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Chapter 2
Linear Differential Equations of Second and Higher Orders
Recall, Chapter 1, a linear differential equation of 1st-order can be written in the form:
)()()(01
xfyxayxa
A 2nd
-order linear differential equation is of the form)012 ()()()( xfyxayxayxa
In general, a linear differntial equation of order n can be written in the form
)()()()(...)()( 012)1(
1)(
xfyxayxayxayxayxa nnn
n
If ,0)( xan we obtain the standard form of a linear nth
-order differential equation
)()()()(... 012)1(
1
)(
xfyxayxayxayayn
n
n
Examples: The following are linear differential equations
xyxyxorderst 2
sectancos:1
0:22 yyxyxordernd
xxeyyorderth cos:4)4(
If 0)( xf in the equations above, the differential equation is called homogeneous.
Otherwise, it is nonhomogeneous.
For instance, the 2nd
order differntial equation above is homogeneous, but the 1st- and 4
th-
orders above are both nonhomogeneous.
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Contents of Chapter 2:
Section 2.1: Homogeneous Linear DE of 2nd
-order
Section 2.2: Second-Order Homogeneous Differential Equations With Real
Constant Coefficients
Section 2.3: The Complex Case for the Second-Order Homogeneous Differential
Equations With Real Constant Coefficients
Section 2.6: Cauchy-Euler Equations
Section 2.7: Existence and Uniqueness Theorem, Wronskain
Section 2.8: NonHomogeneous Equations
Section 2.9: Undetermined Coefficients Solution Method
Section 2.10 Solution by Variation of Parameters Methods
Section 2.12 Modeling of Electric Ciruits
Section 2.13: Higher-Order Linear Differential Equations
Section 2.14: Higehr-Order Linear Homogeneous Differential Equations
With Constant Coefficients
Section 2.15: Higher-Order Nonhomogeneous Linear Differential Equations
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Section 2.1: Homogeneous Linear DE of 2nd
-orderThe standard form for homogeneous linear DE of 2
nd-order is
0)()( yxqyxpy (1)
A homogeneous solution, hy , of (1) on an interval bxa
, is any function )(xhy
defined and continuous on the interval ),( baI and satisfies the eqn (1) above for all
Ix , i.e.,0)()()()()( xhxqxhxpxh
Example 1: Each of the functions xey 1 andx
ey2 satisfies the homogeneous
differential equation
0 yy (2)
So, both of 1y and 2y are homogeneous solutions.
Note also that xxxx eeee 42&4,2 are all homogeneous solutions to the DE in (2).
Actually, for any arbitrary constants 21 & cc , the linear combinationxx
ecec 21 ofx
e
and xe is a homogeneous soluion to the DE in (2).
Theorem 1 (Homogeneous Eqn (vs) Linearity and Superposition)For the homogeneous linear DE (1), any linear combination 2211 ycyc of two solutions
1y and 2y on an open interval Iis also a solution of the DE. So, sums and constant
multiples of solutions are also solutions.
Proof: Substituting 2211 ycyc into eqn (1) above:
))(())(()( 221122112211 ycycxqycycxpycyc
))()(())()(( 22221111 yxqyxpycyxqyxpyc
0)0()0( 21 cc Done
This is the concept of linearity and superposition in case of homogeneaous solutions ofthe homogeneous DEs.
This is not true for non linear or non homogeneous as in the following examples.
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Example 2: Both xy cos11 and xy sin12 are solutions to the linear nonhomogeneou
DE: 1 yy . But, 12y and 21 yy are not solutions.
Example 3:2
1 xy and 12 y are both solutions to the nonlinearhomogeneous DE0 yxyy . But, neither 1y nor 21 yy is a solution.
General and Particular Solutions:
For the homogeneous linear DE (1), the general solution, denoted gy , is any solution in
the form 2211 ycycyg , where 1y and 2y are two solutions and 1c and 2c are any arbitrary
constants. This general solution gy is also called homogeneous solution .hy
If 1cand 2c
are specified values, say 2 & 4 for instance, the solution 21 42 yy is then
called a particular solution, .py
Initial-Value Problem (IVP):
If initial conditions, say 1000 )(&)( yxyyxy , at some point 0x are given then the eqn
1000 )(&)(0)()( , yxyyxyyxqyxpy is called an initial-value problem (IVP).
Example 4: The initial-vlaue problem (IVP) 2)0(,4)0(,0 yyyy
has the general solution xxhg ececyy 21 .
To find the values of 1c and 2c , we substitute the initial conditions in the solution gy as
follow: 214)0( ccy and 212)0( ccy . From these equations we compute
3&1 21 cc . Thusxx
p eey 3 is the only particular solution that satisfies the IVP.
Basis Solution:
Definition Two functions f & g are said to be linearly independent on a set S if0)()( 21 xgcxfc is satisfied (for all x in S) only if 021 cc . Otherwise, f & g are
linearly dependent.
For example, if 0)( xf then f& g are linearly dependent for any function g and all x in
domain g since choosing 0&0 21 cc , we obtain 00)0()(.0)(. 11 cxgxfc .
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On the other hand, xxf cos)( and xxg sin)( are linearly independent on any set since if
01 c then 0sin.cos. 21 xcxc implies that xc
cx sin.cos
1
2 , i.e., xcos is a constant
multiple of xsin , which is not true.
So, two functions f& g are linearly dependent on a set S if one of them is a constant
multiple of the other one.
Definition The set }{ 21, yy of solutions of the DE 0)()( yxqyxpy is said to be abasis solution of the DE on a set S if 1y and 2y are linearly inedependent on S.
Example 5: The set }{ , xx ee
is a basis solution to the DE 0 yy since each of xx ee
&
a solution and xx ee & are linearly independet ( cee
e xx
x
2 for any constant c).
Example 6: Since xx sin&cos are linearly independet and since each of them is a
solution to the DE 0 yy , then }{ sin,cos xx is a basis solution to the given DE.
Remark: A basis solution of a homogeneous differential eqn is not unique. For if the set
}{ 21, yyof solutions of the DE 0)()( yxqyxpy
is a basis of the DE then }{ 2211 , ykyk
is also a basis solution to the DE for any nonzero constants21 & kk .
How do we find a basis solution to the homogeneous 2nd
-order differential equation
0)()( yxqyxpy ?
For now, we assume we have one solution 1y . We now use Reduction of order Method
obtain a second solution 2y that is linearly independent from 1y .
Reduction of order method:
Find a function u such that 12 uyy is a solution to the DE above. Since 112 yuyuy and1112 2 yuyuyuy we substitute into the equation we obtain:
0)())(()2( 11111 uyxqyuyuxpyuyuyu
0)2)(())()(( 111111 yuyyxpuyxqyxpyu
0)/)(2( 11 uyyxpu
Let uv . Then the eqn above reduces to the separable 1st-order differential eqn:
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)/2)(( 11 yyxpv
v
Integrating both sides with respect ot x, we obtain:
1ln2)(ln ydxxpv
dxxpey
vu )(2
1
1
Hence dxey
udxxp
)(
2
1
1and dxe
yyuyy
dxxp
)(
2
1
112
1.
