chapter 1 - differential equations
TRANSCRIPT
-
8/20/2019 Chapter 1 - Differential Equations
1/27
CHAPTER 1 DIFFERENTIAL EQUATIONS
1.1 INTRODUCTION
1.1.1 DIFFERENTIAL EQUATIONS
• What is the meaning of differential equations?
• In sciences and engineering, mathematical models are developed to aid in
understanding of physical phenomena.
EXA!LE 1
Any equations containingn
n
d y
dx, we called differential equations. The differential equations
can be written as (i 2 0dy
x ydx
− = 2 0 xy y′ − =
(ii2
2
2 sin 0
d y xy y x
dx− = 2 sin 0 xyy y x′′ − =
1 | P a g e
So"e of a##li$ations in%ol%ed in t&is 'orld(
free fall of ob!ect
radioactive decay
electric circuit
management
rate of change in temperature
mi"ing problem in tan#
DEFINITION
An equation containing the derivatives of one or more
dependent variables, with respect to one or more independent
variables, is said to be Differential Equations )DE*.
$athematical models often yield an equation that
contains some derivatives of an un#nown function
which is called differential equations.
-
8/20/2019 Chapter 1 - Differential Equations
2/27
• In order to tal# further about differential equations, we shall classify differential
equations by t+#e, order and linearit+.
EXA!LE ,
%rdinary &ifferential 'quations (%&' An equation that involve onl+ ordinar+ deri%ati%es
of one or more dependent variable with respect to a sin-le inde#endent %ariale. )uch
that
EXA!LE /
*artial &ifferential 'quations (*&' An equation that involve #artial deri%ati%es of one or
more dependent variable with respect to t'o or "ore t&an one inde#endent %ariale.
)uch that
1.1., ORDER OF DIFFERENTIAL EQUATIONS
%rder of &ifferential 'quations is determined by the highest derivative.
2 | P a g e
T+#es of differential equations(%rdinary &ifferential 'quations (%&'
*artial &ifferential 'quations (*&'
(i2
2 0
d x dxa kx
dt dt + + = + )ingle independent variable, t
(ii 2dy
x ydx
= − + + )ingle independent variable, x
(i) 2u u
x y x y
∂ ∂− = −
∂ ∂+ $ultiple independent variable, x
and y
ii 3V V
r ∂ ∂
+ = + $ulti le inde endent variable, r and h
Order of differential equations(
irst %rder
)econd %rder
-
8/20/2019 Chapter 1 - Differential Equations
3/27
EXA!LE 0
Order Ea"#le
irst 2 0dy
x ydx − =
2 0 xy y
′ − =
)econd
2
2
2
ydx
yd x −
0" 2
=− y xy
EXA!LE 2
-lassify each of the following as an ordinary differential equation (%&' or a partial
differential equations (*&' and give the order
(i -ompetition between two species, ecology+(2 3 )
(1 3 )
dy y x
dx x y
−=
−.
(ii aplace/s equation, heat, aerodynamics+
2 2
2 2 0
d u d u
dx dy+ = .
)olution
T+#es of differential equations Order of differential equations
(i %&' irst order
(ii *&' )econd %rder
1.1./ LINEARIT3 OF DIFFERENTIAL EQUATIONS
A linear differential equations is any differential equation that can be written in the
following form ( ) ( ) ( ) ( ) ( )1 1
1 1 01 1...
n n
n nn n
d y d y d ya x a x a x a x y F x
dx dx dx
−
− −+ + + + = .
Where ( ) ( ) ( ) ( )1 1 0, ,..., ,n na x a x a x a x− are depending on independent variable, it can be 0ero
or non10ero functions, constant or non1constant functions, linear or non1linear functions.
3 | P a g e
Order of differential equations(inear 2onlinear
-
8/20/2019 Chapter 1 - Differential Equations
4/27
• Criteria of linear differential equations(
i the dependent variable and its derivatives occur to the first #o'er onl+
'"ample2 3
2
d y
dx
ii no #rodu$ts in%ol%in- t&e de#endent %ariale with its derivatives (in other word
each coefficient depends only on the independent variable, "
'"ampledy
ydx
iii no nonlinear fun$tions of t&e de#endent %ariale as trigonometric, quadratic,
e"ponential, etc
'"ample sin y (trigonometric functions, 2 y (quadratic equations, ye (e"ponential
A nonlinear ordinar+ differential equation is simply one that is not linear. 2onlinear
functions of the dependent variable or its derivatives, such as sin y or ye , cannot appear in
the linear equation.
