differential equations mth 242 lecture # 13 dr. manshoor ahmed
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Summary (Recall)
• Differential Operator, which is a linear operator.
• Differential equation in linear operator form.
• Auxiliary equation in terms of differential operator form.
• Annihilator operator.
• Annihilator operators of different functions.
• Solution non-homogeneous equation with annihilator operator.
So that we obtain
This is a variable separable equation.
Thus,
Integration gives
.
As .
Therefore,
or .
11
duy f x
dx
1
1
f xdu dx
y x
11
( )Pdxf x
u x dx e f x dxy
dxxfPdxePdxey p ..
1
1
f xu dx
y x
1 1 py u x y x
Method for Second Order Equation:We consider the second order linear non-homogeneous DE
Divide by , to get equation in the standard form
,
where
are continuous on some interval I.
Consider the associated homogeneous DE,
.
xgyxayxayxa 012
)(2 xa
xfyxQyxPy
01
2 2 2
( )( ) ( )( ) , ( ) , ( )
( ) ( ) ( )
a xa x g xP x Q x f x
a x a x a x
0 yxQyxPy
Complementary function:
Complementary function is
So that
Particular Integral
For finding a particular solution , we replace the parameters
and in the complementary function with the unknown variables
and . So that the assumed particular integral is
xycxycyc 2211
1 1 1
2 2 2
0
0
y P x y Q x y
y P x y Q x y
py 1c
2c
xu1 2u x
1 1 2 2py u x y x u x y x
Since we seek to determine two unknown functions and ,
we need two equations involving these unknowns. One of
these two equations results from substituting the assumed in the
given differential equation. We impose the other equation to
simplify the first derivative and thereby the 2nd derivative of
To avoid 2nd derivatives of and , we impose the condition
Then,
So that
Therefore,
1 u
py
2 u
py
2211221122221111 yuyuyuyuyuyuuyyuy p
1 u 2 u
02211 yuyu
2211 yuyuy p
22221111 yuyuyuyuy p
2211̀2211
22221111
yQuyQuyPuyPu
yuyuyuyuyQyPy ppp
Substituting in the given non-homogeneous differential equation yields
or
Now making use of the relations
we get,
Hence and must be functions that satisfy the equations
By using the Cramer’s rule, the solution of this set of equations is given by
)( 2211̀221122221111 xfyQuyQuyPuyPuyuyuyuyu
)(][] [ 21122221111 xfyuyuQyyPyuyQyPyu
1 1 1
2 2 2
0
0
y P x y Q x y
y P x y Q x y
xfyuyu 22111 u 2 u
1 1 2 2
1 1 2 2
0
u y u y
u y u y f x
1 21 2,
W Wu u
W W
where and denote the following determinants
The determinant can be identified as the Wronskian of the
solutions . Since the solutions are linearly
independent on I. Therefore
Now integrating the expressions for we obtain the value of and , hence the particular solution of the non-homogeneous
linear DE.
Summary of the Method
To solve the second order non-homogeneous linear DE
using the variation of parameters, we need to perform the following steps:
1 2,W W andW
W
2 11 21 2
2 11 2
0 0, ,
y yy yW W W
f x y y f xy y
21 and yy 21 and yy
. ,0, 21 IxxyxyW
1 2u andu
1 u 2 u
,012 xgyayaya
Step 1 We find the complementary function by solving the associated
homogeneous differential equation
Step 2 If the complementary function of the equation is given by
then are two linearly independent solutions of the
homogeneous differential equation. Computing Wronskian.
Step 3 By dividing with , we transform the given non-homogeneous
equation into the standard form
and we identify the function f(x).
0012 yayaya
2211 ycyccy 21 and yy
21
21
yy
yyW
2a
xfyxQyxPy
Step 4 We now construct the determinants given by
Step 5 Next we determine the derivatives of the unknown variables
through the relations
Step 6 Integrate the derivatives to find the unknown variables
. So that
Step 7 Write a particular solution of the given non-homogeneous
equation as
1 2W andW
2 11 2
2 1
0 0,
( ) ( )
y yW W
f x y y f x
1 2u andu
W
Wu
W
Wu 2
21
1 ,
1 2u andu 1 2u andu
1 21 2 ,
W Wu d x u d x
W W
2211 yuyupy
Step 8 The general solution of the differential equation is then given by
Constants of IntegrationWe don’t need to introduce the constants of integration, when
computing the indefinite integrals in step 6 to find the unknown functions of . For, if we do introduce these constants, then
So that the general solution of the given non-homogeneous differential equation is
or
22112211 yuyuycycpycyy
1 1 1 2 1 2( ) ( )py u a y u b y
1 1 1 2 1 2 1 1 2 2y c a y c b y u y u y
2121112211 ybuyauycycyyy pc
1 2u andu
If we replace with and with , we obtain
This does not provide anything new and is similar to the general solution found in step 8, namely
Example 1Solve
Solution:
Step 1 To find the complementary function
Put
11 ac 1C 2 1c b 2C
22112211 yuyuyCyCy
1 1 2 2 1 1 2 2y c y c y u y u y
24 4 1 .xy y y x e
044 yyy
mxemymxmeymxey 2,,
Then the auxiliary equation is
Repeated real roots of the auxiliary equation
Step 2 By the inspection of the complementary function , we make the
identification
Therefore
2
2
4 4 0
2 0, 2, 2
m m
m m
2 21 2 x x
cy c e c xe
cy
xx xeyey 22
21 and
xeexee
xeexeeyyW x
xxx
xxxx
,0
22, , 4
222
2222
21
Step 3 The given differential equation is
Since this equation is already in the standard form
Therefore, we identify the function as
Step 4 We now construct the determinants
xexyyy 2144
xfyxQyxPy
xexxf 2 1
24
1 2 2 2
24
2 2 2
01
1 2
01
2 1
xx
x x x
xx
x x
xeW x xe
x e xe e
eW x e
e x e
Step 5 We determine the derivatives of the functions in this step
Step 6 Integrating the last two expressions, we obtain
Remember! We don’t have to add the constants of integration.
