week 7 [compatibility mode]
DESCRIPTION
KNF1023TRANSCRIPT
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Prepared By
Annie ak Joseph
Prepared ByAnnie ak Joseph Session 2007/2008
KNF1023Engineering
Mathematics II
Second Order ODEs
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Learning Objectives
Explain about Euler-Cauchy ODEs
Discuss about Second order
inhomogeneous ODEs
Explain about Particular Solution by
Guesswork
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Euler-Cauchy ODEs
� An Euler-Cauchy ODE is one of the form
Where are given constants.
� To look for linearly independent solutions of this
ODE, try
where λ is a constant yet to be determined.
� Differentiating, we have
2 "( ) '( ) ( ) 0ax y x bxy x cy x+ + =
candba ,0≠
y xλ=
( )1 2' " 1y x and y xλ λλ λ λ− −= = −
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Euler-Cauchy ODEs
Substituting into the ODE gives
Hence, the value of the constant λ can be
determined from the quadratic equation above. We
consider the following cases.
( )
( )( )
( )
( )
2 2 1
2
1 0
1 0
1 0
0
ax x bx x cx
x a b c
a b c
a b a c
λ λ λ
λ
λ λ λ
λ λ λ
λ λ λ
λ λ
− −⋅ − + ⋅ + =
⇒ − + + =
⇒ − + + =
⇒ + − + =
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Case (a):
In this case, we can find two distinct real values for
as given by
Thus, we have two solutions for the Euler-Cauchy
ODE, namely
[ ]2
4 0b a ac− − >
[ ] [ ]2
1
4
2
b a b a ac
aλ λ
− − + − −= =
[ ] [ ]2
2
4
2
b a b a ac
aλ λ
− − − − −= =
1 2
1 2,y x y xλ λ= =
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These solutions are linearly independent, since
For this particular case where and
are real, the general solution of the Euler-Cauchy
ODE is given by
where A and B are arbitrary constants.
1
1 2
2
11 2
2
( tan ) ( )y x
x cons t asy x
λλ λ
λλ λ−= = ≠ ≠
1 2λ λ≠1 2andλ λ
1 2y Ax Bxλ λ= +
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Example 1
Solve the ODE subject to
This is an Euler-Cauchy ODE. So try
Substituting into the ODE, we obtain
( ) 21 1"', −− −=== λλλ λλλ xyandxyxy
02'4"2 =++ yxyyx
1)1(')1( == yy
( )
( )( )
2,1
021
023
0241
2
−=−=
=++
=++
=++−
λλ
λλ
λλ
λλλ
2 2 1( ( 1) ) 4 ( ) 2 0x x x x xλ λ λλ λ λ− −− + + =
0]24)1([ =++− λλλλx
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y = Ax-1 + Bx-2
Where A and B are arbitrary constant.
Differentiating the general solution gives
Putting the given conditions into the general
solution, we have
32 2' −− −−= BxAxy
12;1)1('
1;1)1(
=−−=
=+=
BAy
BAy
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Solving for A and B, we obtain A=3 and B=-2.
The required particular solution is
21 23 −− −= xxy
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Case (b):
In this case, the quadratic equation
has only one solution.
Thus, in trying we manage to find only
one solution for the Euler-Cauchy ODE. We need
two linearly independent solutions to construct the
general solution of the ODE. To find another
solution, let us try
[ ]2
4 0b a ac− − =
2 ( ) 0a b a cλ λ+ − + =
[ ]1
2
b a
aλ λ
− −= =
y xλ=
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� Here, is a function yet to be determined.