EXA!LE 4&etermine each of the following equation with order and linearity
(i2
3
2 0
d y y
dx+ =
(ii2
3
2
d y y x
dx+ =
(iii2
2 cos
d y dy y x
dx dx− =
(iv sin dy
x y xdx
+ =
(v 3dx x t
dt + =
(vi2
2
2 0
d y y
dx+ =
(vii sin 0dy ydx
+ =
Solution :
Order of differential equations Linearit+
(i)econd order 2onlinear because the dependent
variable occur to the power of three,3 y
4 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
5/27
(ii)econd order inear because the dependent
variable, y occur to the first power
(iii )econd order 2onlinear because
dy
y dx
(iv
(v
(vi
(vii
1., T5E SOLUTION OF DIFFERENTIAL EQUATIONS
• What is t&e solution of differential equations6
EXA!LE 7
3erify that x y e= is a solution of the differential equation
dy y
dx= .
Solution :
If x y e= , then xdy
edx
= .
dy y
dx= , for all values of x .
Therefore, x y e= is the solution for
dy y
dx= .
5 | P a g e
The solution of a differential equation is a relationship
between the deendent and indeendent variables
such that the differential equation is satisfied for all
values of the independent variable over a specified
domain.
-
8/20/2019 Chapter 1 - Differential Equations
6/27
EXA!LE 8
3erify that x y Ce= is a solution of dy ydx =, where C is any constant.
Solution:
If x y Ce= , then x
dyCe
dx= .
dy y
dx= , for all values of x with C for any constant.
C is called an arbitrary constant.
Therefore, x y Ce= is the -eneral solution of
dy y
dx= .
If 0, 4; x y= =
04
4 (1)
4
x y Ce
Ce
C
C
=
===
Therefore, 4 x y e= . This is called a #arti$ular solution.
1.,.1 CONDITION
1.,.1.1 Initial Condition
)olution to the differential equation on an interval, I that satisfies at 0 x the n initial
condition.
0( )o y x y=
1( )ody
x ydx
= or 1'( )o y x y=
6 | P a g e
Conditions
Initial-onditions
4oundary-onditions
-
8/20/2019 Chapter 1 - Differential Equations
7/27
1
0 11 ( )
n
nn
d y x y
dx
−
−− = Where 0 x I ∈ and 1 2 -1, , , n y y y… are given constant.
A differential equation together with an initial condition is called an Initial 9alue !role"
)I9!* . )uch that 0 0( , ) , ( ) y f x y y x y′ = = ;
• With x as independent variable (instead of t
• 0 x and 0 y are given values
• The initial condition ( )0 0 y x y= is used to determine the value of A and 4 in the
general solution• 3alues of a function and its derivative at the same point• 2umber of initial conditions for a given differential equation depends upon the order
of the differential equation.
EXA!LE :
The following e"ample are the Initial 3alue *roblem (I3*
(i , (0) 1 y xy y′ = − =
(ii 2 1 34 12 3 0 , (4) , (4)
8 64 x y xy y y y′′ ′ ′+ + = = = −
1.,.1., ;oundar+ Condition
)olution to the differential equations on an interval, I that specified at two distinct points j x
and k x where ( ) , ( ) j j k k y x y y x y= = .
igure 5
A differential equation together with a boundary condition is called an ;oundar+ 9alue
!role" );9!*. )uch that ( , ) , ( ) , ( ) j j k k y f x y y x y y x y′ = = = ;
• 3alues of a function and its derivative not at the same point
7 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
8/27
• 2umber of boundary conditions for a given differential equations does not depends
upon the order of the differential equations
EXA!LE 1<
The following e"amples are the 4oundary 3alue *roblem (43*(i 2 2 0 , (0) 1, ( 2) 0 y y y y y π ′′ ′+ + = = =
(ii 23 8 3 0 , ( 3) 1, (3) 1 y y y y y e′′ ′− − = − = =
1.,., FIRST ORDER DIFFERENTIAL EQUATIONS
In this section, we will learn how to solve the first order differential equations. 4ut in order to
do that, we need to understand what type of equations we are dealing with.
T&e t+#es of "et&od that we are going to discuss are
1* Se#arale equation
,* Inte-ral Fa$tor
1.,.,.1 Se#arale Equation
A first order differential equations of the form, ( ) ( )dy
g x h ydx = is said to be seara!le or have
to searate t"e #aria!les.
EXA!LE 11
6ow to separate the variables of2 3 4 x ydy y xe
dx
+= ?
Solution:
irstly, we #now that ( , )dy f x ydx = .
Then, factori0e2 3 4 x ydy y xe
dx
+= to be 3 2 4( , ) ( )( ) x y f x y xe y e=
8 | P a g e
( )h y( ) g x
6ow to solve the differential equations by using
se#arale equation "et&od? '"plain generally.
