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First Order Differential Equations

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Page 1
Outline

Motivation

Basic Concepts

Solving of First-Order Differential Equations


Direct Integration

Separable Differential Equations

substitution Methods
Linear Differential Equations
Page 2
Motivation
Page 3
Basic Concepts
Page 4
Basic Concepts
Page 5
Basic Concepts
Page 6
First-Order Differential Equations
Page 7
Solving of First-Order Differential Equations



Solving a differential equation means finding an
equation with no derivatives that satisfies the given
differential equation.
Solving a differential equation always involves one
or more integration steps.
It is important to be able to identify the type of
differential equation we are dealing with before we
attempt to solve it.
Page 8
Direct Integration
Page 9
Separable Differential Equations
Page 10
Separable Differential Equations
Page 11
Exercises
Verify the solution
xy'
y 
 2y
x
2
Page 12
Explicit & Implicit Solution

Explicit Solution
y  h(x )

Implicit Solution
H ( x, y )  0
Page 13
General solution & Particular solution
y ' 
y
c


cosx
sinx
 c
 3 ,  2 ,....
Page 14
General solution & Particular solution
y   ky
y (t )  ce
y (t )  2e
kt
kt
Page 15
Exercises
Example:
9 yy   4 x  0
Sol:
Page 16
Exercises
Example:
y  1  y
2
Sol:
Page 17
Exercises
Example:
y   ky
Sol:
Page 18
Exercises
Example:
y   2 xy , y (0)  1
Sol:
Page 19
Exercises
Substitution Method:
A differential equation of the form
y
y  g ( )
x
can be transformed into a separable differential
equation
Page 20
Substitution Method
Substitution Method:
y  ux
y   u x  u
u x  u  g ( u )
 u x  g ( u )  u
du
dx


g (u )  u
x
Page 21
Substitution Method
Example:
Sol:
2 xyy   y  x
2
2
2 xy y   y 2  x 2
y2
x2
1 y
x
 y 

 (  )
2 xy
2 xy
2 x
y
1
1
 u x  u  ( u  )
2
u
dx
2 udu



1  u2
x
1
 ln( 1  u 2 )   ln x  c 1  ln
 c1
x
c
 1  u2 
x
2
c
 y
 1   
x
 x 
 x 2  y 2  cx
Page 22
Substitution Method
Exercise 1
y   1  0.01 y 2
y   xy / 2
xy   y 2  y
xy ' y  0, y ( 2)  2
Page 23
Linear Differential Equations

Def: A first-order differential equation is said to be
linear if it can be written
y  p( x ) y  r ( x )

If r(x) = 0, this equation is said to be homogeneous
Page 24
Linear Differential Equations

How to solve first-order linear homogeneous ODE ?
y  p( x ) y  0
Sol:
dy
 p( x ) y  0
dx
dy

  p( x )dx
y
 ln y    p( x )dx  c
1
 ye
  p ( x ) dx  c1
e
  p ( x ) dx c1
e  ce
  p ( x ) dx
Page 25
Linear Differential Equations
Example:
y  y  0
Sol:
p ( x ) dx

y ( x )  ce

 ce 
 (  1 ) dx
 ce
x  c1
 ce e
x
c1
 c2e x
Page 26
Linear Differential Equations

How to solve first-order linear nonhomogeneous ODE ?
Sol:
y  p( x ) y  r( x )
dy
 p( x ) y  r ( x )
dx
 ( p ( x ) y  r ( x ))dx  dy  0
1 dF 1


 ( Py  Qx )  ( p( x ) y  r ( x ))  p( x )
F dx Q
y
Page 27
Linear Differential Equations
Sol:
p ( x ) dx

F ( x)  e
p ( x ) dx
p ( x ) dx
p ( x ) dx



e
( y   py )  ( e
y )  e
r
p ( x ) dx
p ( x ) dx


e
y  e
rdx  c
 y( x)  e 
 p ( x ) dx
 e  p ( x ) dx rdx  c 
 

Page 28
Linear Differential Equations
y  y  e
Example:
2x
Sol:
y( x)  e 
 p ( x ) dx
e 
 ( 1) dx

 e  p ( x ) dx rdx  c 
 

 e  ( 1) dx e 2 x dx  c 
 

 e x  e  x e 2 x dx  c

 ex ex  c


 ce x  e 2 x
Page 29
Linear Differential Equations
Example:
y  2y  e (3 sin2x  2 cos2x)
'
x
Page 30
Bernoulli, Jocob
Bernoulli, Jocob
1654-1705
Page 31
Linear Differential Equations

Def: Bernoulli equations
y  p( x ) y  g ( x ) y
a

If a = 0, Bernoulli Eq. => First Order Linear Eq.

If a <> 0, let u = y1-a
u  (1  a ) pu  (1  a ) g
Page 32
Linear Differential Equations
y   Ay   By
Example:
2
Sol:
u  y1a  y12  y 1
 u   y 2 y    y 2 (  By 2  Ay )  B  Ay 1
 u  Au  B
ue
 Ax
 Be
Ax

dx  c  e
 Ax
B
 B Ax

 Ax
 A e dx  c   ce  A
1
1
y 
u ce  Ax  B
A
Page 33
Linear Differential Equations
Exercise 3
y  y  4
y   ky  e  kx
y   3 y  sin x, y ( )
2
2

y  2y  y
y   xy  xy 1
Page 34
Summary
Separable
g ( y ) dy  f ( x ) dx

Substitution
g ( u ) du  f ( x ) dx

Exact
M ( x , y ) dx  N ( x , y ) dy  0

Integrating Factor FPdx  FQdy  0

y  p( x) y  r( x)

Linear
Bernoulli
y   p ( x ) y  g ( x ) y a 
Page 35
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