� Substitution into the ODE, we obtain
( )
1
1 1
1 1 1
1
1
2 1
1 1 1
( ) ( )
'( ) ( ) '( )
"( ) 1 ( ) 2 '( ) "( )
y x x u x
y x x u x x u x
y x x u x x u x x u x
λ
λ λ
λ λ λ
λ
λ λ λ
−
− −
= ⋅
= ⋅ + ⋅
= − ⋅ + ⋅ + ⋅
( )u x
( ) 1 1 1
1 1 1
2 12
1 1 1
1
1
1 ( ) 2 '( ) "( )
( ) '( ) ( ) 0
ax x u x x u x x u x
bx x u x x u x cx u x
λ λ λ
λ λ λ
λ λ λ
λ
− −
−
− ⋅ + ⋅ + ⋅
+ ⋅ + ⋅ + ⋅ =
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Since and this further
reduces to
To solve for let so that
This leads to
( )1 0a b cλ λ λ− + + =[ ]
12
b a
aλ
−= −
'( ) "( ) 0u x xu x+ =
( )u x ( ) '( )v x u x= 0dv
v xdx
+ =
1ln( ) ln( ) ln
1
1
ln( )
dv dx
v x
v xx
vx
du
dx x
u x
= −
⇒ = − =
⇒ =
⇒ =
⇒ =
∫ ∫
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The solution of the Euler-Cauchy ODE which
we are looking for is therefore given by
The solutions and are linearly
independent to each other.
where A and B are arbitrary constants.
1 ln( )y x xλ= ⋅
1y xλ= 1 ln( )y x x
λ= ⋅
1 1 ln( )y Ax Bx xλ λ= +
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Example 2
Solve the ODE subject to
This is an Euler-CODE. So let us try
Substitution into the ODE gives
( )1 2, ' , " 1y x y x y xλ λ λλ λ λ− −= = = −
2 " 3 ' 0x y xy y+ + =
(1) 1 ( ) 2y and y e= =
( )
( )
2
2
1 3 1 0
2 1 0
1 0
1 is the only solution
λ λ λ
λ λ
λ
λ
− + + =
+ + =
+ =
= −
0]13)1([ =++− λλλλx
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Thus, general solution of the ODE is
Where A and B are arbitrary constants.
Using the given conditions, we have
The required particular solution is therefore
1 1 ln( )y Ax Bx x− −= +
(1) 1; 1
1 ln( )( ) 2; 2; 2 1
y A
B ey e B e
e e
= =
= + = = −
( )1 12 1 ln( )y x e x x− −= + −
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Case C:
In this case, the quadratic equation has two complex solutions given
by.
If we proceed on, without bothering about the facts
that and are complex, we can construct the
general solution of the Euler-Cauchy ODE as
[ ]2
4 0b a ac− − <
2 ( ) 0a b a cλ λ+ − + =
[ ] [ ]
[ ] [ ]
2
1
2
2
4
2 2
4
2 2
i b a acb a
a a
i b a acb a
a a
λ λ
λ λ
− −−= = − +
− −−= = − −
1λ 2λ
1 2y A x B xλ λ= +
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Our only problem is to interpret what
means when is complex and also to find a way to
calculate this complex power. Once again we can
resort to the theory of complex functions to resolve
this problem. We can use the following results
1 are any numbers (real or complex) and x is a real number
2 are real numbers
3 is any real numbers
xλ
λ
z w z wx x x if z and w
+ = ⋅
ln( ) 0iy iy xx e if x and y= >
cos( ) sin( )ixe x i x if x
± = ±
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Example 3
Solve the ODE
subject to
2 "( ) 3 '( ) 2 ( ) 0x y x xy x y x+ + =
(1) '(1) 1y y= =
( )1 2, ' , " 1y x y x y xλ λ λλ λ λ− −= = = −
( )2
1 3 2 0
2 2 0
1 , 1i i
λ λ λ
λ λ
λ
− + + =
+ + =
= − + − −
1 1i iy Ax Bx
− + − −= +
( ) 0]231[ =++− λλλλx
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( )( )( )
[ ]
1 1
1
1 ln( ) ln( )
1
1
cos ln( ) sin(ln( ) (cos(ln( )) sin(ln( ))
cos(ln( )) sin(ln( ))
i i
i i
i x i x
y Ax Bx
x Ax Bx
x Ae Be
x A x i x B x i x
x C x D x
− + − −
− −
− −
−
−
= +
= +
= +
= + + −
= +
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where are arbitrary
constants.