-
8/20/2019 Chapter 1 - Differential Equations
9/27
7iven that ( ) ( )dy
g x h ydx
= can be rewritten to isolate the variables x and y on the opposite
side of the equation. It is such as,
1( )
( )dy g x dx
h y=
Then we integrate both sides from the above equation+
1( )
( )dy g x dx
h y=∫ ∫
And we obtain+
1 2( ) ( )
( ) ( )
H y C G x C
H y G x C
+ = +
= + Where 1 2 C C C = + .
Therefore, ( ) ( ) H y G x C = + is the -eneral solution for ( ) ( )dy
g x h ydx
= .
EXA!LE 1,
)olve the nonlinear equation 25dy x
dx y
−= .
Solution:• Step 1 )eparate the variables of equation
2 ( 5) y dy x dx= −
• Step 2 Integrate both sides of the equation2 ( 5) y dy x dx= −∫ ∫
• Step 3 The general solution is3 2
53 2
y x x c= − + .
EXA!LE 1/
)olve the equation (1 ) 0 x dy ydx+ − = .
Solution:• Step 1 )eparate the variables of equation
9 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
10/27
(1 )
1 1
1
x dy ydx
dy dx y x
+ =
=+
• Step 2 Integrate both sides of the equation
1 1
1
ln ln(1 )
dy dx y x
y x c
=+
= + +
∫ ∫
• Step 3 The general solution is
1
(1 )
c
c
ye
x
y x e
=+= +
1.,.,., Inte-ral fa$tor
The linear first order differential equations can be e"pressed in the form of
1 0( ) ( ) ( )dy
a x a x y b xdx
+ = , Where ( )1a x , ( )0a x and ( )b x depends only on the independent
variable x .
or e"ample, the equation2(sin ) (cos ) sin
dy x x y x x
dx+ =
irst, we have to divide all terms in equation 1 0( ) ( ) ( )dy
a x a x y b xdx
+ = with ( )1a x ;
01
1 1 1
( )( ) ( )( ) ( ) ( )
a xa x dy b x ya x dx a x a x
+ =
)econd, from the above, it can be rewritten as+
( ) ( )dy
P x y Q xdx
+ =
Where0
1
( )( )
( )
a x P x
a x= and
1
( )( )
( )
b xQ x
a x= .
Third, we need to determine the integrating factor, ( )v x ; ( )( ) P x dx
v x e∫ = .
10 | P a g e
( )1a x
6ow to solve the differential equations by using inte-ral
fa$tor "et&od? '"plain generally.
( )0a x ( )b x
Standard Form
-
8/20/2019 Chapter 1 - Differential Equations
11/27
-
8/20/2019 Chapter 1 - Differential Equations
12/27
1 1
2 2( )dx x
v x e e∫ = =
• Step 3 $ultiply ( )1
2 x
v x e= to both sides of standard form equation+
1 1 1
2 2 21 32 2
x x x
dye e y edx + =
1 1
2 23
( )2
x xd e y e
dx=
• Step 4 Integrate both sides from equation in Step 31 1
2 2
1 1
2 2
3( )
2
3
x x
x x
d e y dx e dx
dx
e y e C
=
= +
∫ ∫
• Step 5 The general solution1
2
3 x
C y
e
= +.
EXA!LE 12
)olve the linear first %&'3 xdy y e
dx+ =
)olution• Step 1 )ince the above equation is in the standard form then, you don=t &a%e to
$&an-e an+t&in-.
• Step 2 To find the integral factor let ( ) 1 P x = into ( )( ) P x dx
v x e∫ = . Then,1
( ) dx xv x e e∫ = =
• Step 3 $ultiply ( ) xv x e= to both sides of standard form equation+
3. x x x xdy
e e y e edx
+ =
4( ) x xd
e y edx
=
• Step 4 Integrate both sides of the equation in Step 3
12 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
13/27
4
4
( )
1
4
x x
x x
d e y dx e dx
dx
e y e C
=
= +
∫ ∫
Step 5 The general solution is31
4
x
x
C y ee
= + .
1.,.,./ Con$lusion $athematical modeling is the technique of representing real world problem which is
comple", involving multiple variables and some interrelated processes. This method can be
used in the study of growth population, radioactive decay, economics problems, changes in
temperature, mi"tures, chemical reactions, biological reactions, mechanics, velocity of a
falling ob!ect, electric circuits etc.
