mathematic 1st order differrentation
TRANSCRIPT
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DIFFERENTIAL
EQUATION(D.E.)
2
dy
dx
d y
dx,
2
2
INTRODUCTION
etc, is called a differential equation (D.E.).
What is a differential equation ?
An equation wh ich contains
04:Example xdt
dx
xdx
dyx
dx
yd2cos42
2
3
dy
dxx 2
If we are asked to solve the f irst order D.E.
, then what are we trying to find ?
We aret rying to f ind a function of y
in term o f x )(xfywhich satisfies the D.E.
dy
dxx 2
where x is the independent variable,
y is the dependent variable.
dy
dxx 2Given
Dependent variable
Independent variable
4
Example1 : Solve dy
dxx 2
Solution : Given
y x C 2
Solving a D.E. involves some form
of integration.
dy
dxx 2
integrate
5
THE CLASSIFICATION OF D.E.
The are many techniques for solvingdifferential equations
Different methods being applicable todifferent kinds of equation.
It is necessary to understand the variouscategories and classifications of differentialbefore apply the various method to solve
the equation.
6
1. Ordinary and Partial D.E
xdx
dfx
dx
fd2cos4:Example
2
2
Differential equation may involve either
ordinary or partial derivatives.
Thoseequations involving ordinaryderivatives are called ordinary differential
equations
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While those equations involving
partial derivatives are called partial
differential equations
yxy
f
x
fExample 24:
2
8
2. The order of a D.E
yxy
f
x
f24
2
is a firstorderpartial D.E
The order of a differential equation is thedegree of the highest derivative that occurs
in the equation.
Examples :
9
yxy
fx
x
f
yx
f24 2
2
2
2
3
is a 3rdorder
partial D.E.
x
dx
dfx
dx
fd2cos4
2
2
is a 2ndorderordinary D.E.
04
2
dt
dx
dt
dx i s a 1st orderordinary D.E.
The order of an equation is not affected by
any powerto which the derivatives may be raised.
Example :
10
The variables with respect to which
differentiation occurs are called
independent variables
3. Independent and dependent variables
While those that are differentiated
are dependent variables.
11
DependentVariables
(upper)
IndependentVariables
(lower)
yxy
f
x
f24 2
tdt
dyt
dt
yd2cos4
2
2
f
y
yx,
t
12
The standard way of arranging terms in aD.E. so that all terms containing thedependent variable occur on the left-handside of the equality sign, and those termsthat involve only the independent variableand constant terms occur on theright-hand side.
4. Homogeneous &
non-homogeneous D.E.
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When linear differential equation arearranged in this way, those in whichthe right hand side is zero are calledhomogeneous equations and those in
which i t i s n o n - z e r o a r en o n - h o m o g e n e o u s e q u a t i o n s
02
2
cydx
dyb
dx
yda
)(2
2
xfcydx
dyb
dx
yda
Homogeneous
Non-
homogeneous
14
Determine whether the following D.Eis homogeneous or non-homogeneous
04) xdt
dxa
Solution : Given 04 xdt
dx
Dependentvariable
Checking whether the given D.E is arrangedin standard way
*** All the derivatives and term
with dependent variable (x) must be at theleft hand side.
dt
dx
15
Given 04 xdt
dx
Dependentvariable
Right Hand side =0
Answer : Homogeneous D.E.
16
Determine whether the following D.Eis homogeneous or non-homogeneous
Solution : Rearrange in standard way
*** All the derivatives and termwith dependent variable (y) must be atthe left hand side.
dx
dy
dx
yd,
2
2
02cos42
2
xdxdyx
dxyd
17
02cos42
2
xdx
dyx
dx
yd
Dependentvariable
Independentvariable
xdx
dyx
dx
yd2cos4
2
2
Right Hand side
Answer : Non-homogeneous D.E.
0
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Solving 1st Order
Differential Equation
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ii) Exact product type
Types of 1st order D.E.
i) D.E. with variable separable (SOVA)
iii) The integrating factor type
iv) Solve by using substitution method
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Type 1 : D.E. with Variable Separable(SOVA)
When a first order D.E is written as :
)().( ygxhdx
dy
Such differential equations of this formare separable i.e.
dxxhyg
dy
)()(
21
Next, integrate both sides
Thus, we get
dxxhdyyg
)()(
1
CxHyG )()((general solution of the D.E.)
