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4.1 Preliminary Theory Homogeneous LinearEquations
Homogeneous linearnthorder DE
( ) ( 1)1 2 1 0( ) ( ) ... ( ) ( ) ( ) 0
n nn na x y a x y a x y a x y a x y
+ + + + + =
NonHomogeneous linearnthorder DE
( ) ( 1)1 2 1 0( ) ( ) ... ( ) ( ) ( ) ( )
n nn na x y a x y a x y a x y a x y g x
+ + + + + =

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Linear Dependence / Linear Independence
Definition 1
A set of functions f1(x), f2(x), , fn(x) is said to be
linearly dependent on an interval I if there exist
1,
2, ..,
n,
for every x in the interval.
In other words, the set of functions is said to be
linearly independent if
1 1 2 2( ) ( ) ... ( ) 0n nc f x c f x c f x+ + + =
1 2 ... 0nc c c= = = =

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Linear Dependence / Linear Independence
Definition 2
A set of two functions f1(x) and f2(x) is said to be
linearly dependent on an interval I when one
.
In other words, the set of functions is said to belinearly independent if
1
2
( )constant
( )
f x
f x=
1
2
( )constant
( )
f x
f x

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Example 1
Determine whether the functions are linear
independent on the interval (,):2 3
1 2 3
2 2
1) ( ) , ( ) , ( )
f x x f x x f x x= = =
= = =
1 2
, ,
3) ( ) 2 , ( ) 2f x x f x x= + = +

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Wronskian
Suppose each of the functions f1(x), f2(x), , fn(x)
possesses at leastn
1 derivatives. The determinant
21
...
...
nfff
where the primes denote derivatives, is called the
Wronskian of the functions.
)1()1(2
)1(1
21
...
::::,...,,
=
nn
nn
n
fff

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Linear Independence Using Wronskian
Theorem
Let y1, y2,.,yn be n solutions of the homogeneouslinear nthorder differential equation on an interval I.
Then the solutions are linearl inde endent on I if
and only if :
for every x in the interval
1 2( , ,..., ) 0nW y y y

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Fundamental set of solutions
Definition
Any set y1, y2,.,yn of n linearly independent solutions
of the homogeneous linear nthorder differential
of solutions on the interval.

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General Solution
Theorem
Let y1, y2,.,yn be a fundamental set of solutions of the
homogeneous linear nthorder differential equation on
an interva.
en t e genera so ution o t e equation
on the interval is
where ci , i = 1,2,..,n are arbitrary constants.
1 1 2 2( ) ( ) ... ( )n nc y x c y x c y x= + + +

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Example 2
Verify that the given functions form a fundamental set
of solutions of the DE on the indicated interval. Form
the general solution.
3 4'' ' x x , , ,

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Solution for Example 2
Step 1:
Verify that the given functions are solutions for the DE
3 4'' ' 12 0; , , ( , )x xy y y e e =
3 3 31 1 1
3 3 3
Let: 3 9
Substitute into the DE: '' ' 12 0
x x x
x x x
y e y e y e
y y y
= = =
=
42
0 0 (verified)
Let:
e e e
y e
=
=
=4 4
2 2
4 4 4
4 16
Substitute into the DE: '' ' 12 0
16 (4 ) 12( ) 0
0 0 (verified)
x x x
x x x
y e y e
y y y
e e e
= =
=
=
=

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Solution for Example 2
Step 2 :
Verify that the solutions are linearly independent on I
3 4'' ' 12 0; , , ( , )x xy y y e e =
3 43 4
3 4( , )
3 4
x xx x
x x
e eW e e
e e
=
3 4
7 0
and are linearly independent on ( , )
x
x x
e e
e
e e
=
=
Therefore e3x and e4x form a fundamental set of
solutions. Hence the general solution is:
y = c1y1 + c2y2 = c1e3x + c2e4x

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Example 3
Verify that the given functions form a fundamental set
of solutions of the DE on the indicated interval. Form
the general solution.
2 '' ' , , ,

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4.2 Reduction of Order
Suppose y1(x) is a solution of the homogeneous linear
secondorder DE:
2 1 0( ) ( ) ( ) 0a x y a x y a x y + + =
We can construct a second solution y2(x) by the
method of reduction of order so that y1(x) and y2(x)
are linearly independent on I.

