Chapter One
First Order Ordinary Differential Equation
1. Differential Equations
Definition: A differential equation is an equation that contains a function and one or
more of its derivatives. If the function has only one independent variable, then it is an
ordinary differential equation(ODE). Otherwise, it is a partial differential equation
(PDE).
The following are examples of differential equations
 2 u u

 u  e t  u ' 'u 'u  e t
a)
2
t
t
w w

0
 wx  wt  0
b)
x t
-
Ordinary differential equation(ODE)
-
Dependent Variable:
Independent Variable:
-
Partial differential equation(PDE)
Dependent Variable:
Independent Variable:
Order and Degree of Differential Equation
Definition: The order of a differential equation is the order of the highest ordered
derivative that appears in the given equation. The degree of a differential equation is the
exponent of the highest derivative that occurs in the DE, after the DE is expressed as a
polynomial of the dependent variable and its derivative.
Examples:
2
a)
b)
 dy 
2
   y  sin x
 dx 
d 2u
 u 2  cos t
2
dt
Order: 1 Degree: 2
Order: 2 Degree: 1
2w 2w
c)

0
x 2 y 2
d)
u   2u  2u 



t  x 2 y 2 
Order: 2 Degree: 1
3
Order: 2 Degree: 3
Exercise: Find the order and degree of the DEs
5
a.
 y' 2  y' ' '1
Ans :. Order : 3 Degree : 2
b.
y ' '  y  e y ''
Ans : Order : 2 no degree : 3
Remark: In this course we only Consider ODE
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Linear Differential Equations:
 Dependent variables and their derivative are of degree 1
 Each coefficient depends only on the independent variable
 A DE is linear if it has the form
an  x 
dny
d n1 y
d y



a
x
 ...a1  x 
 a0 ( x ) y  g ( x )
n1
n
n 1
dx
dx
dx
Examples:
a)
dy
 y  sin x
dx
b)
d2y
 y  sin 2 x
dx 2
c)
d5y
 sin 2 ( x) y  tan 2 x
dx 5
Nonlinear Differential Equations
 Dependent variables and their derivatives are not of degree 1.
Examples:
dy
a) ( ) 2  y  sin x
dx
d3y 2 d2y 3
y
 ex
b) ( 3 )  ( 2 )  2
dx
dx
x 1
Initial & Boundary Value Problems
Initial conditions: will be given on specified given point
Boundary conditions: will be given on some points
Examples:
a)
b)
y (0)  1; y ' (0)  2
y (1)  5; y ' (2)  2
 Initial condition
 Boundary condition
Initial Value Problems (IVP)
d2y
dy
 2  y  sin x  IVP
2
dx
dx
y (0)  1; y ' (0)  2
 Initial condition
Boundary Value Problems (BVP)
d2y
dy
 2  y  sin x  BVP
2
dx
dx
y (1)  5; y ' (2)  2
 Boundary conditionn
Solution of a Differential Equation
A solution of a differential equation is a function defined explicitly or implicitly by an equation
that satisfies the given equation. The general solution represents all the possible solutions of the
given equation, while a particular solution is a solution of DE that is free of arbitrary
parameters.
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Examples:
a)
y  A cos x  B sin x
 solution of
b)
y  ce x , c  IR
 solution of
c)
( x  1) 2  ( y  3) 2  c 2 ; c  IR
 solution of
y' ' y  0
dy
y
dx
dy
( x  1)

dx
y 3
2
Exercise: Show that y  ce x is the solution of the following DE y '  2 xy
Forming a Differential Equation
Example 1: Find the differential equation for y  x 
A
, by eliminating the constant A.
x
Solution:
Example 1: Form a suitable DE using y  A cos x  B sin x
Solution: y '   A sin x  B cos x  y ' '   A cos x  B sin x  ( y )  y ' '   y
Exercise: Obtain a DE of all circles of radius a and center at (h, k)
1.1 First Order Ordinary Differential Equations (ODE)
Types of first order ODE:
-
Separable equation
Exact equation
Bernoulli Equation
- Homogenous equation
- Linear equation
1.1.1 Separable Equation
General form
f ( x, y ) 
dy
 u ( x)v( y )
dx
Hence this become a separable equation if it can be written as
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dy
 u ( x)dx
v( y )
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Method of Solution: Integrate both sides of equation 
dy
 u ( x)dx
v( y ) 
Example 1: Solve the initial value problem
dy y cos x

