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    4.1 Preliminary Theory Homogeneous LinearEquations

    Homogeneous linearnth-order DE

    ( ) ( 1)1 2 1 0( ) ( ) ... ( ) ( ) ( ) 0

    n nn na x y a x y a x y a x y a x y

    + + + + + =

    Non-Homogeneous linearnth-order 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 nth-order 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 nth-order 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 nth-order 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 e-3x and e4x form a fundamental set of

    solutions. Hence the general solution is:

    y = c1y1 + c2y2 = c1e-3x + 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

    second-order 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 BA-3-71

    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 non-homogeneous

    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 non-homogeneous 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

    non-homogeneous 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 non-homogeneous DE by variation of

    parameters

    ( ) ( ) ( )p x y q x y f x + + =

    Given a second order non-homogeneous 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 non-homogeneous 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 non-homogeneous 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 non-homogeneous 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 Cauchy-Euler Equation (C-E)

    It is a linear DE with variable coefficients.

    A C-E equation has the following form:

    22 1 0

    ( )... ( )

    n nna x y a x y a xy a y g x + + + + =

    The C-E 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 Second-order Cauchy-Euler 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 Cauchy-Euler (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 Cauchy-Euler 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 Cauchy-Euler

    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 Cauchy-Euler

    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 Cauchy-Euler

    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 Cauchy-Euler 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): non-homogeneous Cauchy-Euler

    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): non-homogeneous Cauchy-Euler

    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): non-homogeneous Cauchy-Euler

    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): non-homogeneous Cauchy-Euler

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    Solution (d): non-homogeneous Cauchy-Euler

    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): non-homogeneous Cauchy-Euler

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    Solution (d): non-homogeneous Cauchy-Euler

    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= + = + +