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Formula for Reduction of Order
Consider the homogeneous second order linear DE
If iven the first solution x , then b reduction of
( ) ( ) 0y p x y q x y + + =
Reference: A First Course in Differential Equations with Modelling Applications, (9th Ed) Dennis G. Zill, Brooks/Cole Publishing Company
order method:
( )
2 1 21
( ) ( )( )
p x dxe
y x y x dxy x
=

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Example 1
Given that y1(x) = e2x is a solution of the DE on (,):
y 4y + 4y = 0
Use a formula of reduction oforder to find the second
solution y2(x). Hence, write the general solution of the
differential equation.

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Example 2
Given that y1(x) = sin 3x is a solution of the DE on (,):
y + 9y = 0
Use a formula of reduction of order to find the second
solution y2(x). Hence, write the general solution of the
differential equation.

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Example 3
Given that y1(x) = x4 is a solution of the DE on (0,):
x2y 7xy + 16y = 0
Use a formula of reduction oforder to find the second
solution y2(x). Hence, write the general solution of the
differential equation.

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4.3 Homogeneous Linear DE with ConstantCoefficients
Consider the homogeneous linear second order DE :
where a, b, c are constant coefficients.
0 (1)ay by cy + + =
Try a solution for the DE of the form:
Calculate:
mxe=
2
,mx mx
me y m e = =

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Substitute y, y and y into the DE (1):
0 (1)ay by cy + + =
2
2( ) ( ) 0
( ) 0
mx mx mx
mxa m e b me cee am bm c
+ + =
+ + =
mx ,
This equation is called an auxiliary equation of the DE (1)
where m is the root of quadratic equation (2).
2 0 (2)am bm c+ + =

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In general, if given 2nd
order linear homogeneous DE:
The corresponding auxiliary equation for the DE (1) is
given by:2 0 (2)am bm c+ + =
0 (1)ay by cy + + =
Therefore is a solution of the DE (1) if and only
if m satisfies auxiliary equation (2).
mxe=

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The two roots of equation (2) can be obtained by
factoring method or by quadratic formula:
Roots of auxiliary equation
2 4b b ac
2 0 (2)am bm c+ + =
where the roots may be:
real and distinct roots
real but repeated roots
Complex conjugate roots
2m a=

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The two solutions are:
The general solution is:
Real and distinct roots : (m1 , m2)
1 21 2,
m x m xe y e= =
1 21 1 2 2 1 2
m x m xc y c y c e c e= + = +
Real and repeated roots : (m = m1 = m2)
The two solutions are:
The general solution is:
1 2,mx mxe y xe= =
1 1 2 2 1 2mx mxc y c y c e c xe= + = +

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The two solutions are:
The general solution is:
Complex conjugate roots : (m = ii)
1 2cos , sinx xe x y e x = =
1 1 2 2 1 2cos sinx xc y c y c e x c e x = + = +

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Example 1
Find the general solution of the DE:
) '' 4 ' 5 0a y y y+ + =
) '' 10 ' 25 0b y y y + =
''

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Higher Order linear homogeneous DE
Consider nth order linear homogeneous DE:
The corresponding auxiliary equation is:
( ) ( 1)
1 2 1 0..... 0n n
n na y a y a y a y a y
+ + + + + =
The general solution for the DE:
1 1 2 2 ..... n ny c y c y c y= + + +
1 2 1 0..... 0n na m a m a m a m a
+ + + + + =

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Example 2
Find the general solution of the DE:
) ''' 4 '' 5 ' 0a y y y =
(4)) 2 '' 0b y y y + =
=

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Assignment#1 (5%)
Due: 19 Nov 2012, Monday , 5.00pm at BA371
Find the general solution of the homogeneous DE:
1) 81 0
2) 3 4 12 0
y y
y y y y
+ =
+ =
(5) (4)
(8) (6)
3) 5 36 0
4) 5 2 10 5 0
5) 0
y y y
y y y y y y
y y
=
+ + + =
+ =
www.mnoraini.weebly.com