, y ( 0)  1
dx 1  2 y 2
Solution:
i.
1 2y2
Separate the functions: (
)dy  (cos x)dx
y
ii.
Integrate both sides:  (
Use your Calculus
knowledge to solve
this problem
1 2y2
)dy   (cos x)dx
y
Answer: ln y  y2  sinx c
iii.
general solution
Use the initial condition given, y (0)  1
ln 1  12  sin 0  c  c  1  ln y  y 2  sin x  1
Particular solution
Note: Some DE may not appear separable initially but through appropriate substitutions, the
DE can be separable.
dy
 ( x  y ) 2 can be reduced to a separable equation by using
dx
substitution z  x  y . Then obtain the solution for the original DE.
Example 2: Show that the DE
Solutions:
i.
ii.
iii.
Differentiate both sides of the substitution wrt x
z  x  y .....................................(*)
dz
dy
dy dz
 1


 1.........(**)
dx
dx
dx dx
Insert (*) and (*) into the DE
dz
dz
1  z 2 
 z 2  1.........(* * *)
dx
dx
1
Write (***) into separable form: 2
dz  dx
z 1
iv. Integrate the separable eqn.
1
 z 1 dz   dx
2
Final answer:
y  tan( x  c)  c
Exercise:
1. Solve the following equations
dy
a. x
 cot x
dx
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dy
 (1  y 2 )  0, y (0)  0
dx
c.
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b.
dy
y

dx x( x  1)
d.
2. Using substitution z  xy , convert x
xy
dy
 4 x
dx
dy
 y  2 x 1  x 2 y 2 to a separable equation.
dx
Hence solve the original equation.
1.1.2 Homogenous Equation
Suppose f ( x, y ) 
dy
. Then f ( x, y ) is homogenous if f (x, y )  f ( x, y ) ,   IR
dx
Method of Solution :
i.
Determine whether the equation homogenous or not
dy
dv
Use substitution y  vx and
 v  x in the original DE
dx
dx
Separate the variable x and v
Integrate both sides
Use initial condition (if given) to find the constant value
ii.
iii.
iv.
v.
Separable
equation method
Example 1: Determine whether the DE is homogenous or not
dy
x2  y2

dx ( x  y )( x  y )
a.
dy
x2  y2
f ( x, y ) 

dx ( x  y )( x  y )
Solution:
( x ) 2  ( y ) 2
2 ( x 2  y 2 )
f (  x,  y ) 

 f ( x, y )
(x  y )(x  y ) 2 (( x  y )( x  y ))
 this differential equation is homogenous
dy y
  x2  y2
Ans: non-homogenous
dx x
Example 2: Solve the homogenous equation ( y 2  xy)dx  x 2 dy  0
f ( x, y ) 
b.
Solution:
i.
ii.
dy y 2  xy

………………………(1)
dx
x2
test for homogeneity: f (x.y )  f ( x, y )
Rearrange the DE:
 this differential equation is homogenous
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iii.
Substitute: y  vx and
dy
dv
vx
dx
dx
into (1)
dv (vx) 2  x(vx)
dv
vx

 v2  v  x
 v2
2
dx
dx
x
Solve the problem using the separable equation method: 
iv.
Final answer : x  Ae
1
1
dv   dx
2
v
x
x
y
Note: Non-homogenous DE can be reduced to a homogenous DE by using substitution.
Example 3: Find the solution for this non-homogenous equation
dy
y2

dx x  y  5
by using the following substitutions x  X  3, y  Y  2
Solutions:
I.
Differentiate (2) and (3): dx  dX , dy  dY
II.
and substitute them into (1),
III.
Test for homogeneity, f (x, y )  f ( x, y )
IV.
Use the substitutions Y  VX and
V.
dV
Y
VX
V
dV
V
V2
VX



