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Chapter 3
Second Order Linear Differential Equations
Sec 3.1 : Homogeneous equation
A linear second order differential equations is written as
When d(x) = 0, the equation is called homogeneous, otherwise it is called nonhomogeneous.
,
For the study of these equations sometimes we will consider the standard forms given by
)()(')('' xgyxqyxpy
0)(')('' yxqyxpy
Linear differential operators(ways to write the linear 2nd ODE):
In mathematics a function that ``transforms a function into a different function'' is called an operator.
(1) L -Operator
Let y'' + p(x)y' + q(x)y = g(x)
be a second order linear differential equation. Then we call the operator
L(y) = y'' + p(x)y' + q(x)y
the corresponding linear operator(Differential operator). Thus in this chapter we want to find solutions to
the equation
L(y) = g(x) y(x0) = y0 y'(x0) = y'0
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Proof:
(1)
(2) the proof of (2) is exercise #28 sec 4.2
Note: Any operator that satisfies (1) & (2) is called linear operator. If (1) & (2) fails to hold, the operator is
nonlinear
(2) D- Operator
Recall from calculus that
dx
dyDy ,
2
22
dy
ydyD , and in general
n
nn
dy
ydyD . Using this notations, we can express L as
])[(][ 22 yqpDDqypDyyDyL
for example, if
,3'4''][ yyyyL
then we can write
Linearity of the differential operator L :
Let L(y) = y'' + p(x)y' + q(x)y. if y , 1y and 2y are any twice-differeniable functions on the interval I
and if c is any constant, then
(1) )()()( 2121 yLyLyyL ,
(2) )()( ycLcyL
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])[34(34][ 22yDDyDyyDyL
when p and q are constants we can treat qypDyyD 2 as polynomial , so for the above example
we can write
yDDyDDyDyyDyL )3)(1(])[34(34][ 22
Exercises:
# 3.1.1 Let yyyL ''][2 compute
(a) ][sin2 xL (b) ][2
2 xL (c) ][2r
xL (d) ]2[2
2x
eL
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Sec 3.2: Solution of Homogeneous Linear Second Order Differential
Equations
Theory of Solutions
Here we will investigate solutions to homogeneous differential equations. Consider the homogeneous linear
differential equation
We have the following results:
(1)
The above theorem makes clear the importance of being able to
decide whether or not a set of solutions )(1 xy and )(2 xy of
Are linearly independent.
Methods to check if two functions are Linearly independent:
(1) Linear Dependence and Indepedence using Definition :
Definition:(Linear dependence)
Two functions 22 yandy are said to be linearly independent on the interval I, if
there exist nonzero constants 22 candc such that for all x in I
02211 ycyc
General Solution for a Second-Order Homogeneous Linear Equation: Theorem: The 2nd order homogeneous equation
0)(')('')( yxcyxbyxa (1)
always has exactly two linearly independent solutions )(1 xy and )(2 xy . The general
solution to equation (1) is given by the formula
2211)( ycycxy
where the functions 21, yy is the pair of linearly independent solutions to
equation (1).
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f 22 yandy are linearly
independent then 021 cc is the only solution for
02211 ycyc
Another way of stating this definition is ,
ratio test as follows:
Two functions 22 yandy are linearly dependent in I if cy
y
2
1
Two functions 22 yandy are linearly independent in I if cy
y
2
1
Remark: Proportionality of two functions is equivalent to their linear dependence.
(2) Linear Dependence and Independence using Wronskian:
It is not easy to use the above definition for higher order, Instead we can use the wronskian.
The Wronskian : for two functions
Let and be two differentiable functions. and are linearly dependent(proportional) if and only if
cy
y
2
1 . an equivalent criteria for linearly dependent can be derived from the followingt:
and are linearly dependent
So cy
y
2
1
Differentiate both sides 02
1
y
y
.
is called the Wronskian of of and , that is
Wronskian of and is
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Therefore, we have the following:
Now we have the following conclusion:
Let and be two differentiable functions. The Wronskian , associated to and , is the
function
We have the following important properties:
(1) If and are two solutions of the equation y'' + p(x)y' + q(x)y = 0, then
In this case, we say that and are linearly independent.
(2) If and are two linearly independent solutions of the equation y'' + p(x)y' + q(x)y = 0, then
any solution y is given by
for some constant and . In this case, the set is called the fundamental set of solutions.
Exercises:
# 3.1.3 Determine whether the given solutions of the following DE are linearly independent
(a) xyxyyy 3cos3sin09'' 21
(b) xx
eyeyyy 20'' 21
(c) xx
eyeyyyy2
21 302''
There is a fascinating relationship between second order linear differential equations and the
Wronskian. This relationship is stated below.