Since .constant1 )(2
11
2
dxeyy
y dxxp , then }{ 21, yy is a basis solution and so the general
solution is then 2211 ycycyg .
Example 7: Find a basis solution to the following DE with xy 1 is one solution:
02 yyxyx
Solution: Write the DE in standard form: 0)/1()/1( 2 yxyxy
.0,)/1()( xxee dxxdxxp Hence, xdxx
xu ln
12
, xxuyy ln12 and
)ln(ln 1111 xccxxxcxcyy gh
Problems:
(6) Solve xxyxyyyx /sin,02 1
Answer: xxcxcyxxuyy h /)cossin(/cos 2112
(4) Solve yyx 32 Hint: Substiute yz , then yz and the eqution then becomes: zzx 32
Answer: 22/5
1 cxcy
(5) Solve 2)(2 yyy
Hint: Substiuteyz
, thenz
dy
dz
dx
dy
dy
ydy
and the eqution then becomes:2
2zdy
dzyz
Answer:21
1
cxcy
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Section 2.2: Second-Order Homogeneous Differential Equations With Real
Constant Coefficients
Consider the 2nd
-order linear homogeneous Differential Equation
0 cyybya (1)
Where a, b and c are real constants.
Idea of Solution:Recall a first-orderlinear homogeneous DE
0 kyy has the exponential solution kxey , where kis constant.
So, to find the solution of (1), lets try rxey , where ris constant to be determined.
Substituting rxey into (1), we get the auxiliary (or characteristic) quadratic equation
02
cbrar (2)The roots of this quadratic equation are:
acbb
aracbb
ar 4
2
2
1&4
2
2
121
And we have two solutionsxr
ey 11 andxr
ey 22 (3)From these two solutions, we obtain a basis solution and hence the general solution .gy
Types of Solutions:
There are three types of solutions depending on status of the the roots 21 & rr . We have
three cases of the roots:
(i) Case 1: 21 & rr are real and distinct(ii) Case 2: 21 rr and real(iii) Case 3: 21 & rr are complex and distinct
We now discuss the general solution in each case.
Case 1: Distinct Real Roots
Rrr 21
This case occurs if the discriminant 042 acb , hence the roots 21 & rr are real and distin
and the two solutions xrey 11 andxr
ey 22 are linearly independent
( tconseyy xrr tan/ )(2121 ).
So, in this case, the basis is },{ 21xrxr
ee and the general soultion is xrxr ececg
y 21 21 .
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Example 1: Solve the IVP: .1)0(,1)0(,034 yyyyy Solution: Let rxey . Then we hav 0342 rr The roots are 21 13 rr which are real, hence
xxecec
gy
23
1
Substituting 1)0(,1)0( yy , we obtain 2121 31&1 cccc
And the values of 21 &cc are 121 &0 cc , so the solution isx
ey
Case 2: Double Real Roots Rrr 21
This case occurs if the discriminant 042 acb , hence we have double real roots:
a
brr
221 .
Hence, the two solutions abxeyy 2/21 are dependent.
Question: How do we obtain a basis of soltuion in this case?
Answer: We apply the Rduction-of-Order method (discussed in Section 2.1) on the DE0)/()/( yacyaby
with a solution abxey 2/1 is given. Let 12 uyy . Then dxe
yu
dxxp
)(
21
1, where
abxp /)( . So, xdxee
u abxabx
/
/
1, 12 xyy and the general solution is then
xrg exccy
1)( 21
Example 2: Solve the IVP: .1)0(,1)0(,096 yyyyy Solution: Let rxey . Then, the auxiliary (characteristic) equation is 0962 rr .The roots are 21 3 rr , a real double root. Hence, by case 2 above,
xxxg exccxececy
321
32
31 )(
Substituting 1)0(,1)0( yy , we obtain 211 31&1 ccc
And from which we have 22 c , so the solution isxexy 3)21(
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Section 2.3: The Complex Case for the Second-Order Homogeneous Differential
Equations With Real Constant Coefficients
We are solving the 2nd
-order linear homogeneous Differential Equation
0 cyybya (1)
Where a, b and c are real constants.
Substituting rxey into (1), we get
acbb
aracbb
ar 4
2
2
1&4
2
2
121
Case 3: Complex Conjugates Roots Crr 12
This case occurs if the discriminant 042 acb . Definea
b
2 and abac 2/4 2 .
Since the coefficients a, b and c are real constants then the roots are complex conjugates:
irrir 121 & , where1
i and the two solutions ,xix
eey
1 andxix
eey 2 are linearly independent ( tcons
xieyy tan2
/ 21 ) . But they are complex.
Question: How do we obtain a basis of real soltuions in this case?
Answer:Using Eulers Identities: xixeix sincos and xixeix
sincos
Hence, xee
ixix
cos2
and xi
eeixix
sin2
To find a real basis, note that )21
(xixix
g ececey
Choosing 2
121 cc , we obtain xey xp cos1
.
Choosingi
ci
c2
1&
2
121 , then xey
xp
sin
2
.
Note that21
&pp
yy are real, linearly independent and homogeneous solutions of (1). So,
the basis is }sin,cos{ xx
exx
e
and the general soultion is
)sin2
cos1
( xcxcx
eyg
Example 3: Solve the IVP: .1)0(,1)0(,052 yyyyy
Solution: Letrx
ey . Then we hav 0522
rr The roots are irir 21
2&21
1 . So, 1 and 2 , then
)2sin2
2cos1
( xcxceg
y x
Substituting 1)0(,1)0( yy , we obtain 211 21&1 ccc . Hence, 21 1 cc , so the
solution is )2sin2(cos xxey x .
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Section 2.7 Existence and Uniqueness Theorem, Wronskain
In this section, we give a general theory for the homogeneous linear 2nd
-order differential
equation of the form
0)()(01
yxayxay (1)
where )(&)( 10 xaxa are continuous functions on a given open interval I.
Consider solving the initial-value prorblem:
1001 )0(&)0(,0)()( yyyyyxayxay (2)
Theorem 1: (Existence and Uniqueness Theorem)
If )(&)( 10 xaxa are continuous functions on some open interval Iand if Ix 0 , then the
initial-value problem (2) has a unique solution )(xy on the interval I.