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4.4 The Method of Undetermined Coefficients Superposition Approach
The general solution of nth order nonhomogeneous
linear DE:
( ) ( 1)1 1 0... ( ) (1)
n nn na y a y a y a y g x
+ + + + =
is given by: ( ) ( ) ( )c px y x y x= +
where yc(x) is a complementary function of DE (1) that
can be obtained by solving the associated homogeneous( ) ( 1)
1 1 0... 0n n
n na y a y a y a y
+ + + + =

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and yp(x) is a particular solution of DE (1)
The particular solution yp(x) can be determined by: The method of undetermined coefficients
Superposition approach Annihilator approach
Variation of parameters

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To find the particular solution, yp for the DE:
can be used if and onl if x is either/combination of
The Method of Undetermined Coefficients
( ) ( 1)1 1 0... ( ) (1)
n nn na y a y a y a y g x
+ + + + =
these functions:
Constant, linear, polynomials, sine, cosine functions,
exponential functions.

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For DE:
Choose a trial x that is similar to the function x
The Method of Undetermined Coefficients
( ) ( 1)1 1 0... ( ) (1)
n nn na y a y a y a y g x
+ + + + =
and involving unknown coefficients to be determined
by substituting the choice of yp(x) into (1)

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List of trial yp(x) by Superposition Approach
Form of
1.
2.
3.
4.
)(xgPy
75 +x BAx +
23 2 x CBxAx ++2
13 +xx DCxBxAx +++23
constant)(any4
5.
6.
7.
8.
9.
10.
x4sin xBx 4sin4cos +
xBxA 4sin4cos +x4cos
xe
5 xAe5
xxexeBAx + )(
xe x 3cos2 xBexAe xx 3sin3cos 22 +
xx 4sin5 2 xFExDxxCBxAx 4sin)(4cos)( 22 +++++
4

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Important rules to determine the final yp(x) by
superposition approach
Rule 1:
If g(x) consists of m different functions,then1 2( ) ( ) ( ) .... ( )kg x g x g x g x= + + +
Rule 2:
No duplicate terms between yc(x) and yp(x).
If exist, yp(x) must be multiplied by xn where n is the
smallest positive integer that eliminate the duplication.
1 2....
kp p p px y x y x y x=

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Example 1
Solve the given nonhomogeneous linear DE by
superposition approach:
y 10y + 25y = 30x + 3

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Solution Example 1
y 10y + 25y = 30x + 3
Step 1: Find yc(x):y 10y + 25y = 0
Roots: m1 = m2 = 5 (repeated)
yc(x) = c1e5x + c2xe5x

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Solution Example 1
y 10y + 25y = 30x + 3
Step 2: Find yp(x) using superposition approach
g(x) = 30x + 3Trial: yp(x) = Ax + B
c p ,
Final: yp(x) = Ax + B
Solve all constants in yp(x) :yp(x) = Ax + B , yp(x) = A , yp(x) = 0

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Solution Example 1  continue
Substitute into the given DE:
y 10y + 25y = 30x + 3
0 10A + 25(Ax + B) = 30x + 3(10A + 25B) + 25Ax = 30x + 3
10A + 25B = 3 . (1)
25A = 30 .. (2)
Solve (1) & (2): A= 6/5, B = 3/5yp(x) = (6/5)x + 3/5

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Solution Example 1  continue
General solution is:
y(x) = yc(x) + yp(x)
y(x) = c1e5x + c2xe5x + (6/5)x + 3/5

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Example 2
Determine the form of particular solution for the given
nonhomogeneous linear DE by superposition
approach. (Do not solve the constants)
2 2 x x
4
2 4 2 xy y x xe
=
= +( )
)

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4.6 Variation of Parameters
Applicable for
Linear differential equations with constant or variable
coefficients.
inear i erentia equations w ere g x is any unctions.
( ) ( 1)1 1 0( ) ( ) ... ( ) ( ) ( ) (1)
n nn na x y a x y a x y a x y g x
+ + + + =

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Solving the nonhomogeneous DE by variation of
parameters
( ) ( ) ( )p x y q x y f x + + =
Given a second order nonhomogeneous linear differential
equation in the form:
Step 1: Find yc by solving :
1 1 2 2c c y c y= +
( ) ( ) 0y p x y q x y + + =

