This document discusses methods for solving first order differential equations. It introduces seven methods: variable separable, homogeneous differential equations, exact differential equations, linear differential equations, and nonlinear differential equations. It provides examples of using separation of variables and the method of homogeneous equations. It also discusses the conditions for an equation to be exact and provides steps for solving exact differential equations.

2. ADVANCE ENGINEERING
MATHEMATICS
O.D.E OF FIRST ORDER

3. Method of Solution: 1) Variable Seperable
2) Homogenous Differential Eq.
3) Non Homogenous Differential Eq.
4) Exact Differential Eq.
5) Non Exact Differential Eq.
6) Linear Differential Eq.
7) Non Linear Differential Eq.
Differential Equation of first Order and first Degree

4. Differential Equation of first
Order and first Degree
where f (x, y) is the function of x and y.
A differential equation of the first order and first degree contains
independent variable x, dependent variable y and its derivative
 
dy dy dy
i.e. = ƒ x, y or ƒ x, y, = 0
dx dx dx
 
 
 
   2 dy x + y
For example : xy y +1 dy = x +1 dx, = ,
dx x - y
dy
+ y=sinx etc.
dx

5. Separation of
Variables
 
dy
= ƒ x
dx
Differential equation of the form
 dy = ƒ x dx
 dy = ƒ x dx +C
 
[where C is an arbitrary constant] y = ƒ x dx +C


6. Separation of Variables
 
1
dx = dy
ƒ y

[where C is an arbitrary constant]
 
dy
= ƒ y
dx
Differential equation of the form
 
 
dx 1
= , provided f y 0
dy ƒ y
 
 
1
dx = dy+C
ƒ y

 
 
1
x = dy +C
ƒ y



7. Separation of
Variables
 
 
dy
= p x dx
q y

[where C is an arbitrary constant]
   
dy
=p x q y
dx
Differential equation of the form
 
 
1
dy = p x dx+C
q y

 

8. Example
dy
Solve the differential equation = x -1+ xy - y.
dx
dy
Solution: The given differential equation is = x -1+xy - y.
dx
   
dy
= x -1 +y x -1
dx

  
dy
= x -1 y+1
dx

   
dy
= x -1 dx Variable separable form
y+1


9. Solution Cont.
Integrating both sides, we get
 
dy
= x -1 dx
y+1 
2
e
x
log y+1 = - x+C
2


10. Example
2
2
dy 1- y
Solve the differential equation + = 0.
dx 1- x
2
2
dy 1- y
Solution: The given differential equation is + = 0.
dx 1- x
2
2
dy 1- y
= -
dx 1- x

 
2 2
1 1
dy = - dx Variable separable form
1- y 1- x

2 2
1 1
dy = - dx
1- y 1- x 

11. Solution Cont.
2 2
1 1
dy = - dx
1- y 1- x 
Integrating both sides, we get
-1 -1
sin y = -sin x +C
-1 -1
sin y +sin x = C

12. Example
     2 2
Solve the differential equation
1+ x 1+ y dx + 1+ y 1+ x dy = 0.
     2 2
Solution: The given differential equation is
1+x 1+y dx+ 1+y 1+x dy =0.
     2 2
1+x 1+y dx =- 1+y 1+x dy
 2 2
1+ y1+ x
dx = - dy Variable separable form
1+ x 1+ y
  
        

13. Solution Cont.
Integrating both sides, we get
2 2
1+ y1+ x
dx = - dy
1+ x 1+ y
  
        
2 2 2 2
1 1 2x 1 1 2y
dx+ dx =- dy- dy
2 21+x 1+x 1+y 1+y

   
-1 2 -1 2
e e
1 1
tan x+ log 1+x =-tan y - log 1+y +C
2 2

 
 
 e
ƒ x
dx =log ƒ x
ƒ x
 
 
  Q

14. Solution Cont.
-1 -1 2 2
e e
1 1
tan x + tan y + log 1+ x + log 1+ y = C
2 2

  -1 2 2
e
x+y 1
tan + log 1+x 1+y = C
1- xy 2
 
  
 

15. Homogeneous Function
A function f (x, y) in x and y is called a
homogenous function, if the degrees of each term
are equal.
Examples:
  2 2
g x, y = x - xy + y (is a homogenous function of
degree 2)
  3 2 2
f x, y = x +3x y +2y x is a homogenous function of
degree 3

16. Method of Solution
(2) Substitute y =vx and
dy dv
= v+x
dx dx
in the equation.
(3) The equation reduces to the form
dv
v+x =F(v)
dx
(4) Separate the variables of v and x.
(5) Integrate both sides to obtain the solution in terms
of v and x.
(6) Replace v by
y
x
to get the solution
f(x, y)dy
=
dx g(x, y)
(1) Write the differential equation in the form

17. Example
dy
Solve the differential equation x = x+ y.
dx
dy
Solution: The given differential equation is x = x+y.
dx
dy x+y
= …(i)
dx x

It is a homogeneous differential equation of degree 1.
 
dy dv
Putting y = vx and = v+x ... i , we get
dx dx
dv x+vx
v+x =
dx x

18. Solution Cont.
dv
v+x =1+v
dx

dv
x =1
dx

1
dv = dx
x

Integrating both sides, we get
1
dv = dx
x 
ev =log |x|+C
 e
y
= log | x |+C y = vx
x
 Q
ey = xlog |x|+Cx

19. Example
Solve the differential equation: 2 2dy
xy = x - y
dx
 
2 2
2 2dy dy x - y
Solution: We have xy = x - y = ... i
dx dx xy

It is a homogeneous differential equation of degree 2.
dy dv
Substituting y = vx and = v+x in (i), we get
dx dx
2 2 2 2
2
dv x - v x dv 1- v
v + x = x = - v
dx dx vvx


20. Solution
2 2 2
dv 1- v - v dv 1-2v
x = x =
dx v dx v
 
2 2
v dx -4v dx
dv= dv= -4
x x1-2v 1-2v
 
2
-4v dx
dv =-4
x1-2v

 
 2 2
e e 4 4
C C
log 1-2v =log 1-2v =
x x
  
2
e e elog 1-2v =-4log x +log C
[Integrating both sides]

21. Solution Cont.
2
4
y C
1- 2 =
x x
 
  
 
2 2
2 4
x - 2y C
=
x x

   2 2 2
x x -2y =C C can be +ve or -ve
2
4
y C
or 1-2 =-
x x
 
 
 
2 2
2 4
x -2y C
=-
x x
or

22. 1st Order DE – Exact Equation
The differential equation M(x,y)dx + N(x,y)dy = 0 is an exact
equation if :
The solutions are given by the
implicit equation
x
N
y
M





  CyxF ,
where : F/ x = M(x,y) and F/ y = N(x,y)

23. 1. Integrate either M(x,y) with respect to x or
N(x,y) to y.
Assume integrating M(x,y), then :
Solution :
     ydxyxMyxF   ,,
2. Now :
      yxNydxyxM
yy
F
,', 





 
or :
      

 dxyxM
y
yxNy ,,'

25. Example

26. THANKING YOU
1. 160280102021 MADAM AJAY
2. 160280102023 MAKVANA SHYAM
3. 160280102024 MAKWANA KRUNAL
4. 160280102025 MODI HARSHIT
5. 160280102026 NANDANIYA PRATIK
6. 160280102028 OZA JAYNESH
7. 160280102029 PANCHAL DIVYANG
8. 160280102030 PARMAR CHITRANG