Linear Independence of Solutions,WronskianThe Wronskian ),( 21 yyW of two solutions )(&)( 21 xyxy of (1) is the determinant
2121
21
21
21 det),( yyyyyy
yyyyW
Theorem 2: (Linear Independence and Dependence of Solutions)
Suppose )(&)( 10 xaxa are continuous functions on some open interval I. Then two solutions
)(&)( 21 xyxy of (1) on Iare linearly dependent on Iif their Wronskian is zero at some poin
Ix 0 . Furthermore, if 0W at some Ix 0 , then 0W on I; hence if there is Ix 1 at
which 0W , then )(&)( 21 xyxy are linearly independent on I.
Example 1: xxyxxy sin)(&cos)( 21 are solutions of 02 yy , with 0 , on any
interval. Their wronskian is 0cossin
sincosdet)sin,(cos
xx
xxxxW
So, by Theorem above, xx sin&cos are linearly independent on all ofR. Hence, the
general solution to 02 yy is xcxc sincos 21 , where 21 & cc are arbitrary constants.
Example (2):xx
xexyexy )(&)( 21 are solutions of 02 yyy on any interval.
Their wronskian is 0)1(
det),(2
x
xx
xx
xxe
exe
xeexeeW
So, xx xee & are linearly independent on all ofR , and the general solution to 02 yyy
is xexcc )( 21 where 21 & cc are arbitrary constants.
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Theorem 3: (Existence of a General Solution)
If )(&)( 10 xaxa are continuous functions on some open interval I, then the differential
equation 0)()( 01 yxayxay has a general solution on I.
Theorem 4: (General Solution)Suppose that equation (1) has continuous coefficients )(&)( 10 xaxa on some open interval .
Thenevery solution of (1) on Iis of the form )()()( 2211 xycxycxyg , where )(&)( 21 xyxy
form a basis of solutions of (1) on Iand 21 & cc are arbitrary constants.
Hence, (1) does not have singular solutions.
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Section 2.13 Higher-Order Linear Differential Equations
In this section, the concepts and definitions introduced to the linear differential equations
order 2 in Sections (2.1 & 2.7) are generalized to the Linear Differential Equations of
higher-orders (n > 2):
)()()()( 01)1(
1)( ... xfyxayxayxay nn
n (1)If 0)( xf , we obtain the homogeneous differential equation in (2). Otherwise, it is
nonhomogeneous.
0)()()( 01)1(
1)( ... yxaxayxay ynnn (2)
A homogeneous solutionhy of (2) on an interval bxa , is a function )(xhy , defined
and continuous on the interval ),( baI , that satisfies eqn (2) above for all Ix .
Example 1: Each of the functions xey 1 ,x
ey2 and
xxey 3 satisfies the following
3rd-order linear and homogeneous differential equation: 0 yyyy
So, all of 1y , 2y
and 3y are homogeneous solutions.
Note also that xxxx xeeee 42&4,2 are also homogeneous solutions to the above DE
Theorem 1 (Superposition and Linearity Principles)For the homogeneous linear DE (2) , sums and constant multiples of solutions on an open
interval Iare again solutions of (2) on I.
Proof: It is a generalization of the proof in Sec. 2.1 Done
Again, this is not true for non linear or non homogeneous.
Linear Independence and Dependence Definition
The n functions )(...,),(),( 21 xyxyxy n are called linearly independent on an interval Iif
0)(...)(11 xycxyc nn (3)
For all x in Iimplies that all nccc ...,,, 21are zero. Otherwise, they are linearly dependent.
For example, if 0)(1 xy then )(...,),(),( 21 xyxyxy n
are linearly dependent by putting
0...,1 21 nccc Also, )(...,),(),( 21 xyxyxy n are linearly dependent on I iffone of the functions can be writte
as a linear combination of the other (n-1) functions.
For instance, ))(...)((1
)( 221
1 xycxycc
xy nn
iff 01 c
iff )(...,),(),( 21 xyxyxy n are linearly dependent.
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Example 2: The functions 2321 )(&2)(,)( xxyxxyxxy are linearly dependent since
)(.0)(2
1)( 321 xyxyxy
Example 3: Show that the functions3
3
2
21 )(&)(,)( xxyxxyxxy are linearlyindependent on any interval, say ]3,1[I
Solution:
Equation (3) is: 0332
21 xcxcxc Putting 3&2,1 xxx , into (3) we obtain the following system of linear equations:
02793
0842
0
321
321
321
ccc
ccc
ccc
Solving these equations we obtain 0321 ccc . Hence32
&, xxx are linearly independen
on the interval ]3,1[ .
General Solution, Basis and Particular Solutions:
A basis of solution of (2) on an open interval Iis a set }...,,,{ 21 nyyy ofn linearly
independent solutions.
A general solution of (2), denoted ,gy on an open interval Iis any solution of the form
nng ycycycy ...2211 (4)
where }...,,,{ 21 nyyy is a basis (fundamental system) of solutions of (2) and nn cccc &...,,, 121 are any arbitrary constants. This gy is also called homogeneous solution .hy If nn cccc &...,,, 121
in (4) are specified values, the solution is then called a particular
solution py .
Example 4: The functions 32 &, xxx in Example 3 above are linearly independent solution
to the 3rd
-order linear and homogeneous differential equation 0663 23 yyxyxyxon
any interval does not contain 0.
So, a general solution to the above equation is3
3
2
21 xcxcxcyg and (for example)32
523 xxxy p is a particular solution to the DE in Example 4 above.
Initial-Value Problem, Existence and Uniqueness
Consider the IVP:10
)1(1000
01)1(
1)(
)(...,,)(,)(
0)()()( ...
nn
nn
n
yxyyxyyxy
yxayxayxay(5)
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Theorem 2: (Existence and Uniqueness Theorem)
If )(...,,)( 10 xaxa n are continuous functions on some open interval Iand if Ix 0 , then the
initial-value problem (5) has a unique solution )(xy on the interval I.
Example (5) Solve the IVP
4)1(,1)1(,2)1(
066323
yyy
yyxyxyx
Solution: Note that this equation is not in standard form. Its standard form is
066332
yx
yx
yx
y
So, the coefficeints are continuous on any interval not containing 0. Since ,010 x then b
Theorem 2 above, a unique solution exists on the interval where .0x
In Example 4,we saw that the general solution to this eqn. is3
32
21 xcxcxcyg Substituting the given initial conditions, we obtain
32
321
321
624)1(
321)1(
2)1(
ccy
cccy
cccy
Solving for the constants, we obtain 322 xxxy p as the unique solution with x > 0.
Linear Independence of Solutions. Wronskain
The Wronskian ),...,( 1 nyyW ofn solutions )(,...,)(1 xyxy n of (2) is the nth-orderdeterminant
)1()1(1
21
21
1
........
.
.
.......
......
det)...,,(
n
n
n
n
n
n
yy
yyy
yyy
yyW
Theorem 3: (Linear Independence and Dependence of Solutions)
Suppose )(...,,)( 10 xaxa n
are continuous functions on some open interval I. Then n solutio
)(),...,(1 xyxy n of the differentia equation
0)()(...)( 01)1(
1)( yxayxayxay
nn
n (2)
on Iare linearly dependent on Iif their Wronskian is zero at some point Ix 0 .
