continuous functions of x ( or are constants) is called first-order
linear equation.
Example 2.5.1
y 
2
y  ( x  1) 3
x 1
( First-order linear equation)
First Order Linear Equations
Example2.5.2
Solve the equation y  2 y  x  13
x 1
Solution: First we solve the equation y  2 y  0
x 1
dy 
2
ydx  0
x 1
dy
2

dx  0
y x 1
ln y  2 ln( x  1)  C
ln y  ln( x  1)  ln C , where C1  ln C
2
y  C1 ( x  1) 2
Next, we replace C1 by the unknown function u; and substitute y and y into the original
equation and solve for u.
 y  u ( x  1) 2

 y   u ( x  1) 2  2( x  1)u
u ( x  1) 2  2( x  1)u  
2
u ( x  1) 2  ( x  1) 3
x 1
u  x  1
du  ( x  1) dx
 du   ( x  1)dx  c
u
( x  1) 2
C
2
Therefore, the general integral ( or solution) of the equation is
 ( x  1) 2

( x  1) 4
y  
 C  x  12  
 C ( x  1) 2
2
2


Bernoulli’s Equation
2.6 BERNOULLI’S EQUATION
An equation of the form
dy
 P( x) y  Q( x) y n , where P(x) and Q(x) are continuous
dx
functions of x(or constants), and n  0 and n 1 (otherwise a linear equation) is called
Bernoulli’s equation
Example 2.6.1
dy
 xy  x 3 y 3
dx
Solution of Bernoulli’s equation
Example2.6.2: Solve the equation
dy
 xy  x3 y 3
dx
( Bernoulli’s equation)
Bernoulli’s Equation
x2
2
2
Let z  ue . Then z   u e x  e x u.2 x , Substituting these in to equation (2), we get
Solution
Dividing all terms of the equation by y3, we get
y 3
2
dy
 xy  2  x 3
dx
(1)
2
which is a linear equation .Let us find its complete integral
dz
 2 xz  0
dx
dz
 2 xdx  0
z
dz
 z   2 xdx  C
ln z  x 2  ln C1 , where C  ln C1
2
2
du  2 x 3e  x dx
dx
dz
 2 xz  2 x 3 ,
dx
z  C1e x
2
u e x  2 x 3
By making the substitution z  y 2 and dz  2 y 3 dy in to equation (1), we get
dx
2
ue x  2 xue x  2 xue x  2 x 3
(2)
 
u    2 x 3 e  x dx  C   x 2 d e  x  C  x 2 e  x   e  x d ( x 2 )  x 2 e  x  e  x  C
2
2
2
2
2
Therefore, the complete integral of equation (2) is

2

2
2
z  x 2 e  x  e  x  C e x  x 2  1  ce x
2
Consequently, the complete integral of the given equation is
2
y 2  x 2  1  Ce x or y 
1
x 2  1  Ce x
2
2
Exact Differential Equation
2.7 EXACT DIFFERENTIAL EQUATIONS
Definition 2.7 :
The equation
M ( x, y)dx  N ( x, y)dy  0

is called an exact differential equation if M ( x, y) and N ( x, y) are continuous
differentiable functions for which the following relationship is fulfilled
M N

y
x
and M and
y
N
are continuous in some region
x
2
2
Example 2.7.1 2 x dx  y  3x dy  0 is an exact differential equation, because
3
4
y
M 
y
2x
;
y3
N
y 2  3x 2
0
y4
M  6 x N  6 x
 4 ;
 4
y
x
y
y
Exact Differential Equation
Solution of an exact differential equation
u
y 2  3x 2
N
y
y4
Example 2.7.2 : Integrate the differential equation
we find
2x
y 2  3x 2
dx

dy  0
y3
y4
 3x 2
y 2  3x 2



(
y
)

y4
y4
Solution: The left side of the equation is an exact differential of some unknown function
 ( y ) 
u ( x, y ) . Let us find this function.Since
u 2 x

