chapter 2 second order linear odes - university of macaufstitl/calculus2014/chapter2_ppt.pdf · a...
TRANSCRIPT
Chapter 2 Second Order Linear ODEs 2.1 Homogeneous Linear Equations of 2nd Order
A 2nd order linear equation:
)()(')(" xryxqyxpy =++(1)
is called nonhomogeneous. )0)(( ≠xr(1)
0)(')(" =++ yxqyxpy(2)
is called homogeneous.
For instance a nonhomogeneous lin. ODE
xeyy x cos25 −=+′′
A homogeneous lin. ODE
0' =++′′ xyyyxA nonlinear ODE
02 =′+′′ yyy
Superposition Principle for the Homogeneous Linear ODEs:
Ex. 1 xyxy sin and cos ==are solutions of 0" =+ yy
Then xcxcy sincos 21 +=
is also a solution of the above Eq.
---superposition or linearity principle
Theorem 1. If 1y and 2y are solutions of (2), then 2211 ycycy += is a solution of (2).
Proof: 0)()( 111 =+′+′′ yxqyxpy0)()( 222 =+′+′′ yxqyxpy
We have
0)()()()'()"(
)()(
22221111
221122112211
=+′+′′++′+′′=+++++=
+′+′′
qyypycqyypycycycqycycpycyc
yxqyxpy
Ex. 2 Verify by substitution that xyxy sin1 and cos1 +=+=
are solutions of .1" =+ yy
xxyxy sincos2or )cos1(5 ++=+=
are not solutions of !1" =+ yy
Ex. 3 Verify that 1 and 2 == yxy
are solutions of the nonlinear ODE 0=′−′′ yxyy
22 1or xyxy +=−=
are not solutions of 0=′−′′ yxyy
IVP. Basis. General Solution Initial conditions: (4) 1000 )( ,)( KxyKxy =′=
(4) with (2) is called IVP for the second order homogenous linear ODE.
Ex. 4 Solve the initial value problem .5.0)0(' ,0.3)0( ,0" −===+ yyyy
Soluton. Step 1 xcxcy sincos 21 +=
is the general solution Step 2 particular solution
5.0)0(' and 0.3)0( 21 −==== cycy
xxy sin5.0cos0.3 −=
Definition 1. Two solutions 1y and 2y
of (2) are called a basis of (2) iff constant/ 21 ≠yy
Definition 2. Two solutions 1y and 2y
of (2) are a basis of (2), then
2211 ycycy +=
is called a general solution of (2).
A particular solution of (2) is obtained if we assign specific values to 1c and 2cin the general solution.
We call linearly independent if 21 and yy0 ,0 0 212211 ==⇒=+ ccycyc(7)
Definition 1’. A pair solutions 21 and yyof (2) are called a basis of (2) iff 21 and yy
are linear independent.
Ex. 5 cos x and sin x in Ex. 4 form a basis of solutions of y”+y=0 for all x. y=3.0cos x-0.5sin x is a particular solution.
Ex.6 Verify xx ee − and are solutions of .0" =−yy Then solve the IVP
2)0(' ,6)0( −== yy
Solution. ( ) ( ) 0 and 0 =−″
=−″ −− xxxx eeee
xx ececy −+= 21 is general sol.
4 ,22)0(' and 6)0(
21
2121
==⇒−=−==+=
ccccyccy
xx eey −+= 42 is particular sol.
A nontrivial solution of (2) 1y is obtained by some method, how to get the other 2y
to form a basis of (2)?
