Differential Equations MTH 242 Lecture # 11 Dr. Manshoor Ahmed.
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Transcript of Differential Equations MTH 242 Lecture # 11 Dr. Manshoor Ahmed.
Differential EquationsMTH 242
Lecture # 11
Dr. Manshoor Ahmed
Summary (Recall)
• Construction of 2nd solution from a known solution.
• Reduction of order.
• How to solve a homogeneous linear differential with constant coefficients?
• Auxiliary or characteristics equation.
• How to write the general of solution of linear differential with constant coefficients
Solution of non-homogeneous DE
Solution of non-homogeneous DE
The Method of Undetermined Coefficients
pcy
pby
pay ///
Caution!
• In addition to the form of the input function , the educated guess for must take into consideration the functions that make up the complementary function .
• No function in the assumed must be a solution of the associated homogeneous differential equation. This means that the assumed should not contain terms that duplicate terms in .
)(xg
cy
py
py
cy
Trial particular solutions
No Terms Choice for
1.
2.
3.
4.
py
xke xAe
( 0,1,2......)nkx n 20 1 2 ..... n
nA A x A x A x
cos
sin
k x
k x
cos sinA x B x
cos
sin
x
x
ke x
ke x
( cos sin )xe A x B x
Some functions and their particular solutions
1. Form of
2. 1(Any constant)
3.
4.
5.
6.
7.
8.
( )f x py
A
5 7x Ax B
3 1x x 3 2Ax Bx Cx D
sin 4x sin 4 cos 4A x B x
sin 4 cos 4A x B xcos 4x
5xe 5xAe
2 5xx e 2 5( ) xAx Bx C e
9.
10.
11.
12.
5(9 2) xx e 5( ) xAx B C e
3 sin 4xe x 3( cos 4 sin 4 ) xA x B x e
25 sin 4x x2 2( ) cos 4 ( )sin 4Ax Bx C x Ax Bx C x
3 cos 4xxe x3 3( ) cos 4 ( ) cos 4x xAx B e x Cx D e x
If equals a sum?
Suppose that • The input function consists of a sum of terms of the kind listed
in the above table i.e.
• The trial forms corresponding to be
.
Then, the particular solution of the given non-homogeneous
differential equation is
In other words, the form of is a linear combination of all the
Linearly independent functions generated by repeated differentiation
of the input function .
xg
xg
.21 xgxgxgxg m
xgxgxg m , , , 21
mppp yyy ,,,21
mpppp yyyy 21
py
)(xg
Example 1 Solve the non homogenous equation by using Undetermined
Coefficient Method.
Solution:
For complementary solution consider homogenous equation
Auxiliary equation is
24 8y y x
4 0y y
2
2
01 2
4 0
4
2
( cos 2 sin 2 )xc
m
m
m i
y e C x C x
For particular solution we assume
Substituting into the given differential equation
Equating the coefficients of and constant we have
Constant: Solving these equations we get the values of undetermined coefficients
2py Ax Bx C
2py Ax B 2py A
24 8p py y x 2 2
2 2
2 4( ) 8
2 4 4 4 8
A Ax Bx C x
A Ax Bx C x
2 ,x x2x : 4A 8
: 4 0x B 2 4 0A C
Thus
Hence the general solution is
2, 0, 1A B C
22 1py x
c py y y 2
1 2cos 2 sin 2 2 1y C x C x x
Example 2 Solve the non homogenous equation by using Undetermined Coefficient Method.
Solution:
siny y x
Duplication between and ?
• If a function in the assumed is also present in then this function is a solution of the associated homogeneous differential equation. In this case the obvious assumption for the form of
is not correct.
• In this case we suppose that the input function is made up of terms of kinds i.e.
and corresponding to this input function the assumed particular solution
is
py cy
py cypy
n
)()()()( 21 xgxgxgxg n
py
npppp yyyy 21
• If a contain terms that duplicate terms in , then that must be multiplied with , being the least positive integer that
eliminates the duplication.
Example 3Find a particular solution of the following non-homogeneous differential
Equation
Solution:
To find , we solve the associated homogeneous differential equation
We put in the given equation, so that the auxiliary equation is
ipyipy
cynx n
xeyyy 845 ///
cy
045 /// yyymxey
Thus
Since
Therefore
Substituting in the given non-homogeneous differential equation, we
obtain
or
Clearly we have made a wrong assumption for , as we did not
remove the duplication.
4 ,1 0452 mmm
xxc ececy 4
21
xexg 8)(
xp Aey
xexAexAexAe 845
xe80 py
Since is present in .Therefore, it is a solution of the associated
homogeneous differential equation
To avoid this we find a particular solution of the form
We notice that there is no duplication between and this new
Assumption for .
