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KNF1023
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Prepared By
Annie ak Joseph
Prepared ByAnnie ak Joseph Session 2007/2008
KNF1023Engineering
Mathematics II
First Order ODEs
Learning Objectives
Demonstrate the solution of
inhomogeneous 1st order ODE in linear form
Demonstrate the solution of
Homogeneous 1st order ODE in linear form
Demonstrate how to find integrating
factor for non-exact differential equation
Integrating Factor
� The idea of the method in this section is quite
simple. If given an equation
that is not exact, but if multiply it by a suitable
function , the new equation
is exact, so that it can be solved, the function
is then called an integrating factor of equation (1).
( , ) ( , ) 0 (1)P x y dx Q x y dy+ = −−−
),( yxF
0 (2)FPdx FQdy+ = − − −
),( yxF
How to Find Integrating Factors
� Equation (2) with M=FP, N=FQ is exact by the definition of an integrating factor. Hence
now is
That is (subscripts denoting partial derivatives) which is complicated and useless. So we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one—the result may be useful (and may also help you later on). Hence we look for an integrating factor depending only on one variable;
x
N
y
M
∂
∂=
∂
∂( ) ( ) (3)FP FQ
y x
∂ ∂= −−−
∂ ∂
xxyy FQQFFPPF +=+
How to Find Integrating Factors
� fortunately, in many practical cases, there are such factors, as we shall see. Thus, let
Then and so that (3) becomes
Dividing by and reshuffling terms, we have
).(xFF =
0=yFdx
dFFF x == '
xy FQQFFP += '
F
xy FQQFFP += '
F
FQ
F
QFP x
y +='
How to Find Integrating Factors
xy QQdx
dF
FP +=
1
Qdx
dF
FQP xy
1=−
dx
dF
FQP
Qxy
1)(
1=−
∂
∂−
∂
∂=
x
Q
y
P
Qdx
dF
F
11
How to Find Integrating Factors
� Thus, we can write it as
This prove the following theorem.
Rdx
dF
F=
1 1(4)
P QR
Q y x
∂ ∂= − −−−
∂ ∂
How to Find Integrating Factors
* If we assume Then and so that (3) becomes
Dividing by and reshuffling terms, we have
).(yFF = 0=xFdy
dFFF y == '
xy FQFPPF =+'
F
How to Find Integrating Factors
( )
( )
'
1
1(4*)
1 1
1 1,
y
x
x y
x y
x y
x y
FPF PQ
F F
dFQ P P
F dy
dFQ P P
F dy
dFQ P
P F dy
dFR R Q P
F dy P
= +
= +
− = − − −
− =
= = −
Theorem 1 (Integrating factor F(x))
� If (1) is such that the right side of (4), call it R depends only on x, then (1) has an integrating factor F=F(x), which is obtained by integrating (4) and taking exponents on both sides,
Similarly, if F=F(y), then instead of (4) we get
Here And have the companion
( )( ) exp ( ) 5F x R x dx= − − −∫
( )1 1
6dF Q P
F dy P x y
∂ ∂= − − − −
∂ ∂
∂
∂−
∂
∂=
y
P
x
Q
pR
1
THEOREM 2 [Integrating factor F(y)]
� If (1) is such that the right side R of (6) depends only on y, then (1) has an integrating factor F=F(y), which is obtained from (4*) in the form
( )( ) exp ( ) 7F y R y dy= − − −∫
Example 1 (Integrating factor F(x))
� Solve by Theorem 1.
We have hence (4) on the right,
And thus
0)cos()sin(2 22 =+ dyyxydxy
)cos(),sin(2 22yxyQyP ==
[ ]x
yyyyyxy
R3
)cos()cos(4)cos(
1 22
2=−=
( ) 3/3exp)( xdxxxF == ∫
Example 2
xyP 2=
22 (4 3 ) 0,
(0.2) 1.5
xydx y x dy
y
+ + =
= −
234 xyQ +=
2)(
2)26(
2
1
yyThusF
yxx
xyR
=
=−=
� Solve the initial value problem
Here , , the equation is not exact, the right side of (4) depends on both x and y (verify!), but the right side of (6) is
Continue...