Set 12 uyy = then we have
112 '' uyyuy +=′ and
1112 "''2"" uyyuyuy ++= This gives
0)''("''2"
111
111
=+++++
quyuyyupuyyuyu
hence
.0')'2(",0)'"(')'2("
111
111
111
=++=+++
++
upyyyuqypyyu
upyyyu
Let "' ,' uUuU == then
.0)/'2(U' 11 =++ Upyy
dxpey
U
dxpyU
dxpyy
∫−=
−−=
+−=
∫
21
1
11
1
,||ln2||ln
,)/'2(U
dU
dxuuU ∫== U ,' dxUyuyy ∫== 112
(?)/ 12 constdxUuyy ≠== ∫form a basis of (2). 21 and yy
Ex. 7 Find a basis of solutions of
.0'")( 2 =+−− yxyyxxSolution. xy =1 is one of the solution
Set uxuyy == 12 then we have uxuy +=′ '2 and
'2""2 uxuy += This gives
0)'()'2")(( 2 =++−+− uxuxuxuxuxx
hence
0')2(")( 2 =−+− uxuxx Let 'uv =
0)2(')( 2 =−+− vxvxx
dxxx
dxxx
xvdv
−
−=
−−
−=2
112
2
|||1|ln||ln2|1|ln||ln
xxxxv −
=−−=
∫ +==−=−
=x
xvdxuxxx
xv 1||ln ,11122
Finally, we have
.1||ln2 +== xxuxy
,1 xy =with
They form a basis of the solutions.
2.2 2nd-Order Homogeneous Equations with Constant Coefficients
0'" =++ byayy(1) kxey −= is a solution of .0' =+kyy
To try as a solution of (1) the function xey λ=(2)
xey λλ=' and xey λλ2"= (1) becomes
0)( 2 =++ xeba λλλ
We got a characteristic equation of (1)
02 =++ baλλ(3)
2/)4( 22,1 baa −±−=λ(4)
xey 11
λ= and xey 2
2λ=
are solutions of (1).
042 >− ba042 =− ba
042 <− ba
Case I two real roots if
Case II a real double root if
Case III complex roots if
There are 3 cases for the roots.
Case I. Two Distinct Real Roots xey 1
1λ= and
xey 22
λ=
constant/ )(21
21 ≠= − xeyy λλ
The general solution is
xx ececy 2121
λλ +=
Ex. 1 in Ex. 6 of Section 2.1 .0" =−yySolution. The characteristic equation is
012 =−λIt has the roots 1±=λ
Hence the general solution xx ececy −+= 21
Ex. 2 Solve the IVP .5)0(' ,4)0( ,02'" −===−+ yyyyy
Solution. The characteristic equation is 022 =−+ λλ
It has the roots 2 ,1 21 −== λλ
Hence the general solution xx ececy 2
21−+=
xx ececy 221 2' −−=
According to the initial condition:
3 ,152)0(' and 4)0(
21
2121
==⇒−=−==+=
ccccyccy
Case II. Real Double Root
042 =− ba 2/21 a−=== λλλhence only one solution xaey )2/(
1−=
12 uyy =Set 112 '' uyyuy +=′
1112 "''2"" uyyuyuy ++=
0)''("''2"
111
111
=+++++
buyuyyuauyyuyu
.,0"
)2/' (since 0",0')'2("
,0)'"(')'2("
21
111
111
111
111
cxcuu
yayyuuayyyu
byayyuuayyyu
+==
−===++=+++
++
xu =
xaxexyy )2/(12
−==
0/ 12 ≠= xyy
The general solution is xaecxcy )2/(
21 )( −+=
Warning. (7) will not be a solution of (1) for the simple root of (4).
(7)
Ex. 3 Solve 016'8" =++ yyy
Solution. The characteristic equation is 01682 =++ λλ
It has the double root 4−=λand hence the general solution is
xecxcy 421 )( −+=
Ex. 4 Solve the initial value problem 1)0(' ,3)0( ,04'4" ===+− yyyyy
Solution. The characteristic equation is
0442 =+− λλIt has the double root 2=λand hence the general solution is
xexccy 221 )( +=
xx ecexccy 22
221 )(2' ++=
5 ,3,12)0('
,3)0(
21
21
1
−===+=
==
ccccy
cy
and the answer is
xexy 2)53( −=
Case III. Complex Roots
If (2) has a pair of complex roots 4/ ,2/ , 2
2,1 abai −=−=±= ωαωαλ
xiaey )2/(1
ω+−= and xiaey )2/(2
ω−−=
are solutions of (1).