Now
Substituting in the given differential equation, we obtain
or
xAe cy
045 /// yyy
xp Axey
cy
py
xxxxp AeAxeAeAxey 2y , //
p/
.84552 xxxxxx eAxeAeAxeAeAxe
.3883 AeAe xx
So that a particular solution of the given equation is given by
Hence, the general solution of the given equation is
Example 4
Find a particular solution of
Solution:
Consider the associated homogeneous equation
Put
xp ey )38(
xxxc exececy )3/8(4
21
xeyyy /// 2
02 /// yyy
mxey
Then the auxiliary equation is
Roots of the auxiliary equation are real and equal. Therefore,
Since
Therefore, we assume that
This assumption fails because of duplication between and .
We multiply with .
Therefore, we now assume
1 ,1
0)1(12 22
m
mmm
xxc xececy 21
xexg )(
xp Aey
cy py
x
xp Axey
However, the duplication is still there. Therefore, we again multiply with
and assume
Since there is no duplication, this is acceptable form of the trial .
Example 5Solve the initial value problem
SolutionAssociated homogeneous DE
xx
p eAxy 2
py
xp exy 2
2
1
2)(y0,)y(
,sin104/
//
xxyy
Put
Then the auxiliary equation is
The roots of the auxiliary equation are complex. Therefore, the
complementary function is
Since
Therefore, we assume that
0// yy
mxey
imm 012
xcxcyc sincos 21
)()(sin104)( 21 xgxgxxxg
sincos C , 21
xDxyBAxy pp
So that
Clearly, there is duplication of the functions and .To remove this duplication we multiply with . Therefore,we assume that
So that Substituting into the given non-homogeneous differentialequation,we have
Equating constant terms and coefficients of , , we obtain
xcos xsin
2py x
.sincos xC xDxxBAxy p
2 sin cos 2 cos sinpy C x Cx x D x Dx x
xDxBAx cosx2sin C2yy p//
p
xxxDxBAx sin104cosx2sin C2
x xsin xx cos
xDxBAx sincos Cyp
So that
Thus
Hence the general solution of the differential equation is
We now apply the initial conditions to find and
Since
Therefore
02 ,102 ,4 ,0 DCAB
0 ,5 ,0 ,4 DCBA
xxxy p cos54
xxxcxcyyy pc cos5- x4sincos 21
1c 2c
0cos54sincos0)( 21 ccy
1cos,0sin
91 c
Now
Therefore
Hence the solution of the initial value problem is
Example 6
Solve the differential equation
xxxxcxy cos5sin54cossin9 2/
2cos5sin54cossin92)( 2/ cy
7.c2
.cos54sin7cos9 xxxxxy
122696 32/// xexyyy
Solution:The associated homogeneous differential equation is
Put
Then the auxiliary equation is
Thus the complementary function is
Since
We assume that
Corresponding to:
096 /// yyy
mxey
3 ,30962 mmm
xxc xececy 3
23
1
)()(12)2()( 2132 xgxgexxg x
2)( 21 xxg CBxAxy p 2
1
Corresponding to:
Thus the assumed form of the particular solution is
The function in is duplicated between and .
Multiplication with does not remove this duplication. However, if
we multiply with , this duplication is removed.
Thus the operative from of a particular solution is
Then
And
Substituting into the given differential equation and collecting like
term, we obtain
xexg 32 12)(
xp Dey 3
2
xp DeCBxAxy 32
xe32py cy py
x
2py 2x
xp eDxCBxAxy 322
xxp eDxDxeBAxy 323 322
xxxxp eDxDxeDxeDeAy 32333 96622
Equating constant terms and coefficients of and yields
Solving these equations, we have the values of the unknown
coefficients
Thus,
.
Hence, the general solution
xxppp exDeBAxBAAxy 3232/// 12262C962)912(9y6y
2, xx xe3
0912 2,C962 BABA
122 ,6 9 DA
-6D and 32,98,32 CBA
xp exxxy 322 6
3
2
9
8
3
2
.63
2
9
8
3
2yy 3223
23
1pcxxx exxxxececy
Higher –Order EquationThe method of undetermined coefficients can also be used for higher
order equations of the form
with constant coefficients. The only requirement is that consists
of the proper kinds of functions as discussed earlier.
Example 7Solve
Solution:To find the complementary function we solve the associated
homogeneous differential equation
)(............ 011
1
1 xgyadx
dya
dx
yda
dx
yda
n
n
nn
n
n
)(xg
xeyy x cos/////
0///// yy
Put
Then the auxiliary equation is
Or
The auxiliary equation has equal and distinct real roots. Therefore, the
complementary function is
Since
Therefore, we assume that
Clearly, there is no duplication of terms between and .
mxmxmx emymeyey 2,,
0 23 mm
1,0,00)1(2 mmm
xc ecxccy 321
xexg x cos)(
xBexAey xxp sincos
cy py
Summary (Recall)