Furthermore, if 0W at some Ix 0 , then 0W on I; hence if there is Ix 1 at which W
, then )(,...,)(1 xyxy n are linearly independent on I.
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Example 6: 32
32
322
620
321det),,( x
x
xx
xxx
xxxW
By Example 5, the functions
32&,xxx are solutions to the 3
rd
-order linear and homogeneodifferential equation 0663 23 yyxyxyx on any interval not containing 0.
Since, 02 3 xW for any ,0x then the solution set },,{ 32 xxx is liearly independent on
),0( I and hence a basis of solutions of the equation above.
Example 7: Since 04
)2(
)1(det),,(
x
xxx
xxx
xxx
xxx e
exee
exee
xeee
xeeeW
And sincexxx
xeee &,
are solutions to 0 yyyy for all x, then by Theorem above,},,{
xxxxeee is a basis of solutions to the equation.
Theorem 4: (Existence of a General Solution)
If the coefficients )(...,,)( 10 xaxa n of (2) are continuous functions on some open interval I,
then (2) has a general solution on I.
Theorem 5: (General Solution)
Suppose that (2) has continuous coefficients )(...,,)( 10 xaxa n on some open interval . Then
every solution of (2) on Iis of the form )(...)()()( 2211 xycxycxycxy nng where )(,...,)(1 xyxy n form a basis of solutions of (2) on Iand 21 & cc are arbitrary constant
Hence, (2) does not have singular solutions.
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Section 2.14: Higehr-Order Linear Homogeneous Differential Equations
With Constant CoefficientsIn this section, we discuss the solution to the Higehr-Order Homogeneous Differential
Equations with real constant coefficients 001)1(
1)( ... yayayay nnn (1)
Since the coeffecients of (1) are all consatnts, and hence contintuous on R, then,
by Theorem 4 of Section 2.13, a general solution always exists on all ofR.
To find the solution of (1), we imitate the method of in Sections 2.2 (the 2nd
-order
homogeneous equation with constant coefficients). So, we assume an exponential
solution rxey , and we substitue it in eq. (1) we obtain the auxiliary equation:
0011
1 ... ararar nnn (2)
Let nrr ...,,1 be the n roots of (2). Then we obtain the solutionsxr
n
xr n
eyey ...,,1
1 To find a basis of solutions, we look for n linearly independent solutions of the form ey
Remark:Before discussing the cases of the roots, we need the following Vandermond Determinant
of order n
:
distinctarerallifnonzero
jisomeforrrif
rr
rrr
rrr
V
i
ji
ji
nji
nn
n
n
nn
n
,
,,0
)()1(
.
.
1.......11
det
1
2/)1(
11
2
1
1
21
Cases of the Roots:As in Section 2.2, we have three cases to consider:
(iv) Case 1: all roots are real and distinct(v) Case 2: some complex roots(vi) Case 3: some repeated real roots.
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Case 1: Distinct Real Roots jiRrr ji ,
In this case, the n solutions are xrnxr neyey ...,,11 . Using Vandermond determinnat, the
Wronskian of these solutions is nonzero:
xrnn
xrnxrn
xrn
xrxr
xrxrxr
xrxr
n
n
n
n
ererer
ererer
eee
eeW
11
2
1
1
21
21
21
21
1
.
.
.......
det),...,(
0
)()1(
.
.
1.......11
det
1
2/)1()...(
11
2
1
1
21
)...(
21
21
ji
nji
nnxrrr
n
n
nn
n
xrrr
rre
rrr
rrr
e
n
n
So the solutions above form a basis and a general solution in this case is nxr
g ececy ...1
1
for all real x.
Theorem 1: (Basis)
n solutionsxr
nxr n
eyey ...,,1
1 of (1) form a basis of solutions of (1) iff all the n roots
nrr ...,,1 of (2) are distinct.
Theorem 2 (Linear Independence)
Any number of solutions xrkxr keyey ...,,11 of (1) are linearly independent on any open
interval I iff krr ...,,1 are distinct.
Example 1: Solve the differential Equation022 yyyy
Solution: rxey 022 23 rrr The roots of this auxiliary equation are: 2&1,1 and the general solution is then
xxxg ecececy
2321
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Remark To look for the roots, we use the following helpful steps:
- By rational zero test, the rational zeros are obtained from the set }1
2,1{
where 1 &
2 are the divisors of 2 (the constatnt term) and 1 is the divisor of 1 (the leading
coefficient)
- By Descartes rule of sign, there are 2 or no positive real roots and exactly onenegative root.
- Use synthetic division by the possible rational zeros to obtain the roots.
Case 2: Some Complex Conjugate Roots Crr ji Note: If the coefficients 01 ...,, aan are all real, then if there is a complex root, say
ir , then its conjugate ir is also a root, i.e., complex roots occurd as pairs.
As in Section 2.3, the two real solutions that correspond to the complex conjugate roots
ir are xe x cos and xe x sin .
Example 2: Solve the IVP: 299)0(&11)0(,4)0(,0100100 yyyyyyy
Solution: rxeyThe auxiliary equation is 010010023 rrr
-The rational zeros are divisors of 100.
- There are 3 or 1 positive root
- There are no negative roots
- By synthetic division by 1, we obtain a quadratic quation:
010001
10001
100100111
So, 11 r is a real root. The other two roots are the roots of the quadratic equation
01002 r which are the complex conjugate roots ir 103,2
So, the general solution is xcxcecy xg 10sin10cos 321
We now substitute the initial values
21
21
21
100299)0(
1011)0(
4)0(
ccy
ccy
ccy
Solving for the constants 321 &, ccc we obtain xxeyx
g 10sin10cos3
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Case 3 Some Repeated Real Roots mrrr ....21
Ifa real root 1r is repeated m-times ( 1r is a root of multiplicity m), say mrrr ....21 ,
then (as a generalization of the 2nd
-order in this case) we obtain m independent solutions
as follow: xrmxrxr exmyxeyey 111
11 ...,,2
,
Example 3: Solve the DE: 033 yyivyvy
Solution: The auxiliary equation is 033 2345 rrrr
The roots are: 0, 0, 1,1 ,1 and the general solution is xexdxddxccy )()( 221010
Ifa complex root, say
ir and hence its conjugate ir are repeated, say
mrrr ....21 and mrrr ....21 then indepndent solutions are obtained by multiplying
the solutios xe x cos and xe x sin by x for each time the roots are repeated, so we
obtain the solutions,sin...,,sin,sin
cos...,,cos,cos
1
1
xexxxexe
xexxxexe
xmxx
xmxx
For example, if ir 211 is a root of multiplicity 3 of the auxiliary equation of a
homogeneous DE with real constant coefficients, then its conjugate ir 211 is also a
repeated root of multiplicity 3. Hence, the solutions corresponding to these roots are
)2sin)(2cos)((2
2102
210 xxBxBBxxAxAAex
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Section 2.6: Cauchy-Euler Equations
Consider the 2nd
order Cauchy-Euler DE: 02 cyybxyax (1)
where a , b and c are real constants. Writing this equation in standard form, we obtain
02
y
ax
cy
axby
(2)
By Theorem 4 of Section 2.13, since the coeffecients of (2) are all contintuous on }0/{R ,
then a general solution always exists on any interval not containing 0.