X

V  
dX X  Y X  VX V  1
dX V  1
1V
Use the separable equation method to solve the problem
1V
1
  2 dV   dX
X
V
LASTLY, do not forget to
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REMEMBER! Now we use
instead of
dY
dV
V  X
dX
dX
x 3
y 2
replace
y 2  xy
x2
Ans : y  
Ans: y  2  Ae
Exercises: Solve y ' ' 
dY
Y

dX X  Y
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with
x
ln x  c
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1.1.3 Exact Equation
Suppose f ( x, y )  
M ( x, y )
N ( x, y )
dy M(x, y)

 M(x, y)dx N(x, y)dy  0

 

Therefore the first order DE is given by dx
N(x, y) 
u
u
x
Condition for an exact equation:
If
y
M N

y
x
M N
u
u

, then there exist a function u ( x, y ) such that
 M,
N
y
x
x
y
Method of Solution (Method 1):
i.
Write the DE in the form M ( x, y )dx  N ( x, y )dy  0 and test exactness
ii.
If the DE is exact, then M 
M N

y
x
u
u
, N
………………………………(1)
x
y
u ( x, y )   M ( x, y )dx   ( y ) ……..(2)
To find u ( x, y ) , integrate (1) wrt x :


u 

M ( x, y )dx   ' ( y )  N
y y 
iii.
To determine  ( y ) , differentiate (2) wrt y:
iv.
Integrate  ' ( y ) to get  ( y )
v.
Replace  ( y ) into (2). If there are any initial conditions given, substitute the condition
vi.
into the solution.
Write down the solution in the form: u ( x, y )  A, where A is a constant
Method of Solution (Method 2):
i.
Write the DE in the form M ( x, y )dx  N ( x, y )dy  0 and test for the exactness
ii.
If the DE is exact, then M 
iii.
To find u ( x, y ) from, integrate (1) wrt x to get
u
u
, N
……………… ……………………(1, 2)
x
y
u ( x, y )   M ( x, y )dx  1 ( y ) ….. ……………………..(3)
v.
To find u ( x, y ) from, integrate (2) wrt y to get u ( x, y )   N ( x, y )dy   2 ( x)........(4)
vi.
Compare 3 and 4 to get value for 1 ( y ) and  2 ( x) .
vii.
Replace 1 ( y ) into (3) OR  2 ( x) into (4)
viii.
ix.
If there are any initial conditions given, substitute the conditions into the solution.
Write down the solution in the form: u ( x, y )  A, where A is a constant
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Example 1: Solve (2 xy  3)dx  ( x 2  1)dy  0
Solution (Method 1):
i)
Check the exactness
ii)
Find u ( x, y ) ,
M
N
 2x 
 this equation is exact.
y
x
u
 2 xy  3  M ( x, y )..........................................................( 1)
dx
u
 x 2  1  N ( x, y ).............................................................(2)
dy
To find u ( x, y ) , integrate either (1) or (2), let’s say we take (1)
 u   (2 xy  3)x  u ( x, y)  x y  3x   ( y)..........................(3)
2
iv.
u
N
y
Now we differentiate (3) wrt y to compare with
u
 x 2   ' ( y )........................................................................( 4)
y
Now, let’s compare (4) with (2): x 2   ' ( y )  x 2  1   ' ( y )  1
v.
Find  ( y ) ,   ' ( y )    1dy   ( y )   y  B
vi.
Now that we found  ( y ) , our new u ( x, y ) should looks like this:
u ( x, y )  x 2 y  3 x  y  B
vii.
Write the solution in the form u ( x, y )  A,
u ( x, y )  x 2 y  3 x  y  B  A  x 2 y  3 x  y  C , where C  A  B is a constant
Exercise : Try to solve Example 1 by using Method 2
Answer:
M
N
 2x 
 this equation is exact.
y
x
i.
Since
ii.
Find u ( x, y ) ,
u
 2 xy  3  M ( x, y )...........................................................( 1)
dx
u
 x 2  1  N ( x, y )..............................................................(2)
dy
To find u ( x, y ) , integrate both (1) and (2),
 u   (2 xy  3)x  u ( x, y)  x y  3x   ( y)...................(3)
2
1
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 u   ( x  1)y  u ( x, y)  x y  y   ( x)...................(4)
2
2
2
iii.
Compare u ( x, y ) to determine the value of 1 ( y ) and  2 ( x)
x 2 y  3 x  1 ( y )  x 2 y  y   2 ( x)
Hence 1 ( y )   y and  2 ( x)  3 x
iv.
Replace 1 ( y ) into (3) OR  2 ( x) into (4)
u ( x, y )  x 2 y  y  3 x
v.
Write the solution in the form u ( x, y )  A
x 2 y  y  3x  A
Note: Suppose non-exact equation can be turned into exact equation by multiplying it
with a function say  ( x, y ) . Then  ( x, y ) is called an integrating factor.
Theorem: Consider the DE
M ( x, y )dx  N ( x, y )dy  0 which is not exact but
M ( x, y )dx  N ( x, y )dy  0 is exact, then
1
1
( M y  N x ) are independent of y and say
( M y  N x )  g ( x) then
N
N
 If  and
 (x)  e 
g ( x ) dx
1
1
( M y  N x ) are independent of x and say
( M y  N x )  h( y ) then
M
M
 If  and
 ( y)  e 
Proof: If M ( x, y )dx  N ( x, y )dy  0 exact, then