Differentiating, we have
Using the given conditions, we have
( )C A B and D i A B= + = −
( )( ) ( )( ) ( )( ) ( )( )1 1 1 2' sin ln cos ln cos ln sin lny x Cx x Dx x x C x D x− − − − = − + − +
(1) 1; 1
'(1) 1; 1; 2
y C
y D C D
= =
= − = =
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Thus, the required particular solution is
( )( ) ( )( )1 cos ln 2sin lny x x x− = +
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THEORY FOR INHOMOGENEOUS ODEs
Consider the 2nd order linear inhomogeneous
ODE
Let be any particular solution of
the inhomogeneous ODE. Then
To solve the inhomogeneous ODE, let us make the
substitution
" ( ) ' ( ) ( )y f x y g x y r x+ + =
( )p py y x=
" '( ) ( ) ( )p p py f x y g x y r x+ + =
( ) ( )p
y y x Y x= +
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On substituting into the ODE, we obtain
Thus, we obtain a 2nd order linear
homogeneous ODE in
" '
" '
( ) "( ) ( ) ( ) '( ) ( ) ( ) ( ) ( )
( ) ( ) " ( ) ' ( ) ( )
( ) " ( ) ' ( ) ( )
" ( ) ' ( ) 0
p p p
p p p
y x Y x f x y x Y x g x y x Y x r x
y f x y g x y Y f x Y g x Y r x
r x Y f x Y g x Y r x
Y f x Y g x Y
+ + + + + =
+ + + + + =
+ + + =
+ + =
( )Y Y x=
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To summarise, assuming that we can find a
particular solution for the 2nd order linear
inhomogeneous ODE and also that we can solve the
homogeneous ODE in for its general solution of the
inhomogeneous ODE is given by
{ {
homhom
( ) ( )p
general solution of thea particular solutioncorresponding ogeneous ODEof in ogeneous ODE
y y x Y x= +
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PARTICULAR SOLUTIONS BY GUESSWORK: Example
Solve the ODE subject to
Firstly let us solve the corresponding homogeneous
ODE
This is a 2nd order linear ODE with constant
coefficients. Let us try
" 3 ' 2 0Y Y Y+ + =
"( ) 3 ' 2 2exp(5 )y x y y x+ + =
(0) '(0) 0y y= =
2, ' , "x x xY e Y e Y e
λ λ λλ λ= = =
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Then, substitution into the homogeneous ODE
Gives
The general solution of the homogeneous ODE is
( )( )
2 3 2 0
1 2 0
1, 2
λ λ
λ λ
λ
+ + =
+ + =
= − −
2x xY Ae Be
− −= +
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Now, to construct the general solution of the
inhomogeneous ODE we
need to find one particular solution of the ODE. We
will try to look for one by guesswork. The right
hand side of the ODE, namely the term
suggests that we try a particular solution of the
Form
where the constant is to be selected to satisfy the
ODE
exp(5 )py xα=
"( ) 3 ' 2 2exp(5 )y x y y x+ + =
2exp(5 )x
α
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Differentiating, we have
Substituting into the ODE, we obtain
Hence, is a particular solution of the
inhomogeneous ODE.
"' 5 exp(5 ), 25 exp(5 )p py x y xα α= =
25 exp(5 ) 15 exp(5 ) 2 exp(5 ) 2exp(5 )
42 exp(5 ) 2exp(5 )
1
21
x x x x
x x
α α α
α
α
+ + =
⇒ =
⇒ =
1exp(5 )
21p
y x=
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The general solution of the inhomogeneous
ODE is
1(0) 0;
21
5'(0) 0; 2
21
y A B
y A B
= + = −
= + =
2 1
7 3B and A= = −
1 1 2exp(5 ) exp( ) exp( 2 )
21 3 7y x x x= − − + −
( ) xxBeAexy
25exp21
1 −− ++=
( ) )2exp(2)exp(5exp21
5'xBxAxy −−−−=
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Prepared By
Annie ak Joseph
Prepared ByAnnie ak Joseph Session 2008/2009