Step 7 Therefore, a particular solution of then given differential
equation is
1 2u andu
1
1
1
4
42
2
24
41
1
xe
ex
W
Wu
xxe
xex
W
Wu
x
x
x
x
.2
)1(
23
)(
2
2
232
1
xx
dxxu
xxdxxxu
xxexxxe
xxpy 2
2
22
2
2
3
3
or
Step 8
Hence, the general solution of the given differential equation is
Example 2
Solve
Solution:
Step 1
To find the complementary function we solve the associated homogeneous differential equation
The auxiliary equation is
xexx
py 22
2
6
3
3 22
1 22 2
6 2x x xx xy y y c e c xe e
c p
.3csc364 xyy
090364 yyyy
imm 3092
Roots of the auxiliary equation are complex. Therefore, the complementary function is
Step 2 From the complementary function, we identify
as two linearly independent solutions of the associated
homogeneous equation. Therefore
Step 3 By dividing with , we put the given equation in the following
standard form
So that we identify the function as
xcxccy 3sin3cos 21
3sin ,3cos 21 xyxy
33cos33sin3
3sin3cos3sin,3cos
xx
xxxxW
.3csc4
19 xyy
xxf 3csc4
1
Step 4 We now construct the determinants
Step 5 Therefore, the derivatives are given by
Step 6 Integrating the last two equations w.r.t , we obtain
Here, no constants of integration are used.
xuxu 3sinln36
1 and
12
121
1
2
0 sin 31 1
csc3 sin 314 4csc3 3cos3
4
cos3 01 cos3
14 sin 33sin 3 csc3
4
xW x x
x x
xx
Wxx x
x
x
W
Wu
W
Wu
3sin
3cos
12
1 ,
12
1 22
11
1 2W andW
1 2u andu
Step 7 The particular solution of the non-homogeneous equation is
Step 8 Hence, the general solution of the given differential equation is
Example 3Solve
Solution: Step 1 For the complementary function consider the associated
homogeneous equation
To solve this equation we put
1 1cos3 sin 3 ln sin 3
12 36y x x x xp
xxxxxcxcpycyy 3sinln3sin36
13cos
12
13sin3cos 21
.1
xyy
0 yy
mxmxmx emyemyey 2, ,
Then the auxiliary equation is:
The roots of the auxiliary equation are real and distinct. Therefore, the complementary function is
Step 2 From the complementary function we find
The functions are two linearly independent solutions of the homogeneous equation. The Wronskian of these solutions is
Step 3 The given equation is already in the standard form
1012 mm
xecxeccy 21
xeyxey 21 ,
2
,
xx
xxxx
ee
eeeeW
y p x y Q x y f x
21 and yy
Here
Step 4 We now form the determinants
Step 5 Therefore, the derivatives of the unknown functions and are
given by
Step 6 We integrate these two equations to find the unknown functions
xxf
1)(
)/1( /1
0 W
)/1( /1
0 W
2
1
xexe
e
xeex
e
xx
x
xx
x
11
22
1/
2 2
1/
2 2
x x
x x
e xW eu
W x
e xW eu
W x
1 2
1 1,
2 2
x xe eu dx u dx
x x
1 2u andu
The integrals defining cannot be expressed in terms of the elementary functions and it is customary to write such integral as:
Step 7 A particular solution of the non-homogeneous equations is
Step 8 Hence, the general solution of the given differential equation is
1 2u andu
1 2
1 1,
2 2
x xt t
xx
e eu dt u dt
t t
x
x
x
x
tx
tx
p dtt
eedt
t
eey
2
1
2
1
x
x
tx
x
x
txxx dt
t
eedt
t
eeececpycyy
2
1
2
121
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