h ( y ) dy



M 
N
( M ) 
( N ) 
M 

N
y
x
y
y
x
x


  (M y  N x ) 
N M
x
y
Case1: If  and
1
( M y  N x ) are independent of y , then
N
(M y Nx )
(M y  N x )
x


1
 (M y  N x ) 
N 0 
 x       e N
x
N

Case 2: If  and
1
( M y  N x ) are independent of x , then
M
(M y Nx )
(M y  N x )

x

1
M
 (M y  N x )  0  M
 
 x       e
y
M

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Example 2: Show that the equation ( xy  y 2  y )dx  ( x  2 y )dy  0 is not exact. And by
using integrating factor,  ( x, y )  e x solve the equation.
Solution:
i.
ii.
Show that it is not exact
M
N
M N
 x  2 y  1,
 1  Since

,
y
x
y
x
this equation is not exact.
Multiply  ( x, y ) into the DE to make the equation exact
( xy  y 2  y )(e x )dx  ( x  2 y )(e x )dy  0


M
N
M
N
 xe x  2e x y  e x 
 the equation is exact.
y
x
iii.
Check the exactness again:
iv.
u
 ( xy  y 2  y )e x  M ( x, y ) ..............................................(1)
x
Find u ( x, y ) ,
u
 ( x  2 y )e x  N ( x, y ) ....................................................(2)
y
To find u ( x, y ) , integrate either (1) or (2), let’s say we take (2)
 u   ( x  2 y)(e )y  u ( x, y)  ( xy  y )(e )   ( x)..............................(3)
x
v.
2
x
u 
 (( xy  y 2 )(e x )   ( x))  ( xy  y 2  y )(e x )   ' ( x).....(4)
Find  (x) , x x
Now, let’s compare (4) with (1)
( xy  y 2  y )(e x )   ' ( x)  ( xy  y 2  y )e x   ' ( x)  0   ( x)  B
vi.
Write the solution in the form u ( x, y )  A,
u ( x, y )  ( xy  y 2  y )(e x )  B  A
u ( x, y )  ( xy  y 2  y )(e x )  C , where C  A  B
Example 3: Find an integrating factor for the DE (3 x 2  y 2 )dy  2 xydx  0
Solution: M ( x, y )  2 xy and N ( x, y )  3x 2  y 2
M
N
 2 x ,
 6 x  2 x  6 x  the equation is not exact.
y
x
But
1
1
4
(M y  N x ) 
(2 x  6 x)  which is independent of x
M
2 xy
y
Therefore  ( y )  e
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
4
 y dy
 y  4 . Hence  2 xy 3 dx  y 4 ( y 2  3 x 2 )dy  0 is exact DE.
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Exercises :
1. Try solving Example 2 by using method 2.