2y
� Is an integrating factor by (7). Multiplication by gives the exact equation
Which we can write as
0)34(2 2233 =++ dyyxydxxy
223
3
34
2
yxyN
xyM
+=
=
Continue...
� As we know that
So to get , we use
( ) cyxu =,
( ) cyxu =, ( )∫ += ykMdxu
( )ykdxxyu += ∫32
( )ykyxu += 32
Continue...
� To get , we differential with respect to ,from there we get
( )yk uy
3
22322
4
343
ydy
dkso
yxyNdy
dkyx
y
u
=
+==+=∂
∂
*4cyk +=
cyyxu =+= 432
ccyyxu =++= *432
5.1,2.0 −== yx
9275.4324 =+ yxy
Continue...
1st Order ODEs In Linear Form
� A first order differential equation is linear if it has the form
� If the right side r(x) is zero for all x in the interval in which we consider the equation (written r(x)≡0), the equation is said to be homogeneous other it is said to be nonhomogeous.
( ) ( )dy
f x y r xdx
+ =
Homogeneous 1st Order ODEs In Linear Form
� Linear ODE is said to be homogeneous if the function r(x) is given by r(x)=0 for all x. That is, a homogeneous 1st order ODE is given by
( ) 0dy
f x ydx
+ =
yxfdx
dy)(−=
dxxfdyy
)(1
−=
Homogeneous 1st Order ODEs In Linear Form
∫∫ −= dxxfdyy
)(1
ln( ) ( )y f x dx= −∫
Gives us the general solution of the homogeneous 1st order
ODE above.
exp( ( ) )y f x dx= −∫
Inhomogeneous 1st Order ODEs in Linear Form (Method 1:use of integrating factor)
1. is a general form of the
linear DE.
2. Here f and r are function of x or constants
3.
4. Integrating factor =I.f=
5. Solution is
( ) ( )xryxfdx
dy=+
∫ fdx
∫ fdx
e
( ) ( ) CdxfIrfIy += ∫ ..
Example
xey
dx
dy 25 =+
( ) ( )xryxfdx
dy=+
xerf
2,5 ==
xfdx
eefI5. =∫=
The above DE is of the form
( ) ( ) CdxfIrfIy += ∫ ..
Continue...
( )
xx
xx
xx
xxx
Ceey
Ce
ye
Cdxeye
Cdxeeey
52
75
75
525
7
1
7
−+=
+=
+=
+=
∫
∫
� In the 1st step, we solve the corresponding homogeneous ODE, i.e
� Let us say that we obtain as particular solution for the above homogeneous ODE. We will use it in the 2nd step below to construct a general solution for the original inhomogeneous ODE.
( ) 0dy
f x ydx
+ =
)(xyy h=
Method 2: Variation Parameter
Method 2: Variation Parameter
( )xyxy h=)(� In the 2nd step, for the general solution of the
inhomogeneous ODE, we let . v(x) and substitute it into ODE to obtain a 1st order separable ODE in v(x).
25
5 0
5
5
xdyy e
dx
dyy
dx
dyy
dx
dydx
y
+ =
+ =
= −
= −∫ ∫
Example
Continue...
( )
( )
5
5
5 ' 5
ln 5
.
. 5 .
x
x
x x
y x
y e
y e v x
dye v e v x
dx
−
−
− −
= −
=
=
= −
( )5 ' 5 5 2
5 ' 2
5 2
2
5
. 5 . 5. .x x x x
x x
x x
x
x
e v e v x e v e
e v e
dve e
dx
dv e
dx e
− − −
−
−
−
− + =
=
=
=
Continue…
7
7
75
25
7
( )7
7
x
x
xx
xx
dve
dx
ev c
ey e c
ey ce
−
−
=
= +
= +
= +
Continue…
Prepared By
Annie ak Joseph
Prepared ByAnnie ak Joseph Session 2007/2008