chapter 2 second order linear odes - university of macaufstitl/calculus2014/chapter2_ppt.pdf · a...
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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.
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For instance a nonhomogeneous lin. ODE
xeyy x cos25 −=+′′
A homogeneous lin. ODE
0' =++′′ xyyyxA nonlinear ODE
02 =′+′′ yyy
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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
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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
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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
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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
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IVP. Basis. General Solution Initial conditions: (4) 1000 )( ,)( KxyKxy =′=
(4) with (2) is called IVP for the second order homogenous linear ODE.
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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 −=
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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).
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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.
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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.
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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.
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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
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.0')'2(",0)'"(')'2("
111
111
111
=++=+++
++
upyyyuqypyyu
upyyyu
Let "' ,' uUuU == then
.0)/'2(U' 11 =++ Upyy
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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
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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
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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
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Finally, we have
.1||ln2 +== xxuxy
,1 xy =with
They form a basis of the solutions.
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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
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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).
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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.
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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 +=
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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
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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−+=
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xx ececy 221 2' −−=
According to the initial condition:
3 ,152)0(' and 4)0(
21
2121
==⇒−=−==+=
ccccyccy
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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
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.,0"
)2/' (since 0",0')'2("
,0)'"(')'2("
21
111
111
111
111
cxcuu
yayyuuayyyu
byayyuuayyyu
+==
−===++=+++
++
xu =
xaxexyy )2/(12
−==
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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)
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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 )( −+=
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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 )( +=
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xx ecexccy 22
221 )(2' ++=
5 ,3,12)0('
,3)0(
21
21
1
−===+=
==
ccccy
cy
and the answer is
xexy 2)53( −=
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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
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)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)
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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 += −
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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−=
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Ex. 6 A general solution of 0 where,0" 2 ≠=+ ωω yy
is
xBxAy ωω sincos +=
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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 −=
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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).
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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)
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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
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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
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hence
.ln||ln|'|ln ,'"
)2/)1(' (since "
,0')'2("
,0)'"(
')'2("
11
112
1112
1112
1112
xuxuuxu
yaxyxu'yyxu
uayxyxyxubyaxyyxu
xuayxyyxu
=−=−=
−=−=
=++
=+++
++
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Thus ,ln12 xyy = where constxyy ≠= ln/ 12
The general solution is
2/)1(21 )ln( axxccy −+=
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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 +=
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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).
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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 ±=
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Hence the general solution is
)]ln4sin()ln4cos([2.0 xBxAxy +=
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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)
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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 −==
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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.
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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.
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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.
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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
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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.
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(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
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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.
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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.
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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).
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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.
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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 =
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Thus a particular solution
)()()(* 2211 xyCxyCxy += of (1) satisfies (10). From this and the uniqueness theorem 1 we conclude that everywhere on I. )()(* xYxy =
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2.7 Nonhomogeneous Equations
The nonhomogeneous linear equation
)()(')(" xryxqyxpy =++(1)
Its corresponding homogeneous equation is
0)(')(" =++ yxqyxpy(2)
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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
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(b) Denote that then ,* yYy +=
)()(0)'"()'"()()'()"(
**)'("*)(
xrxrqypyyqYpYYyYqyYpyY
qyypy
=+=+++++=+++++=
++
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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 +=
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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
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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 +=
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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
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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
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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
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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
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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(' −+−=
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.)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 +−−−= −
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5.105.1)0(' 22 =⇒=−= ccyThis gives the answer xexxy 5.12 )55.11( −−+=
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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 += −
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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
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.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 +++= −
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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 ++= −
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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
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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
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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
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⇒=+++++++
=+++++++
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
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Pages 103-104c
Continued
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Pages 103-104d