Two functions 1y and 2y are linearly dependent if 0),( 21 yyW
0),( 21 yyWif dependentinlinearly are 2yand 1yTwo functions
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Theorem: Abel's Theorem
Let y1 and y2 be solutions on the differential equation
L(y) = y'' + p(x)y' + q(x)y = 0
where p and q are continuous on [a,b]. Then the Wronskian is given by
dxxp
CexyyW)(
21 ))(,(
where C is a constant depending on only y1 and y2, but not on x. The Wronskian is
either zero for all x in [a,b] or always positive, or always negative.
Proof :
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Methods for solving second order Linear
Homogeneous DE:
Section 3.3: Method (1) )(For Constant coefficients)
Characteristic equation (Auxiliary equation)
A second order homogeneous equation with constant coefficients is written as
where a, b and c are constant. This type of equation is very useful in many applied problems (physics,
electrical engineering, etc..). Let us summarize the steps to follow in order to find the general solution:
if 0a , we get the first order equation of the same family
0' cyby
and this is first order seperable differential equation which has a solution of the form
bcAeyx
/,
In the same way we can think that , x
ey will be a solution of the second order equation
so, by substituting x
ey ,'
xey ,
2 xey in the left-hand side of differential equation,
we get
02 xxx
ceebea
now, dividing both sides by ,x
ey , we obtain
the characteristic equation (Auxiliary equation)
02 cba
This is a quadratic equation. Let 1 and 2 be its roots we have
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a
acbb
2
4,
2
21
The actual form of the solution is strongly dependent on whether the roots are real versus
complex or distinct versus repeated. In fact three special cases can be identified based on
whether the term inside the radical is positive, negative, or zero. There are three cases as follows
Three Special Cases
Case I. Real Distinct Roots:
If 1 and 2 are distinct real numbers (this happens if ), then the general solution is
xxececy 21
21
Exercises 3.3.2 Find the general solution of 04'3'' yyy
Case II. Repeated Roots:
If 21 (which happens if ), then the general solution is
xxxececy 11
21
Exercises 3.3.4 Find the general solution of 0'4''4 yyy
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Case III. Complex Conjugate Roots: (section 2.3)
If 1 and 2 are complex numbers (which happens if ),
Then a
baci
a
b
2
4
2,
2
21
And the general solution is
where
Example: Find the general solution 2y''+2y'+3y=0
Exercises 3.3.12 Solve 016'' yy y(0)=1 y’(0)=0
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Justification of Case 3:
xixi ececy )(
2
)(
1
using Euler identity: xixeix sincos
the solution can be written as
)]sin(cos)sin(cos[ 21 xixcxixcey x
]sin)(cos)[( 2121 xiccxcce x
)sincos( 21 xCxCe x where 211 ccC and iccC )( 212
Proof of Euler identity:
From Maclaurin series expansion
.....!4
)(
!3
)(
!2
)(1
432
ixixix
ixeix
.....!5!4!3!2
15432
ixxixxix
....!5!3
sin53
xx
xx
...!6!4!2
1cos642
xxxx
From the above expansions , we conclude the following
xixeix sincos
which is Euler identit
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Method (2): Euler Equations
An Euler-Cauchy equation is of the form
(EC) 0'''2 cyxyyax x>0
where a, b and c are constant numbers. Let us consider the change of variable
x = et.
Then we have
xdt
dy
dx
dt
dt
dy
dx
dy 1 and
222
2
22
2
2
2 1111
xdt
dy
xdt
yd
xdt
dy
xdx
dt
dt
yd
dx
yd
)(1
2
2
2 dt
dy
dt
yd
x
Substitute in (EC)
The equation (EC) reduces to the new equation
0)(2
2
cydt
dyab
dt
yda
We recognize a second order differential equation with constant coefficients. Therefore, we use the previous
sections to solve it. We summarize below all the cases:
(1)
Write down the characteristic equation
0)(2 caba
(2)
If the roots r1 and r2 are distinct real numbers, then the general solution of (EC) is given by
21
21)(
xcxcxy
(3)
If the roots 1 and 2 are equal ( 21 ), then the general solution of (EC) is
xxcxcxy ln)( 11
21
02
02
(4)
If the roots 1 and 2 are complex numbers, then the general solution of (EC) is
xxcxxcy lnsinlncos 21
Short cut for using Euler method:
We assume a solution of the form
xy
Therefore,
1'
xy and 2
)1('' xy
Upon substitution into the original ODE, we get
0))1(( xcba
and dividing by
x gives the characteristic equation,
0)(2 caba .
Which is similar to equation in step (1) above, so we can continue finding the general solution as
in steps 2,3,4.
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Example: Find the general solution to 0)(6)('2)(''2 xyxxyxyx x>0
Solution: First we recognize that the equation is an Euler equation, with a=1, b=2 and c=-6.
Exercise # 3.3.14 solve 04'''2 yxyyx x>0
Example: Find the general solution of
0)(14)(')1(10)('')1(2 xyxyxxyx x>-1, Ans:
72
21 )1()1( xcxc
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Method(3): For variable coefficients
integration using one known solution(Reduction of order method)
____________________________________________________________________________________
This technique is very important since it helps one to find a second solution independent from a known one.