Remark : Solving a 1st order D.E.involves some form of integration
22
EXAMPLE 1 : Solve the differential equation
3 5 2ydydx
x
STEP 1 : SEPARATE THE VARIABLES
In the form of
3 5 2ydy x dx
TYPE 1 : SOVA
Solution :
dxxhdyyg )()(
23
dxxydy 253
This is the general solution of the differential
equation because value C not yet determine.
STEP 2 : INTEGRATE BOTH SIDES
Cxy
3
5
2
3 32
24
Simplify answer :
Cxy
3
5
2
3 32
6
Cxy
35
236
32
cxy 6109 32
Axy 32 109 (where A =6C)
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EXAMPLE 2 :
Given the following differential equation,solve for s .
s
dt
dst )3(
Solution :
26
27
29
NOTE :
If value of C is to be evaluated,i.e. the i n i t i a l c o n d i t i o n is given,then the solution of the D.E. is calledthe p a r t i c u l a r s o l u t i o n
Example 3 : Find the part icular solut ion
of the D.E.05 y
dx
dy
for the initial condition x=0, y=1
30
STEP 1 : Rearrange and separate the variables
Solution :
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EXAMPLE 4
2 1 42 2x dy
dxx y
If y = 1 when x = 0, f ind the particular so lut ion
by expressing y in terms of x .
Find the general solution of the D.E.
35
dxyxdyx 22 412
)1(2)4( 22
x
dxx
y
dy
STEP 1 : SEPARATE THE VARIABLES
Solution :
22 412Given yxdx
dyx
dx
x
xdy
y )1(2)4(
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STEP 2 : INTEGRATE BOTH SIDES
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39 42
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12ln4
12ln
4
1Cyy
Therefore, the general solution is :
dx
x
xdy
y )1(2)4(
1:From
22
partialfraction substitutionmethod
2
21ln
4
1Cx
46
Cxy
y
)1)(2(
2ln
2
DCBACD
AB
BAB
A
BAAB
lnlnlnlnln
lnlnln
lnln)ln(
Simplify Answer by usinglaw of logarithmic
2
2
1 1ln4
12ln
4
12ln
4
1CxCyy
12
2
1ln4
1
2ln4
1
2ln4
1
CCxyy
Let
)(412 CCC
Generalsolution
)(41ln2ln2ln12
2 CCxyy
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Find value C to get particular solution,
substitute the initial condit ion into t he general solution
Cxy
y
)1)(2(
2ln
2
When 1,0 yx
C
)10)(12(
12ln
3lnC
48
3)1)(2(
22
xy
y
( substitute C=ln3 into general solution)
3ln)1)(2(
2ln
2
xy
y
Take exponential both sides
3ln)1)(2(
2ln
2
ee xy
y
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To express y in terms of x
3)1)(2(
22
xy
y
)1)(2(32 2 xyy
yyxxy 33662 22
26633 22 xyxyy
Crossmultiplication
Rearrange y in theleft hand side
expand
50
2
2
34
46
x
xy
(particular solution
where y in terms of x)
26634 22 xyxy
266)34( 22 xyx
26633 22 xyxyy
Factorize y
51
FIRST ORDER D.E.
(Engineering Problems)
SOVA Method
52
EXAMPLE 1 : Materials
l
is given by : )( lxwdx
dM
where w is the constant load. Find an expression
M in term of x
.
The bending moment , M , of a beam of length
53
)( lxwdx
dM
Solution :
dxlxwdM )(
Step 1 : Separate the variables
)( lxwdx
dM
dependent variable (M)
independent variable ( x) constants (w, l)
Given
54
Step 2 : Integrate both sides
dxlxwdM )(
M
Clxx
wM
2
2
dxlxw )(
Take out coefficient
expression M in term of x
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The velocity, atuv of an object
dtdsv where
s is the displacement ,u is the initial velocity
a is the acceleration.
0t 0, s
show that the displacement ,2
2
1atuts
EXAMPLE 2 : Mechanics
is defined as :
Given that when
56
atuv
Solution :
-------equation (i)
dt
dsv -------equation (ii)
Substitute equation (ii) in equation (i)
atu
dt
ds
Given
Given
57 dtatuds
atudt
ds
Step 1 : Separate the variables
Solution :
atudt
ds
dependent variable (s)
independent variable ( t) constants (u, a)
Given
58
dtatuds
Ct
auts
2
2
(general solutio n)
Step 2 : Integrate both sides
Find value of C by using the given initial condition
0C
Substitute ,0t 0s into the general solution
Cau
2
0)0(0
2
C 000
59
Ct
auts
2
2
02
2
at
uts
2
2atuts ANSWER :
Substitute C=0 into the general solution
(SHOWN !)