2. Determine whether the following equation is exact. If it is, then solve it.
a. (2 x  y )dx  ( x  2 y )dy  0
b. (cos x cos y  2 x)dx  (sin x sin y  2 y )dy  0
3. Find the integration factor for the DEs
Ans :  ( x)  x 3
2 sin( y 2 )dx  xy cos( y 2 )dy  0
4. Solve
a. 2 xydx  ( x 2  1)dy  0
Ans : x 2 y  y  c
b. 2 x  y 3  (3 xy 2  e  2 y ) y '  0
Ans : x 2  y 3 x 
NOTE: Must be +
here!!
1.1.4 Linear First Order Differential Equation
The general form of the first order linear DE is given by 1
e 2 y
c0
2
dy
 p ( x) y  q ( x )
dx
Method of Solution 1:
i. Determine the value of p (x) and q (x) such the coefficient of
ii. Calculate the integrating factor:  ( x)  e 
dy
is 1.
dx
p ( x ) dx
d
(  ( x))   ( x)q( x)   ( x) y    ( x)q( x)dx
dx
1
iv. The general solution is given by: y 
 ( x)q( x)dx
 ( x) 
iii. Write the equation in the form of:
Method of Solution 2:
To find the general solution multiply the DE by e 
p ( x ) dx
p ( x ) dx
p ( x ) dx
d
( ye 
)  q ( x )e 
dx
Is the general
p
(
x
)
dx
p
(
x
)
dx

p
(
x
) dx 
 p ( x ) dx )dx 
ye 
  ( q ( x )e 
)dx  y  e 
(
q
(
x
)
e

solution


e
AASTU
p ( x ) dx
y ' e 
p ( x ) dx
p ( x ) y  q ( x )e 
p ( x ) dx
11

Banchi K.
Example 1: Solve this first order DE
dy 1  x
ex

y
dx
x
x
Solution:
i.
Determine p (x) and q (x) : p( x) 
ii.
Find integrating factor,  ( x)  e 
iii.
Write down the equation:
1 x
ex
, q( x) 
x
x
p ( x ) dx
1 x
e
 x dx
 e (ln x  x )  xe x

d
ex
1 e 2x
( xe x y )  ( xe x )( )  xe x y   e 2 x dx  y  x 
 c
dx
x
xe  2

Final answer: y 
iv.
ex
c
 x
2 x xe
Note: Non-linear DE can be converted into linear DE by using the right substitution.
Example 2: Using, z  y 2 convert the following non-linear DE into linear DE and solve it
x2 y
dy
 xy 2  1 ; y (1)  1
dx
Solutions:
i.
ii.
dy
and replace into the nonlinear equation
dx
dz
dy
1 dz
dy
 2y 
y
dx
dx
2 dx
dx
dy
1 dz dy
x2 y
 xy 2  1  x 2
 xz  1
dx
2 dx dx
Differentiate z  y 2 to get
Change the equation into the general form of linear equation & determine p and q :
dz
2
2
dz 2
2
 ( 2 )( xz )  ( 2 )1   ( ) z  2
dx
dx 
x
x
x
x