To find the real solutions, we can obtain from (5) that
)sin(cos
),sin(cos2/)2/(
2
2/)2/(1
xixeeyxixeey
axxia
axxia
ωω
ωωω
ω
−==
+==−−−
−+−
and hence that xe ax ωcos2/− and
xe ax ωsin2/− are solutions of (1).
The corresponding general solution is
)sincos(2/ xBxAey ax ωω += −(9)
Ex. 5. Solve the initial value problem 3)0(' ,0)0( ,004.9'4.0" ===++ yyyyy
Solution. The characteristic equation is
004.94.02 =++ λλIt has the roots ,32.02,1 i±−=λand a general solution is
)3sin3cos(2.0 xBxAey x += −
The 1st initial condition gives
0)0( == Ay
It concludes that .3sin2.0 xBey x−=
)3cos33sin2.0(' 2.0 xxBey x +−= −
From this and the 2nd initial condition we get .133)0(' =⇒== BBy The answer is
xey x 3sin2.0−=
Ex. 6 A general solution of 0 where,0" 2 ≠=+ ωω yy
is
xBxAy ωω sincos +=
Ex. 7 Solve the boundary value problem
3)2/( ,3)0( ,0" −===+ πyyyy
Solution. The general solution is known as xBxAy sincos +=
So we have 3)2/( ,3)0( −==== ByAy π
The answer is xxy sin3cos3 −=
2.5 Euler-Cauchy Equation
0'"2 =++ byaxyyx(1)
is called the Euler-Cauchy equation. Let mxy = is a solution of (1), we find that
0)1( 122 =++− −− mmm bxaxmxxmmx or
0)1(2 =+−+ bmam(3)
(3) is called the characteristic equation of (1).
Case I. Two Distinct Real Roots If (3) has two distinct real roots 2,1mwe get a basis of solutions 21
21 and mm xyxy ==
constant/ 2121 ≠= −mmxyy
and a corresponding general solution of (1) is as follows
2121
mm xcxcy +=(4)
Ex. 1 Solve the Euler-Cauchy equation 05.0'5.1"2 =−+ yxyyx
Solution. The characteristic equation is 05.05.02 =−+ mm
?) 05.05.1(not 2 =−+ mmThe roots are m=0.5 and -1, this gives
xy =1 and xy /12 =and the general solution
0 ,/21 >∀+= xxcxcy
Case II. Real Double Root
If (3) has a double root, then 2/)1(2,1 am −=
hence only one solution .2/)1(1
axy −=
Set ,12 uyy = then we have
112 '' uyyuy +=′
1112 "''2"" uyyuyuy ++= This gives
0)''()"''2"(
111
1112
=+++++buyuyyuax
uyyuyux
hence
.ln||ln|'|ln ,'"
)2/)1(' (since "
,0')'2("
,0)'"(
')'2("
11
112
1112
1112
1112
xuxuuxu
yaxyxu'yyxu
uayxyxyxubyaxyyxu
xuayxyyxu
=−=−=
−=−=
=++
=+++
++
Thus ,ln12 xyy = where constxyy ≠= ln/ 12
The general solution is
2/)1(21 )ln( axxccy −+=
Ex. 2 Solve 09'5"2 =+− yxyyx
Solution. The characteristic equation 0962 =+− mm
has the double root m=3
Hence the general solution is
321 )ln( xxccy +=
Case III. Complex Conjugate Roots
If (3) has a pair of complex roots im νµ ±=2,1
then xii exxxy ln2,1
νµνµ ±± == are solutions of (1). To find the real solutions, we have
)]lnsin()ln[cos(
)],lnsin()ln[cos(
2
1
xixxyxixxy
νν
ννµ
µ
−=
+=
and hence that )lncos( xx νµ and
)lnsin( xx νµ are solutions of (1).