,
x y 3
1
1
,  ( y)    C
y
y2
u ( x, y ) 
x2 1
 C
y3 y
Thus the complete integral of the original equation is
it follows that
2x
x2
u   3 dx   ( y )  3   ( y )
y
y
x2 1
  C1
y3 y
(1)
where,  ( y ) is an as yet undetermined function of y. Differentiating equation (1) with
respect to y and noting that
Integrate the following exact differential equations:
x3
 yx  y 2  C
3
1.
x  y dx  ( x  2 y)dy  0 ;
Ans.
2.
y  3x dx  (4 y  x)dy  0 ;
Ans. 2 y
2
2
2
 xy  x 3  C
INTEGRATING FACTORS
INTEGRATING FACTORS
Let the left side of the equation
M ( x, y )dx  N ( x, y )dy  0
not be an exact differential. It is sometimes possible to choose a function  ( x, y) such
that after multiplying all terms of the equation by it, the left side of the equation is
converted in to an exact differential. The general solution of the equation thus obtained
coincides with the general solution of the original equation; the function
called the integrating factor of the equation.
 ( x, y) is
INTEGRATING FACTORS
Example2.8.1 : The left side of the equation 2ydx + xdy = 0 is not an exact deferential.
But multiplying by x yields x(2ydx + xdy) = d(x2y)
How to find an integrating factor.
N M


x
y is not dependent on x but only on y, then the integrating
1. If the expression
M
factor  can be found from the differential equation
N M

 ln 
y
y

y
M
2. If the expression
M N

y
x
N
is not dependent on y but only on x, then the integrating
factor  can be found from the equation
M N

 ln 
y x

x
N
INTEGRATING FACTORS
Example 2.8.2 : Solve the equation
Solution: Since
( y  xy 2 )dx  xdy  0
M
N
 1  2 xy ,
 1
y
x
Solving this equation, we find its complete integral
x x2

C  0
y 2
it is not an exact differential equation. Since
N M

 1  1  2 xy  2
x
y


M
y  xy 2
y
y
the equation permits of an integrating factor dependent only on y. We find it.
d ln   2

dy
y
After multiplying the equation by
,
ln   2 ln y ,

1
y2
1
, we obtain an exact differential equation
y2
1

x
  x  dx  2 dy  0
y
y

2x
x  2C
2
INTEGRATING FACTORS
Find the integrating factors and solve the following differential equations
1.
x  y dx  xdy  0 ;
2
2. 2 x tan ydx  ( x 2  2 sin y)dy  0
Ans.   1 ; x  y  C
x2
x
Ans. ln u  ln cos y; x 2 sin y  1 cos 2 y  C
2
Integrate the differential equations with variables separable
5. (1  y ) dx  (1  x) dy  0 ;
6.
t  xt  dx
 x  tx  0 ;
dt
7.
d   tan   0
2
2
2
2
Integrate the following homogeneous differential equations:
8.
( x  y)dx  ( y  x)dy  0 ;
9.
xdy  ydx 
10.
2 st  s dt  tds  0 ;
x 2  y 2 dx ;
Integrate the following linear differential equations:
ds
cos t  s sin t  1;
11 .
dt

13. y 

2 3
14. y  x y  xy  1 ;
12. ds  s cos t  1 sin 2t ;
dt
Integrate the Bernoulli equations:
15.
2
n
y  ex xn ;
x
 y ln x  2 ydx  xdy ;
16.. y  y cos x  y 2 cos x(1  sin x ) ,
Integrate the following exact differential equations:
17. ( y 3  x ) y   y ;
18. 
1
y2
1
x2 

dx


dy  0 ;


2
2
x
 ( x  y)
 y ( x  y) 


2
3
2
2
19. 2 3 xy  2 x dx  3(2 x y  y )dy  0 ;
Find the integrating factors and solve the following differential equations


2x
2
20. e  y dx  ydy  0
21.
1  3x sin y dx  x cot ydy  0
2
SUMMARY

Let y  f (x) . Then its first derivative y   f (x) is another function of x. The
geometrical meaning of y   f (x) is that y  is the slope of the tangent line to the
curve y  f (x) at the point (x, y). The physical meaning is that y  is the rate of
change of y at time x.

A first-order differential equation with no initial condition has many solutions and
we call the collections of these solutions a general solution which is dependent on
a single parameter.
If initial conditions are specified, then the differential
equation will have a unique solution.

Equations with variables separable are the simplest type of differential equations.
Other types of differential equations are reduced to equations with variables
separable through different methods.
END OF UNIT ONE