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1 2y y
Step 2 : Find yp by variation of parameters
2(a): Calculate the following Wronskians:
Solving the nonhomogeneous DE by variation of
parameters
1 2
21
2
12
1
0
( )
0
( )
y yy
Wf x y
yW
f x
=
=

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11 1 1
Wu u u dxW
= =
Step 2 (b): find u1 and u2 to obtain the particular solution
Solving the nonhomogeneous DE by variation of
parameters
2 2 2
1 1 2 2p
u u u dxW
y u y u y
= =
= +
Step 3:The general solution for the DE is:
c py y y= +

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Example 1
Solve the given nonhomogeneous linear DE by
variation of parameters.
'' tany x+ =

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Example 1  solution
'' tany x+ =
Step 1: Find yc by solving:
0y y + =
1 2sin cosc
m m
c x c x
+ = =
= +

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Example 1  continue
'' tany x+ =
Step 2: Find yp by variation of parameters
Define:
1 2sin , cos , ( ) tanx y x f x x= = =
Calculate:
1 2 2 2
1 2
sin cossin cos 1
cos sin
y y x xW x x
y y x x
= = = =

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Example 1  continue
'' tany x+ =
Step 2: Find yp by variation of parameters
Calculate:
0 0 cosx1
2
cos tan s n
( ) tan sin
x x x
f x y x x
= = = =
21
2 1
0 sin 0 sinsin tan
( ) cos tan cos
y x xW x x
y f x x x x= = = =

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Example 1  continue
'' tany x+ =
Step 2: Find yp by variation of parameters
Find u1:
1 sinW x = = =
1
1sin cos
Wu x dx x
= =

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Example 1  continue
'' tany x+ =
Step 2: Find yp by variation of parameters
Find u2:
2 2sin / cos sinW x x x2
2 2
2
2
1 cos
sin 1 cossec cos
cos cos
ln sec tan sin
u
W x
x xu dx dx x x dx
x x
u x x x
= = =
= = =
= + +

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1 1 2 2
cos [sin ] [ ln sec tan sin ]cos
py u y u y
x x x x x x
= +
= + + +
=
Therefore:
Example 1  continue
1 2sin cos ln sec tan
c py y y
y c x c x x x
= +
= + +
The general solution for the DE is:

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4.7 CauchyEuler Equation (CE)
It is a linear DE with variable coefficients.
A CE equation has the following form:
22 1 0
( )... ( )
n nna x y a x y a xy a y g x + + + + =
The CE equation has a special characteristic such that
the degree of the monomial coefficient matches
the order k of differentiationk
k
dx
yd
kx

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Solution for Secondorder CauchyEuler Equation
Consider the homogeneous linear second order DE :
where a, b, c are constant coefficients.
2 0 (1)ax y bxy cy + + =
Try a solution for the CauchyEuler (1) of the form:
Calculate:
mx=
1 2, ( 1)m mmx y m m x = =

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Substitute y, y and y into the DE (1):
2 2 1[ ( 1) ] [ ] 0
( 1) 0
m m m
m m m
m
ax m m x bx mx cx
am m x bmx cx
+ + =
+ + =
=
2 0 (1)ax y bxy cy + + =
Since: for real value of x, therefore:
This equation is called an auxiliary equation of the CE (1).
2
( 1) 0 (2)
( ) 0
am m bm c
am b a m c
+ + =
+ + =
0mx

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In general, the auxiliary equation for the CauchyEuler DE:
is given by:
2 0 (1)ax y bxy cy + + =
2
( 1) 0 or
0 2
am m bm c
am b a m c
+ + =
+ + =
The roots of the equation can be real and distinct, real but
repeated or complex conjugates.
Therefore y = xm is a solution of (1) if m is the root of
auxiliary equation (2).