The corresponding general solution is
)]lnsin()lncos([ xBxAxy ννµ +=
Ex. 3 Solve 004.16'6.0"2 =++ yxyyx
Solution. The characteristic equation 004.164.02 =+− mm
has two complex conjugate roots
im 42.02,1 ±=
Hence the general solution is
)]ln4sin()ln4cos([2.0 xBxAxy +=
2.6 Existence and Uniqueness Theory Wronskian A general homogeneous linear equation
0)(')(" =++ yxqyxpy(1)
with continuous coefficients. The existence of a general solution
2211 ycycy +=and two initial conditions
1000 )(' ,)( KxyKxy ==
(2)
(3)
Theorem 1. If p(x) and q(x) are continuous on some open interval I, , then the initial value problem consisting of (1) and (3) has a unique solution on I.
Ix ∈0
Wronski determinant (or Wronskian) of two solutions of (1) as and 21 yy
122121
2121 ''
''),( yyyy
yyyy
yyW −==
We call linearly independent on I if 21 and yy0 ,0 on 0 212211 ==⇒=+ ccIycyc(4)
We call linearly dependent if 21 and yy
(4) holds for constants c1 and c2 not both 0.
Theorem 2. If p(x) and q(x) are continuous on some open interval I. Then two solutions of (1) on I are linearly dependent iff and 21 yy
. ,0))(),(( 00201 IxxyxyW ∈=
Furthermore, if . ,0))(),(( 00201 IxxyxyW ∈=
; ,0))(),((then 21 IxxyxyW ∈∀≡ hence if
, ,0))(),(( 11211 IxxyxyW ∈≠ then and 21 yy
are linearly independent on I.
Proof: (a) If and 21 yy are linearly dependent
on I, then there exists k, s.t. 21 kyy = and
0''''''
),( 222222
22
21
2121 ≡−=== kyyyky
ykyyky
yyyy
yyW
(b) Conversely, we assume that
IxxyxyW ∈= 00201 ,0))(),(( and show that
and 21 yy are linearly dependent on I.
We consider the linear equations
0)(')(',0)()(
022011
022011
=+=+
xykxykxykxyk
in the unknowns and 21 kk
(7)
(7) is homogeneous.
Its determinant is exactly 0))(),(( 0201 =xyxyW
hence the system has a nontrivial solution and 21 kk
Then we have a function )()()( 2211 xykxykxy +=
By Theorem 1 in Sec. 2.1, y(x) is a solution of (1) on I. We know that 0)(' ,0)( 00 == xyxy
It concludes by Theorem 1 that . ,0 Ixy ∈∀≡
Ixxykxyk ∈∀=+ ,0)()( 2211
this means linear dependence of and 21 yyon I.
(c) If 0))(),(( 0201 =xyxyW then and 21 yyare linear dependent, and then by (a),
.0),( 21 ≡yyW Hence 0))(),(( 1211 ≠xyxyW
implies linear independence of and 21 yy
Ex. 1 xy ωcos1 = and xy ωsin2 =
are solutions of .0" 2 =+ yy ω The Wronskian is
ωωωωωωω
=−
=xx
xxyyW
cossinsincos
),( 21
They are linearly independent iff .0≠ω0=ωIf then ,02 =y
hence they are linear dependence.
Ex. 2 A general solution of 0'2" =+− yyy
on any interval is .)( 21xexccy +=
The Wronskian is
0)1(
),( 2 ≠=+
= xxx
xxxx e
exexee
xeeW
Thus xe and xxe are linearly independent.
Theorem 3 If p(x) and q(x) of (1) are continuous on an open interval I, then (1) has a general solution on I.
Proof. By Theorem 1, (1) has solutions and 21 yy on I satisfying the initial conditions:
1)(' ,0)( 0101 == xyxy and 0)(' ,1)( 0202 == xyxy
Then and 21 yy are linearly independent on I. Thus they form a basis of (1) and then
2211 ycycy += is a general solution of (1).
Theorem 4 If p(x) and q(x) of (1) are continuous on an open interval I, then for every solution y=Y(x) of (1) is of the form
)()()( 2211 xyCxyCxY +=
where and 21 yy form a basis of (1) on I
1C 2Cand , are suitable constants.