So, a solution of (2)( hence of (1)), exists on ),0( or )0,( .
Idea of Solution:
We look for a function fsuch that fxfx 2& are both constant multiples off. For
instance, rxxf )( satifies frrfxrffx )1(& 2 .
So we assume the solution of the formr
xy where r is a constant to be determined.Substituting rxy into (1), we get the characteristic (auxiliary) quadratic equation
0)1( cbrrar (3)
Or 0)(2 crabar (4)
The roots 21 &rr of this quadratic equation are:
acababa
r 4)()(2
1
12 and acabab
ar 4)()(
2
1
22
And we have two solutions 11r
xy and 22r
xy (5)
From these two solutions, we obtain a basis solution and hence the general solution .gy
Types of Solutions:
There are three types of solutions depending on the the roots 21 &rr . We have three cases
of the roots:
Case 1: 21 & rr are distinct and real
Case 2:21rr and real
Case 3: 21 & rr are distinct but complex conjugates.
Note that if 21 &rr are distinct, then the two solutions:1
1
r
xy and2
2
r
xy areindependent since consatntx
y
y rr 12
1
2 (
(The Wronskian is 0)( 11212
11
21
21
21
rr
rr
rr
xrrxrxr
xxW (all 0x )).
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Case 1: Distinct Real Roots Rrr 21
In this case, the two solutions 11r
xy and 22r
xy are independent (by the Wronskian) and
hence the general solution is 21 21rr
g xcxcy
Example 1: Solve 025.22 yyxyx
Solution: Let rxy . Then we hav 025.32 rr The roots are 21 2/14 rr which are real, hence
2/141 2
xcxcg
y
Case 2: Double Real Roots Rrr 21
This case occurs if the discriminant 04)( 2 acab , hence we have double real roots:
a
abrr
2
)(21
. Hence, the two solutions are dependent ( aabxyy 2/)(21
).
To obtain a basis of solution in this case, we apply the Rduction-of-Order method(Section 2.1) on the DE 0)/()/( 2 yaxcyaxby , where a solution aabxy 2/)(1
is give
Let 12 uyy . Then xdey
udxxp
)(
21
1, where )/()( axbxp .
So, ,ln/)(
/
xdxx
xu
aab
ab
xyy ln12 and the general solution is then1)ln( 21r
g xxccy
In this case, the Wronskian of the solutions is then 0//ln
ln 21
111
11
xyxyxyy
xyy
Example 2: Solve 0432 yyxyx
Solution: Let rxy . The auxiliary equation is 0442 rr . The roots are 21 2 rr , a
real double root. So, 221 )ln( xxccyg .
Case 3: Complex Conjugates Roots Crr 12
This case occurs if the discriminant 04)( 2 acab .
Definea
ab
2
and aabac 2/)(4 2 , then the roots are 0& ,21 irir
Note that the roots 221 rrr are complex conjugates and the two solutions1
1r
xy &
22
r
xy are linearly independent since 02),( 1221 xiyyW or ( tconsxyy i tan/ 221
).
But, they are complex: ixxy 1 and i
xxy2 .
To obtain a basis of real soltuions in this case, note that
)lnsin()lncos(&)lnsin()lncos(lnln xixexxixex xiixii
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Then: )lncos()(21 xxx ii and )lnsin()(
2
1xxx ii
i
So, )lncos(1 xxy and )lnsin(2 xxy
are two real and linearly independent
solutions of (1) when the roots are complex conjugates )tan)lncot(/( 21 tconsxyy
Hence, the general soultion is ))lnsin()lncos(( 21 xcxcxyg
Example 3: Solve 01372 yyxyx
Solution: Let rxy . Then we hav 01362 rr The roots are irir 23&23 21 , so, 3 and 2 and
))ln2sin()ln2cos(( 213 xcxcxyg
Example 4: Boundary Value Problem (BVP)Electric Field Between two Concentric Spheres
Find the electrostatic potential )(rvv between two concentric spheres of radiicmrcmr 8&4 21 kept at potentials voltsvvoltv 0&110 21 , respectively.
Solution: Physical InformationThe potential )(rvv satisfies 02 vvr .
This is Cauchy-Euler equation, So, let mrv .
Then we have 02 mm . The roots are .1&0 m
So, the potential is rccrv /)( 21 . Substituting the boundary values:
8/21
0)8(
4/
21
110)4(
ccv
ccv
The solution is then rrv /880110)(
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Section 2.8 NonHomogeneous Equations
In this section, we give a general theory for the non homogeneous linear 2nd
-order
differential equation of the form
)()()(01
xfyxayxay
(1)
where )(&)( 10 xaxa are continuous functions on a given open interval I.
Consider also the homogeneous linear 2nd
-order differential equation of the form:
0)()( 01 yxayxay (2)
Theorem 1: Relations Between Solutions of (1) & (2)
a- If 21 & yy are two solutions of (1) on some open interval I, then 21 yy is a solutionof (2) on Ialso.
b- If py is a solution of (1) and hy is a solution of (2) on some open interval I, thenhp yy is a solution of (1) on I.
Proof: Define yxayxayyL )()(][ 01 .
Note that ][yL is a linear operator, i.e., ][][][ 22112211 yLcyLcycycL a- If 21 & yy are solutions of (1), then )(][][ 21 xfyLyL
Hence, 0)()(][][][ 2121 xfxfyLyLyyL , so 21 yy is a solution of (2) on Ialso
b- If py is a solution of (1) on some open interval Iand hy is a solution of (2) on Ialso, then 0][&)(][
hp
yLxfyL . Hence )(0)(][][][ xfxfyLyLyyLhphp
Definitions: General Solution of the nonhomogeneous eqn (1)
If )()( 2211 xycxycyh is a solution of (2) and if )(xy p is a solution of (1), then
)()()( 2211 xyxycxycyyy pphg (3)
is a general solution of (1).
A Particular Solution of the nonhomogeneous eqn (1) is a solution of the form
)()()( 2211 xyxycxyc p where the constants 21 & cc are specified numbers.
General Solution of (1) Includes All Solutions
Theorem 2: Suppose the coefficients )(),( 10 xaxa and the function )(xf of (1) are all
continuous functions on some open interval I. Thena general solution of (1) exists on I
and every solution of (1) is obtained by assigning suitable values to the arbitrary
constants in the genearl solution (3).