60
Consider a tank full of water which is being
drained out through an outlet. The height
of water in the tank at time is given by)(mH
Hdt
dH )108.2( 3
Given that when ,0t 4H
find an expression for H in terms of t
EXAMPLE 3 : Fluid Mechanics
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Solution :
H
dt
dH)108.2( 3
dtHdH 0028.0
dtH
dH0028.0
Step 1 : Separate the variables
62
Step 2 : Integrate both sides
dtH
dH0028.0
dtdHH 0028.021
CtH
0028.0
21
12
1
CtH 0028.02 2
1
Rewrite
General
solution
63
Find value of C by using the given initial condition
Substitute 4H into the general solution,0t
C
C
042
)0(0028.042 21
CtH 0028.02 21
40028.02 21
tH
CtH 0028.02 21
4C (substitute into the general solution)
64
To get an expression for H in terms of t
.
2
40028.02
1
t
H
2
4
2
0028.02
1
t
H
220014.0 tH
40028.02 21
tH
Square both sides
20014.021
tH
65
mh
Area,
Further Engineering applications
at time t (seconds) is given by :
ghA
A
dt
dh20
0AWhere and A are the
cross-sectional areas of theoutlet and the tank respectively.
0A
1. Apply first order differential equations to
fluid mechanics. The differential equationdescribing the tank filled to a height
Area,0A
66
A cylindrical tank of diameter 3 m is filled with water to a
height of .The water is drained througha c i r c u l a r hole of diameter 0.1 m.
mh
i) Obtain the differential equation relating the height h
of water at time t.
,0t 2h
iii) How long (in minutes) does it take to empty
the tank which is 2 m full ?
EXAMPLE 4 : Fluid Mechanics
ii) Solve the differential equation f or the initial condition,
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SOLUTION (i) : From the formu la ghA
A
dt
dh20
The outlet area , is a circle of diameter 0.1 m0A
So, area of circle
0025.02
1.0 2
2
0
rA
Area,0A
Since we have a cylindrical tank of
diameter 3 m, therefore cross-sectional area of the tank:
25.22
3 2
2
rA
Step 1 : find 0A
Step 2 : find A
68
Substitute ,0025.00 A 25.2A and
81.9g into the formula :
ghA
A
dt
dh20
hdt
dh 81.92
25.2
0025.0
hdt
dh 62.190011.0
Answ er (i) : h
dt
dh00487.0
simplify
69
dthdh 00487.0
dth
dh00487.0
ii) Solve the differential equation for the initial condition,
,0t 2h
SOLUTION (ii) : From the D.E. hdt
dh00487.0
step 1 : Separate the variables
70
Step 2 : Integrate both sides
dth
dh00487.0
dtdhh 00487.021
Cth
00487.0
21
12
1
Cth 00487.02 21
Rewrite
General
solution
71
Find value of C by using the given initial condition
Substitute 2h into the general solution,0t
C
C
022
)0(00487.022 21
828.200487.02 2
1
th
Cth 00487.02 21
828.2C (substitute into the general solution)
72
To get an expression for h in terms of t
.
2
828.200487.02
1
th
2
828.2
2
00487.02
1
t
h
2414.1002435.0 th
828.200487.02 21
th
Square both sides
414.1002435.021
th
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iii) How long (in minutes) does it take to empty
the tank which is 2 m full ?
Solution : If the tank empty, height of water, h=0
2414.1002435.0 th
t = ?h = 0
2414.1002435.00 t
414.1002435.00 t
From
74
0414.1002435.0 t
ondst sec581698.580002435.0
414.1
414.1002435.0 t
Time in minute : min60
581t min68.9
Answer : It take approximately 9.68minutes to empty the tank.
DISCUSSIONDISCUSSIONSolve the following 1st order differential equations :
75
22
22
)1xyx
xyy
dx
dy
2)0(thatgiven,2
3
2
1)2 yy
dx
dy
x
y
dx
dy 1)3
2
xydx
dyxy
22 sec4tan)4