q
p
2
iii.
iv.
AASTU
  dx
p ( x ) dx
1
Find the integrating factor,  ( x)  e 
  ( x)  e x  2
x
d 1
1 2
2
z
2
2
Find y :
( 2 z )  ( 2 )( 2 )  4  2   4 dx  z  x 2 ( 2  c)
dx x
x x
x
x
x
3x
12
Banchi K.
2
 cx 2
3x
Use the initial condition given, y (1)  1
Since z  y 2 ,
v.
12 
y2 
2
5
2 5 2
 c(1) 2  c   y 2 
 x
3(1)
3
3x 3
Exercise: Solve the following DEs
a. xy ' y  2 x , y (1)  0
b. y '5 y  x ,
c. y ' y cos x  sin x
c
1
 G.S and y  x  , 0  x  8  P.S
x
x
5 1
Ans : y  
 ce 5 x  G.S
x 25
1
d. y ' y 
1  e2x
Ans : y  x 
1.1.5. Bernoulli Equation
dy
 b( x) y  c( x) y n ......(1) , where n  0, n  1 .
dx
To reduce the equation to a linear equation, use substitution z  y 1 n ..........(2)
The general form of the Bernoulli equation is:
Method of Solution:
i.
ii.
dy
 b( x) y 1 n  c( x)...........................(3)
dx
dy
Differentiate ( 2 ) wrt x ( to get
)
dx
dy
dy
dz
1 dz
 (1  n) y  n

 y n
..........................................(4)
dx
dx
(1  n) dx
dx
Divide ( 1 ) with y n  y  n
iii.
Replace ( 4 ) into ( 3 )
1 dz
dz
 b( x ) z  c ( x ) 
 (1  n)b( x) z  (1  n)c( x)


(1  n) dx
dx 
p( x)
q(x)
iv.
Solve using the linear equation solution
 Find integrating factor  ( x)  e 
 Solve
AASTU
p ( x ) dx
d
(  ( x ) z )   ( x )q ( x )
dx
13
Banchi K.
Example 1: Solve
dy 1
 y  ex y4
dx 3
Solutions:
i.
Determine n  4
ii.
Divide the DE by y 4 :
iii.
Using substitution, z  y 3
1 dy 1 3
 y  e x .............(2)
4
y dx 3
dy
dz
1 dy
1 dz
 3 y  4
 4

...............................(3)
dx
dx
3 dx
y dx
iv.
v.
vi.
vii.
Replace ( 3 ) into ( 2 ) and write into linear equation form
1 dz 1
dz

 z  ex 
 z  3e x  p ( x)  1, q ( x)  3e x
3 dx 3
dx
 1dx
Find the integrating factor  ( x)  e   e x
Solve the problem
d x
1
(e z )  e  x (3e x )  z    x  3dx  e x 3 x  c 
dx
e
3
3
x
Since z  y ,
y  ce  3 xe x
Exercise:
Answer:
dy 1
 y  xy 2
dx x
1
x2 c


y
3 x
2. xy ' y '  x 2 y 2 , x  0
1
x2 c


y
3 x
1.
3.
dy y
  y2
dx x
1
 x(c  ln x)
y
dy
 y  xy 3
dx
dy 2
5.
 y   x 2 y 2 cos x
dx x
y2 
4. x
1
2 x  cx 2
1
 x 2 (sin x  c)
y
dy
(4 x  5) 2 3
6. 2  (tan x) y 
y
dx
cos x
1
(4 x  5) 3
c


2
12 cos x cos x
y
7. Given the differential equation, (2 x 4 y ) dy  (4 x 3 y 2  x 3 ) dx  0
Show that the equation is exact. Hence solve it.
8. The equation in Question 6 can be rewritten as a Bernoulli equation,
dy
1
2y  4y  . By using the substitution, z  y 2 solve this equation. Check the
dx
y
answer with Question 6.
AASTU
14
Banchi K.