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How to Solve (1) or IVP for (1)?
We find the homogeneous solution hyof equation (2) and find the particular solution py
of (1).
Example 1: Find the general solution of .454 2xeyyy
Solution:
Step 1: Find the homogeneous solution of 054 yyy
Since the coefficients are constants, then the solution is of the form rxh ey .
Substitutinginto the homogeneous equation above , we get the auxiliary
equation 0542 rr . We obtain irir 2&2 11 and so
)sincos( 212 xcxcey xh
.Step 2: We now find the particular solution py of .454
2xeyyy Since xexf 24)( and since the coefficients of the equation are conatant,
then py and its derivatives should be in the form of .2xAe So, let
xp Aey
2 ,
then we have xxxx eAeeAeA 2222 45)2(4)4( .Solving for A, we obtain 17/4A
Step 3: So, the general solution is then xxg excxcey2
212
)17/4()sincos( .
Example 2: Solve the initial value problem IVP
.1)0(&0)0(,454 2 yyeyyy x
Solution:
Step1: xxh ececy5
21 is the homogeneous solution of 054 yyy
Step 2: Again, as in Example (1), we assume the particular solution as xp Aey
2 .
Then we have, xxxx eAeeAeA 2222 145)2(4)4(
Solving for A, we obtain .2A
Step 3: xxxg eececy25
21 2
Step 4: Find the coefficients 21 &cc such that the solution satisfies the initial values.
With ,0)0( y then 20 21 cc And ,1)0( y then 451 21 cc
Solving these two equations for 21 &cc , we obatin .2
1&
2
521 cc
Hence the unique particular solution to the IVP is xxxp eeey25
22
1
2
5 .
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Section 2.9 Undetermined Coefficients Solution MethodIn this section, we solve the nonhomogeneous eqn
)()()( 01 xfyxayxay (1)
by the Undetermined Coefficients Solution Method. This method requires eqn (1) to havetwo properties:
1- The coefficients are constants, i.e., eqn (1) becomes:)(012 xfyayaya (2)
In this case, the homogeneous solution hy is found as in Sections 2.2 & 2.3 by
assuming .rxh ey
2- The function )(xf is a linear combination of bxorandbxbxbxex axr cosh/sinh,cos,sin,,Steps of Solution
Step 1: As in Sec 2.2 & 2.3 find the homogeneous solution )()( 2211 xycxycyh of
0012 yayaya (3)
Step 2: Find the particular solution py of (2) according to the following rules:
(a)Basic Rule:Choose *py
in the same form as )(xf but with undetermined coefficients as in the
following table:
)(xf Assumed *py
nkx 0
11 ... AxAxA
nn
nn
rxke rxAe
bxkcos or bxksin bxBbxA sincos nrxxke rxnn eAxA )...( 0
bxkerx
cos or bxkerx sin rxebxBbxA )sincos(
bxkxn
cos or bxkxn sin bxAxA nn cos)...( 0 bxBxBn
n sin)...( 0
bxekxrxn
cos or bxekx rxn sin bxAxAe nnrx cos)...( 0 bxBxBe
nn
rxsin)...( 0
(b) Modification Rule:
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If the assumed *py or any of its derivatives is already a homogeneous solution of (3) then
multiply *py byrxxx ...,,,
2 where ris the least positive integer such that *pryx is not a
homogeneous solution. The particular is then *pr
p yxy .
(c) Sum Rule (Superposition)If )(...)()()( 21 xfxfxfxf m , then for each )(xfk , we assume a particular solution kpy ( modified if needed) and the particular solution will be the superposition ppp yyy ...1
Example 1: Solve 284 xyy
Solution:
Step1: The homogeneous solution of 04 yy is xcxcyh 2sin2cos 21
Step 2: Since 2)( xxf , then the assumed paticular solution is CBxAxy p 2* .
No modification is needed, so *pp
yy .
To find A, B& C, substitute py in the nonhomogeneous eqn we obtain22
8)(42 xCBxAxA and from which we have 042&04,84 22 CABxxAx
So, 1&0,2 CBA and then )12(2sin2cos 221 xxcxcyg
Example 2: Solve xeyyy 2423
Solution:
Step1: xxh ececy2
21
Step 2: Since xexf 24)( , we assume xp
Aey2* which is a homogeneous solution. So
a modification is needed by multiplying by x: xpp Axexyy2* .
To find A, substitute py in the nonhomogeneous eqn we obtainx
p Axey2 xp Axey
222
xp exAy
2)12(
xp exAy
2)36(3
xp exAy
2)44(
xp exAy
2)44(
We add and obtain: 44 22 AeAe xx . So,xxx
g xeececy22
21 4
Example 3: Solve the IVP 1)0(,0)0(,62 yyeyyy x
Solution:
Step1: xxh xececy 21
Step 2: : Since xexf 6)( , we assume xp Aey* . Again, a modification is needed
since xAe is a homogeneous solution. To modify it, multiply by x . Again
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xAxe
is a homogeneous solution, so multiply by x again and this timex
pp eAxyxy 22 is not a homogeneous solution.
To find A, substitute py in the nonhomogeneous eqn we obtain
xp eAxy 2 xp eAxy 2
xp exxAy
)2( 2
xp exxAy
)42(2 2
xp exxAy
)24( 2
xp exxAy
)24( 2
We add and obtain: 362 AA , so, xg exxccy
)3( 221
Step 3: Find the coefficients 21 &cc
10)0( cy and 211)0( ccy , sox
g exxy )3( 2
Example 4: Solve the IVP 1)0(,0)0(,2cos452 yyxxeyyy x
Solution:
Step1: )2sin2cos( 21 xcxceyx
h
Step 2: xDCxexBAxey xxp 2sin)(2cos)( .
Note that part of *py is a homogeneous solution, namely
)2sin2cos( xDxBe x . So, modify *py by multiplying it by x to obtain
xDxCxexBxAxexyy xxpp 2sin)(2cos)(22
Step 3: Find the constants A, B, C & D by substituting py in the nonhomogeneouseqn. Then ]2sin)(2cos)[( 2
21
2 xcDxCxxcBxAxey xg
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Section 2.15 (a) Undetermined Coefficients Solution Method for Higher Order DEsWe now apply Undetermined Coefficients Solution Method to Higher Order
nonhomogeneous differential equations with the same properties on the coefficients and
f(x).