Proof. A general solution of (1) is given as .2211 ycycy += We choose the suitable
and 21 cc such that )(')(' ),()( 0000 xYxyxYxy ==
),()()( 0022011 xYxycxyc =+ )(')(')(' 0022011 xYxycxyc =+
The Wronskian of and 21 yy is nonzero,
(10)
such that (10) has a unique solution 11 Cc =
and .22 Cc =
Thus a particular solution
)()()(* 2211 xyCxyCxy += of (1) satisfies (10). From this and the uniqueness theorem 1 we conclude that everywhere on I. )()(* xYxy =
2.7 Nonhomogeneous Equations
The nonhomogeneous linear equation
)()(')(" xryxqyxpy =++(1)
Its corresponding homogeneous equation is
0)(')(" =++ yxqyxpy(2)
Theorem 1. (a) If are solutions of (1) on I, then is a solution of (2) on I. (b) If y is a solution of (1) on I, and Y is a
solution of (2) on I, then is a solution of (1) on I.
and 21 yy21 yy −
Yy +
Proof. (a) Denote that then ,21 yyY −=
0)()()'"()'"()()'()"(
'"
222111
212121
≡−=++−++=−+−+−=
++
xrxrqypyyqypyy
yyqyypyyqYpYY
(b) Denote that then ,* yYy +=
)()(0)'"()'"()()'()"(
**)'("*)(
xrxrqypyyqYpYYyYqyYpyY
qyypy
=+=+++++=+++++=
++
Definition. A general solution of (1) on I is )()()( xyxyxy ph +=(3)
where 2211 ycycyh += is a general solution of (2), and )(xy p is any solution of (1).
A particular solution of (1) on I is a solution obtained from (3) by assigning specific values to the arbitrary constants in and 21 cc
.2211 ycycyh +=
Theorem 2. Suppose that the coefficients and r(x) in (1) are continuous on I. Then every solution of (1) on I is obtained from (3) by assigning suitable values to . and 21 cc
Proof. Let y*(x) be any solution of (1) on I. Let (3) be any general solution of (1) on I. Then is a solution of (2) by Theorem 1(a). By theorem 4 in Sec. 2.6, Y is obtained by assigning suitable values to the arbitrary constants From this and the statement follows.
pyyY −= *
. and 21 cc,* pyYy +=
Rules for the method of undetermined coefficients A.Basic Rule. Choosing in the Table. B. Modification Rule. Multiplying by
where j=1 or 2, if or is a solution of the corresponding homogeneous equation (or double roots).
C.Sum Rule. If r(x) is a sum of functions in the table, then is chosen as the sum of corresponding functions.
pyjx
xeγ xe x ωα cos
py
Ex. 1. Solve the equation
Solution. Solve the corresponding homogenous equation. We have
5.1)0(',0)0(,001.0" 2 ===+ yyxyy
,012 =+λ i±=2,1λ and then .sincos xBxAyh +=
We find that 0 is not a root of the characteristic equation. So
012
2 kxkxkyp ++= is chosen. Then
212 2",2 kykxky pp =+=′
.001.0)(2 201
222 xkxkxkk =+++
Compare with the coefficients above, we have
.02 ,0 ,001.0 2012 =+== kkkk Thus
.002.02 ,0 ,001.0 0012 −=−=== kkkk
Hence .002.0001.0 2 −= xyp And
002.0001.0sincos 2 −++= xxBxAy
Setting x=0, y(0)=A-0.002=0, hence A=0.002. y’(0)=B=1.5. Thus gives the answer