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General Solutions of Homogeneous CauchyEuler
Linear Second Order Differential Equation
Homogeneous 2Homogeneous 2ndnd
Order CEOrder CE
AuxiliaryAuxiliary
equationequation
02 =++ cyybxyax
2 ( ) 0+ + =am b a m c
Distinct realDistinct realrootsroots
1 2
21
1 2
1 2
1 2GS:
,m m
mm
m m
y x y x
y c x c x
= =
= +

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General Solutions of Homogeneous CauchyEuler
Linear Second Order Differential Equation
Homogeneous 2Homogeneous 2ndnd
Order CEOrder CE
AuxiliaryAuxiliary
equationequation
02 =++ cyybxyax
2 ( ) 0+ + =am b a m c
Repeated realRepeated realrootsroots 1 2
1 2
1 2GS:
m m
m m
m m m
y x y x x
y c x c x x
= =
= =
= +
, ln
ln
G l S l i f H C h E l

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General Solutions of Homogeneous CauchyEuler
Linear Second Order Differential Equation
Homogeneous 2Homogeneous 2ndnd
Order CEOrder CE
AuxiliaryAuxiliary
equationequation
02 =++ cyybxyax
2 ( ) 0+ + =am b a m c
ComplexComplexconjugates rootsconjugates roots 1 2
1 2
1 2GS:
= + =
= =
= +
,
cos( ln ), sin( ln )
cos( ln ) sin( ln )
m i m i
y x x y x x
y c x x c x x

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Example:
Solve the following CauchyEuler equation:
2
2
(a) 5 3 0
(b) 5 4 0
+ + =
+ + =
x y xy y
x y xy y
2
2
(c) 7 41 0
(d) 2
+ =
+ =
x y xy y
x y xy y x
S l i

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Solution:
2(a) 5 3 0 1 5 3 + + = = = =, ,x y xy y a b c
auxilliary equation:
2
2
0+ + =
( )am b a m c
Roots:
2 4 3 0 3 1 0+ + = + + =( )( )m m m m
1 2
3 11 2
3 11 2
3 1
= =
= =
= +
,
,
m m
y x y x
y c x c x
S l ti

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Solution:
2(b) 5 4 0 1 5 4 + + = = = =, ,x y xy y a b c
auxilliary equation:
2
2
0+ + =
( )am b a m c
Roots:
2 4 4 0 2 2 0+ + = + + =( )( )m m m m
1 2
2 21 2
2 21 2
2 2
= =
= =
= +
,
, ln
ln
m m
y x y x x
y c x c x x
S l ti

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Solution:
2(c) 7 41 0 1 7 41 + = = = =, ,x y xy y a b c
auxilliary equation:
2
2
0+ + =
( )am b a m c
2 8 41 0 4 5 + = = m m m i
4 41 2
4 41 2
4 5
5 5
5 5
= =
= =
= +
,
cos( ln ), sin( ln )
cos( ln ) sin( ln )
y x x y x x
y c x x c x x
Sol tion (d) non homo eneo s Ca ch E ler

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Solution (d): nonhomogeneous CauchyEuler
2 2 + =x y xy y x
Find yc :
2 0 + =x y xy y
2 =
Therefore:
2
21 2
2 1 0
1 0 1
+ =
= = =( )
m m
m m m
1 2= + lncy c x c x x
Solution (d): non homogeneous Cauchy Euler

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Solution (d): nonhomogeneous CauchyEuler
22
1 1 22 + = + =x y xy y x y y yx xx
Find yp using variation of parameters:
Write the DE in the form of:
Define:
1 22
= = =, ln , ( )y x y x x f xx
Solution (d): non homogeneous Cauchy Euler

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Solution (d): nonhomogeneous CauchyEuler
Calculate all the Wronskians:
1 2
1 2
ln
(1 ln ) ln1 1 ln
0 0 ln
y y x x x
W x x x x xy y x
x x
= = = + = +
12
12
1
2 n
( ) 2 / 1 ln
0 02
( ) 1 2 /
W x
f x y x x
y xW
y f x x
= = =
+
= = =
Solution (d): nonhomogeneous CauchyEuler

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Solution (d): nonhomogeneous CauchyEuler
Find u1 and u2 :
1
1
2 21
2ln
2ln2 (ln )
W xu
W x
xu dx udu u x
= =
= = = = x
22 1
2 22ln
Wu u dx x
W x x = = = =
Solution (d): nonhomogeneous CauchyEuler

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Solution (d): nonhomogeneous CauchyEuler
Write the particular solution:
1 1 2 2
2
2
(ln ) 2 ln ( ln )
py u y u y
x x x x x
= +
= +
=
General solution is:
21 2 ln (ln )c py y c x c x x x x= + = + +