)(...0
)1(
1
)( xfyayaya n
n
n
n
(4)where )(xf is a linear combination of bxorandbxbxbxex axr cosh/sinh,cos,sin,,
Remark: Let yayayayL nnn
n 0)1(
1)(
...][
So eqn (4) becomes)(][ xfyL (5)
We give example using Superposition Principle:
If )(...)()()( 21 xfxfxfxf m , then for each )(xfk , we assume a particular solution kpy ( modified if needed) and the particular solution will be the superposition
mpppyyy ...1
Example 1: Without evaluating the undetermined coefficients, write the general solution o
xeexyyy xxiv sin27854223
Solution:
Step 1: Soving 054 yyy iv , we obtain )sincos()( 322
10 xcxcexccyx
h Step 2: To find the form of py , we find the particular solution for each )(xfk .
)(xf Assumed: py Modified: py
38x DCxBxAx 23 )( 232 DCxBxAxx
xe2
7 xEe2 xEe2
xexsin2
2 )cossin(2 xGxFe x )cossin(2 xGxFxe x
So, )cossin()( 22232 xGxFxeEeDCxBxAxxy xxp and phg yyy .
Example 2: Suppose the homogeneous solution and )(xfof (2) are given, write the
general sloution in each of the following cases:
(i) xh ecxxccxxccy 254321 2sin)(2cos)( and xx exexxf 22 2sin102cos)( Then: xxp EexDxCexBxAy
22)2sin2cos()2sin2cos(
Modification is needed for the first and third terms of py The modified py is then
xxp ExexDxCexBxAxy
222)2sin2cos()2sin2cos(
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(ii) xxh HeeGFxExxDxCBAxy 22
)()sincos()(
and xx eexxxxf 52cos)( 223
Then: xxp JeIHxGxeFExDxCxxBxAy22223
)()()sincos(
Modification is needed for all the terms
The modified py
is thenxx
p JxeIHxGxexFExDxCxxxBxAxy2223232
)()()sincos(
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Section 2.10 Solution by Variation of Parameters Methods
In this section, we use the Variation of Parameters Method to solve the
nonhomogeneous DE
)()()(01
xfyxayxay (1)
where )(&)(),( 10 xfxaxa are continuous functions on a given open interval I.
Contrary to the method of Undetermined Coefficients, this method requires no specific
conditions on the coefficients )(),( 10 xaxa nor on the function )(xf other than being
continuous on an open interval I.
How does this method (Variation of Parameters ) work?
Steps of the Solution Method:
Step 1: Find the homogeneous solution )()( 2211 xycxycyh of the homogeneous
equation: 0)()( 01 yxayxay (2)
Step 2: Find the particular solution py as follow:
2211 )()( yxuyxuy p (3)
where 21 &uu are variables (functions of x) to be determined.
The general solution is then phg yyy
Remark(1): This way of finding py resembles the method of finding a 2nd
- order
linearlyindependent solution 12 uyy from a given one 1y .
Computation of 21 &uu :
2211 yuyuyp
)( 22112211 yuyuyuyuyp
In order not to obtain 2nd
- order differential equation in the new variables 21 &uu , we
require0
2211
yuyu (4)Then 22112211 yuyuyuyuyp
Substituting ppp yyy &, in equation (1) as a particular solution and simplifying we obtain:
)()())()((
))()(()()(
2211202122
10111101
xfyuyuyxayxayu
yxayxayuyxayxay ppp
(5)
Since 21& yy are homogeneous solution satisfying (2), then (5) simplifies to
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)(2211 xfyuyu
(6)
We now have two equations relating 21 &uu , namely (4) and (6):
02211 yuyu
)(2211 xfyuyu
Using Cramers Rule, we solve for 21 &uu and obtain:
W
Wu
W
Wu 22
11 & , where 1
1
1
22
2
2
1 )()(
0,)(
)(
0yxf
xfy
yWyxf
yxf
yW
and
21
21
yy
yyW The wronkain of 21 & yy .
Integrating both 21 & uu and substituting 21 &uu in (3), we obtain dxuydxuyyp2
211
Remark(2) Note that equation (1) is put in standard form
)()()( 01 xfyxayxay
Example (1) Solve xyy sec
Solution:
Step1: xcxcyh sincos 21 and hence 1cossin
sincos 1
xx
xxW
Step 2: xuxuyp sincos 21 .
Then xxxyxfxx
xW tansinsec)(
cossec
sin021
and 1cossec)(secsin
0cos12
xxyxf
xx
xW
So, xxdxuxu coslntantan 11 and xdxuu 22 1 And the general solution is xxxxxcxcyg sincos)cos(ln)sincos( 21
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Section 2.15(b) Solution of Nonhomogeneous Higher Order Differential Equations
using Variation of Parameters Method
In this section, we generalize the Method Variation of Parameters to solve the higher
order nonhomogeneous eqn
)()()(...)( 01)1(
1)(
xfyxayxayxayn
nn (1)
Steps of the Solution Method
Step 1: Find the homogeneous solution )(...)(11 xycxycy nnh of the homogeneous
equation
0)()(...)( 01)1(
1)( yxayxayxay
nn
n
(2)
Step 2: Find the particular solutionpy as follow:
nnp yxuyxuy )(...)( 11
(3)
Where nuuu ...,,, 21 are variables (functions ofx) to be determined.
The general solution is then phg yyy
Computation of nuu ...,,1 Given nnp yuyuy ...11 )...(... 1111 nnnnp yuyuyuyuy
In order not to obtain higher order differential equations in the new variables nuu ...,,1 ,
we require 0...11 nn yuyu (4)
We continue computing )1(...,, npp yy and in each step we set:
1...,,1,0...)1()1(
11
nkyuyuk
nnk (5)
Then compute)(n
py and substitue each of)()1(
&...,,,,n
p
n
pppp yyyyy in equation (1), we get
)(
)...())(...)((
.......))(...)((
))(...)(()(...
)1()1(
1101
202122
1011110)(
xf
yuyuyxayxayu
yxayxayu
yxayxayuyxay
n
nn
n
nnnnn
n
nppn
(6)
Using the fact that each of nyyy ...,, 21 is a homogeneous solution satisfying equation (2),
we then obtain )(... )1()1(22)1(
11 xfyuyuyun
nnnn (7)
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So, from equations (4), (5) and (7) we obtain the following sytem of linear equations in
nuu ...,,1 :
0...2211 nn yuyuyu 0...2211 nnyuyuyu
.
.
.
)(...)1(
22
)1(
11
)1(
xfyuyuyun
nnn
n
To solve for each of nuu ...,,1 , we apply Cramers Rule:
W
Wu kk , nk ...,,2,1 (8)
Where Wis the Wronskian of )(),...,(1 xyxy n , namley
)1()1(2
)1(
1
21
21
..
...
...
...
...
det
n
n
nn
n
n
yyy
yyy
yyy
W and
...).(..
.0.
..
0
0
det
)1()1(
1
1
1
n
n
n
n
n
k
yxfy
yy
yy
W
Where the column
)(
.
.
.0
0
xf
replaces the kth column of the matrix of the wronskian W.