Therefore, according to the previous section, in order to find the general solution to y'' + p(x)y' + q(x)y = 0,
we need only to find one (non-zero) solution, .
Let be a non-zero solution of
Then, a second solution independent of can be found as
,
Development of the Method
Given a valid solution to a homogeneous linear 2nd order differential equation, one can obtain a
second linearly independent solution by letting
)()()( 12 xyxuxy
where y1(x) is known and u(x) is to be determined. This technique applied to a 2nd order system
is as follows:
Given the original differential equation
with known solution y1(x), we let
dxxy
exyxy
dxxp
2
1
)(
12)]([
)()(
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or
Now, substitution into the original equation gives
Or
Letting z = u' and noting that the third term vanishes from definition of the original ODE, we
have
Now, separating variables and integrating gives
Or
Finally, since du/dx = z, we have
Example 22 )(;0,06'6'' xxfxyxyyx .Ans:
3 xy
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Sec 3.4 Nonhomogeneous Second Order Linear
Equations
So far we have considered equation of the form
)(''' xgcybyay for the case where 0)( xg ,
When 0)( xg , then 02 cbrar giving 1rr and 2rr and the solution is in general
xrxr
h ececy 21
21 .
In the equation )(''' xgcybyay , the substitution xrxr
h ececy 21
21 would make the left-
hand side zero. Therefore , there must be a further term in the solution which will make the L.H.S equal to
)(xg and not zero.
The complete solution will therefore be of the form
p
xxyececy 21
21
where
py is the extra function yet to be found.
xx
h ececy 21
21
is called the complementary solution or homogeneous solution.
)(xyyp
is called the particular solution.
And General solution = homogeneous solution + particular solution
In the previous sections we discussed how to find . In this section we will discuss two techniques to find
:
(1) Method of Principle of superposition:
(2) Method of variation of parameters
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Method (1) : Principle of superposition:
Let y be a solution of the differential equation
)(''' 1 xfcybyay
and let y
be a solution of
)(''' 2 xfcybyay
Then for any constants
21,cc , the function ycyc
21
is a solution of the differential equation
)()(''' 2211 xfcxfccybyay
Example:
Given that tty cos)(1 is a solution to tyyy sin'''
And 3/)( 2
2
tety is a solution to
teyyy
2'''
Use the superposition principle to find solution to the following differential equations
(a) tyyy sin5'''
(b) t
etyyy23sin'''
(c) t
etyyy218sin4'''
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Method (2) : Variation of Parameters
This method has no prior conditions to be satisfied. Therefore, it may sound more general than the previous
method. We will see that this method depends on integration while the previous one is purely algebraic
which, for some at least, is an advantage. It is easily extended to higher order linear ODEs.
Consider the equation )()(')('' xgyxqyxpy
In order to use the method of variation of parameters we need to know that 21, yy is a set of
fundamental solutions of the associated homogeneous equation y'' + p(x)y' + q(x)y = 0. We know that, in this
case, the general solution of the associated homogeneous equation is . The idea behind
the method of variation of parameters is to look for a particular solution such as
where and are functions. From this, the method got its name.
The functions and can be calculated as follows,
Let’s start with the general 2nd order equation written in standard form,
)(')('' xgyyxpy
with homogeneous solution
2211)( ycycxyh
Now assume yp of the form
2211 )()()( yxuyxuxyp (1)
which gives,
At this point we recognize that an addition constraint, beyond the original balance equation (1),
will be needed because we are trying to find two unknowns, u1 and u2. For simplicity, let’s
choose the following relationship,
0'' 2211 yuyu (2)
which greatly simplifies the above equation for yp’, giving
2211 ''' yuyuy p
22221111 ''''' yuyuyuyuy p
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Now the second derivative becomes
22221111 '''''''''' yuyuyuyuy p
Substitution of these expressions into the original differential equation gives
)('''''''''' 2211221122221111 xgyquyquypuypuyuyuyuyu
Rearranging gives
)('''')'''()'''( 221122221111 xgyuyuqypyyuqypyyu
Therefore, since both the first and second terms vanish (because y1 and y2 are solutions to the
homogeneous equation), we have
)('''' 2211 xgyuyu (3)
The functions and are solutions to the system of equations (2) and (3)
which can be written in matrix form (two equations with two unknowns) gives
gu
u
yy
yy 0
'
'
'' 2
1
21
21
Using Cramer’s rule the solution to this system can be written as
gW
y
y
yy
yy
yg
y
u2
2
21
21
2
2
1
1
0
''
'
0
' and gW
y
y
yy
yy
gy
y
u1'
0
''
'
0
'1
1
21
21
1
1
1
Finally, integrating these latter expressions to give u1 and u2 allows one to write the desired
expression for yp(x) as
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This result is consistent with the general relationships given above.
Exercise 3.4.6 find the general solution of xyy sec'' .
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Exercise # 3.4.10 Solve using variation of parameters
12
3'5'' xyxyyx