Referen$es• 8ill, &.7. (9::5. A Firt !oure in "i##erential $%uation &ith 'odelin Appliation, ;th
ed. 4roo#s. "i##erential $%uation &ith *oundary+alue -roble,
4roo#s
-
8/20/2019 Chapter 1 - Differential Equations
14/27
-
8/20/2019 Chapter 1 - Differential Equations
15/27
a 0sin2
2
=+ ydx
yd
Solution :
• 2onlinear • )econd order
b xe y y y =+− 2')1(
Solution :
c 0'2'' =+− y y y
Solution :
d 04)( =+− xdydx x y
Solution :
e y xdx
yd 4)12(
2
2
=+
Solution :
f 22t x
dt
dx=+
Solution :
g 2
2
2
xudx
ud =+
Solution :
h 32
2
2
2 x ydx
yd =+
Solution :
. ind the solution to given initial>%alue #role" by using separation of variablesmethod.
a 24( 1), ( ) 1
4dx x xdt
π = + = b2
21 ; (1) 21
dy x ydx y
−= =−
15 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
16/27
Solution :
Ste# 1 dt dx x
41
12
=+
Ste# ,
C t x
dt dx x
+=
=+
− ∫ ∫ 4)(tan41
1
1
2
To find t&e %alue of C?
π
π π
π
4
3
4
)4
(4)1(tan
4)(tan
1
1
−=
+=
+=
+=
−
−
C
C
C
C t x
Ste# / )
4
34tan(
)4tan(
π −=
+=
t x
C t x
Solution :
c ; (3) 4dy x
ydx y
= − =
Solution :
d 1)1(;2 −=−−= y xy ydx
dy x
Solution :
e5
2 1; (0)2
dy y y
dt + = =
Solution :
f ; (0) 2dy
xy ydx
= = −
Solution :
16 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
17/27
B. ind the general solution for the following %&' by searation of #aria!les.
a 2)1( += x
dx
dy
Solution :
Step 1 dx x xdy )12(1 2 ++=
Step 2 ∫ ∫ ++= dx x xdy )12(1 2
Step 3 C x x x y +++= 23
3
b 0)1( 2 =−− dx ydy
Solution :
c 02 =+ xydx
dy
Solution :
d
++
=54
32
x
y
dx
dy
Solution :
17 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
18/27
e y xe
dx
dy 23 +=
Solution :
f ydx
dy x 4=
Solution :
g 03cos23sin =+ xdy y xdx
Solution :
h )70( −= Qk dt
dQ
Solution :
>. ind the general solution for the following %&' by using the inte-ral fa$tor "et&od.
a
x
e ydx
dy 3
=+
Solution :
Ste# 1( )ince the equation is already inthe standard form then, no need to doanything.
Ste# , ( xdx ee x
x P
==
=
∫ 1)(
1)(
µ
Ste# / (
[ ] x x
x x x x
e yedxd
ee yedx
dye
4
3.
=
=+
b
xe ydx
dy363
=+
Solution :
18 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
19/27
Ste# 0 (
[ ]
C e ye
dxedx yedx
d
x x
x x
+=
= ∫ ∫
4
4
4
1
Ste# 2 ( x x
eC e y += 3
41
c xe y y =+'
Solution :
d 0cos2 3 =−− x x ydxdy
x
Solution :
e 32 =+ ydx
dy x
Solution :
f xy ydx
dy x 485)2( 2 −−=+
Solution :
19 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
20/27
g 4123 =+ ydxdy
Solution :
h x xdx
dy
412 =+
C. &etermine the general solution of this non linear first %&' using inte(ral fa$tor)
xe ydx
dy5
3
13 =+
!"en0
= x
an#1−= y
20 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
21/27
Solution :
;. In a chemical reaction, hydrogen pero"ide is continuously converted into water and
o"ygen. At time t minutes after the reaction started, the quantity of hydrogen pero"ide
that has not been converted is x cm and the rate at which 9 cm is decreasing
proportional to x .
The problem above can be translated by mathematical word, 2dx
xdt
= − .
(i )olve the above equation by using separation of variables. 7iven that initial
amount of x is 5 where ( )1 0t = . (ii &etermine the solution when it too# minutes for the hydrogen pero"ide to be
reduced to half the original amount where1
32
t = ÷
.
Solution :
21 | P a g e
-
8/20/2019 Chapter 1 - Differential Equations
22/27
)@ul+ ,
-
8/20/2019 Chapter 1 - Differential Equations
23/27
-
8/20/2019 Chapter 1 - Differential Equations
24/27
)@ul+ ,
-
8/20/2019 Chapter 1 - Differential Equations
25/27
)Se#te"er ,
-
8/20/2019 Chapter 1 - Differential Equations
26/27
)@anuar+ ,
-
8/20/2019 Chapter 1 - Differential Equations
27/27
)Se#te"er ,