Integrating each of nuu &...,1 obtained from (8) and substituting nuu ...,,1 in (3), we obtain
dxuydxuyyn
np ...11
Remark(2) Note that equation (1) is put in standard form.
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Example (1) Solve xxyyxyxyx ln663 423
Solution: This is a 3rd
- order Cauchy-Euler differential equation as seen in Section 2.6.
Step 1: Solve the homogeneous equation 0663 23 yyxyxyx
We substitut rxy and obtain the auxiliary equation 06116 23 rrr
Its roots are 1, 2 & 3 and the homogeneous solution is then 33221 xcxcxcyh
Step 2: Compute: 32
32
2
620
321 x
x
xx
xxx
W
Step3: Compute 333
21 xuxuxuyp
Write the equation in standard form: )(ln661
332
xfxxyx
yx
yx
y
Then compute: xx
xxx
xx
xx
W ln
62ln
320
0
52
32
1 , xx
xxx
x
xx
W ln2
6ln0
301
0
42
3
2 , and
xx
xx
xx
xxx
W ln
ln20
32132
32
3
So, 312
3
51
118
)1ln3(
2
ln
2
lnx
xu
xx
x
xx
W
Wu
223
42
24
)1ln2(
2
ln
2
lnx
xu
xx
x
xx
W
Wu
And xx
ux
x
xx
W
Wu
2
)1(ln
2
ln
2
ln33
33
3
So the particular solution is then:
36
)11ln6(
2
)1(ln
4
)1ln2(
18
)1ln3(4444
332211
xxxxxxxxyuyuyuy p
And the general solution is )11ln6(36
)(
43
32
21 xx
xcxcxcyg
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Example (2) Solve 211
43343 xyyxyx
Solution: Again this is 3rd
-order Cauchy-Euler differential equation
Step 1: Solve the homogeneous equation 0334 3 yyxyx
We substitut rxy and obtain the characterestic equation 0311124 23 rrr
Its roots are 1,2
1 &2
3 and the homogeneous solution is then 21
321 cxcxcyh
Step 2: Compute:4
1
4
3
4
10
2
3
2
11
2
1
2
3
2
1
2
1
2
3
2
1
xx
xx
xxx
W
Step 3: To obtain 23
2
1
321 xuxuxuyp , write the equation in standard form:
)(4
3
4
32
5
32xfxy
xy
xy
Then compute:
27
2
1
2
3
2
5
2
1
2
1
2
3
2
1
1
4
3
4
1
2
3
2
10
0
x
xxx
xx
xx
W
2
7
11 4x
W
Wu
2
9
19
8xu
42
2
1
4
30
2
301
0
2
1
2
5
2
1
2
3
x
xx
x
xx
W
422 2x
W
Wu 52
5
2xu
and
33
2
1
4
30
02
11
0
2
5
2
1
2
1
2
1
x
xx
x
xx
W
333 2xW
Wu
432
1xu
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36
So the particular solution is then: 211
2
3
42
1
52
9
90
1.
2
1.
5
2.
9
8xxxxxxxyp
And the general solution is 211
321
90
12
3
2
1
xxcxcxcyg
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37
Section 2.12 Modeling of Electric Ciruits
In this section we study an RLCseries circuit, where, as in Sectin 1.7, R representsresistance, L represents inductance and Crepresents capacitance.
RLC- Series CircuitFind the current in the given RLC- series circuit for a periodic electromotive force
wtEV sin0
Solution: By Kirchhoffs voltage law 0v , we have:
wtEtVdttI
C
RI
dt
dIL sin)()(
10 or
L
wtwEI
LCdt
dI
L
R
dt
Id cos1 02
2
(1)
The general solution to this equation is: phg III (2)
To find hI , we solve the homogeneous differential equation
01
2
2
ILCdt
dI
L
R
dt
Id(3)
With R, L and Cbeing constants, we use undetermined coefficients method. So assume
rt
h eI and find ras the root of the auxiliary equation 0
12
LCrL
R
r
The roots are:LCL
R
L
Rr
1
22
2
2,1
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38
Define
01
,1
01
,1
&2
22
22
LCif
LC
LCif
LC
L
R
Cases of the roots The homogeneous solution
1- Two distinct reals tth ececI )(2)(1 2- One repeated real: th etccI )( 21 3- Two complex roots )sincos( 21 tctceI th
Recall: From Section 1.7: An electrical system is said to be in steady-state if the
variables describing the behaviour are either periodic or constant. Otherwise, transient
Remark If ,0R then 0 and 0lim
t
te . Hence the homogeneous current tends to
zero and the steady state current tend to the particular current .pI
To find pI , we use undetermined coefficients method.
SinceL
wtwE
dt
tdV
L
cos)(1 0 is a cosine function and assuming jwr 2,1 , we let
wtBwtAIp sincos .
Then wtwAwtwBIp sincos
And wtBwwtAwIp sincos 22
Substituting these into equation (1), we obtain
L
wtwEwt
LC
B
L
RwABwwt
LC
A
L
RwBAw
cossin)(cos)( 0
22
From this we have the two linear equations in the unknowns A and B:
0)1(
)1
(
2
02
BC
LwAwR
wEBwRAC
Lw
Using Cramers Rule, we solve forA & B and obtain
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39
22
0
221
10
2212
10
2
2
2
2
0
)(
)(
))((
)(
1
1
10
RS
SE
RwC
wL
wCwLE
RwC
wLw
wCwLEw
CLwwR
wR
C
Lw
CLw
wRwE
A
WherewC
wLS 1 is called reactance. Similarly,
22
0
2
2
02
1
1
0
1
RS
RE
CLwwR
wR
C
Lw
wR
wEC
Lw
B
With this we have: )sincos(222222
0 wtRS
Rwt
RS
S
RS
EIp
Note that
B
A
R
S and 22
22
0 BA
RS
E
.
To write the solution in amplitude/phase form )sin(0 wtIIp , note that the two terms
2222&
RS
R
RS
S
can be represented as the sine and cosine, respectively of an angle
in a right triangle. Then we have:
2222cos,sin
RS
R
RS
S
and pI becomes
)sin()sincoscossin( 022
0
wtIwtwtRS
EIp
WhereR
S
RS
EI
tan&
22
00
The quantity 22 RS is called the impedence and it is equal to the ratio .0
0
I
E
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ExampleFind the current in an RLC- series circuit with R= 8 ohms, L=2 henrys, C=0.1 farad
and tV 5sin160 .
Solution ,011
&2
2
2
LCL
R Hence the roots are complex conjugate
jr 22,1
and the homogeneous solution is then )sincos( 212
tctceI th
To find the steady state current (the particular current) we copmute:
82101
wC
wLS , 1tan&2106464
160
22
00
R
S
RS
EI
So, )5sin(2104
tIp and the general solution is then
)5sin(210)sincos(421
2 ttctceI t