002.0001.0sin5.1cos002.0 2 −++= xxxy
Ex. 2. Solve the initial value problem 0)0(' ,1)0( ,1025.2'3" 5.1 ==−=++ − yyeyyy x
Solution. The characteristic equation is .025.232 =++ λλ It has the double root
.5.1−=λ and a general solution is
.)( 5.121
xexccy −+= At first, xp Cey 5.1−=
by the table. According to the rule, we choose
.5.12 xp eCxy −= Hence x
p exxCy 5.12 )25.1(' −+−=
.)25.262(" 5.12 xp exxCy −+−= Substitution gives
xxxx eeCxexxCexxC 5.15.125.125.12 1025.2)5.12(3)25.262( −−−− −=+−++−
,102 5.15.1 xx eCe −− −= 5−=CThus gives the general solution is
xexxccy 5.1221 )5( −−+=
,1)0( 1 −== cy)5.7105.15.1(' 2
225.1 xxxccey x +−−−= −
5.105.1)0(' 22 =⇒=−= ccyThis gives the answer xexxy 5.12 )55.11( −−+=
Ex. 3 Solve the initial value problem
08.40)0(' ,16.0)0( ,10sin19010cos405'2" 2/
==−+=++
yyxxeyyy x
Solution. The characteristic equation is .0522 =++ λλ It has the roots
,212,1 i±−=λ and a general solution is
).2sin2cos( xBxAey x += −
We choose .10sin10cos2/ xMxKCey x
p ++=
.10cos1010sin102/' 2/ xMxKCey xp +−=
.10sin10010cos1004/" 2/ xMxKCey xp −−=
Substitute this into the given equation.
16.0)514/1( 2/2/ =⇒=++ CeCe xx
.2,01909520,409520
.10sin19010cos4010cos)2095(10sin)2095(
10cos)520100(10sin)520100(
==⇒−=−−=−⇒
−=+−+−−=
+−−+++−
MKMKKM
xxxMKxKM
xKMKxMKM
This gives the general solution
10sin216.0)2sin2cos( 2/ xexBxAey xx +++= −
From the initial conditions, we have
.1008.402008.02)0('10cos2008.0)2cos22sin('
016.016.0)0(2/
=⇒=++=+++−=
=⇒=+=−
BByxexBxBey
AAyxx
We thus obtain the answer
10sin216.02sin10 2/ xexey xx ++= −
2.10 Solution by Variation of Parameters A general nonhomogeneous linear equation
)()(')(" xryxqyxpy =++(1) with continuous ).(),(),( xrxqxpMethod of variation of parameters,
dxrydxryy p ∫∫ +−=Wy
Wy 1
22
1(2) where 21, yy form a basis of the corresponding homogeneous equation of (1)
0)(')(" =++ yxqyxpy(3)
1221 '' yyyyW −= is the Wronskian of 21, yy
Ex. 1. Solve xyy sec" =+Solution. A basis of the homogeneous equation is Thus .sin and cos 21 xyxy ==
.1)sin(sincoscos'' 1221 =−−=−= xxxxyyyyW
xxxxxdxxxxdxxx
dxrydxryyp
sin|cos|lncosseccossinsecsincos
Wy
Wy 1
22
1
+=+−=
+−=
∫∫∫∫
xxcxxcxxxxxcxcy
sin)(cos|)cos|ln( sin|cos|lncossincos
21
21+++=
+++=We then obtain the answer
How to get the method? )()()( xyxyxy ph += with 2211 ycycyh +=
Let be a solution of (1). )()()()( 21 xyxvxyxuy p +=
2121 ''''' vyuyyvyuy p +++=0''Set 21 =+ yvyu then 21 ''' vyuyy p +=
2121 ""''''" vyuyyvyuy p +++=
Substituting to (1), and collecting terms
⇒=+++++++
=+++++++
ryvyuqypyyvqypyyu
rvyuyqvyuypvyuyyvyu
21
222111
21
'212121
'''')'"()'"(
)()'()""''''(
ryvyu =+ 21 '''' with 0'' 21 =+ yvyu
So we have W
ryvW
ryu 12 ' ,' =−=
dxW
ryvdxW
ryu ∫∫ =−= 12 ' ,
dxrydxryy p ∫∫ +−=Wy
Wy 1
22
1
Pages 103-104c
Continued
Pages 103-104d