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Chapter 2 Higher Order Difference and Differential Equations 2.1
Higher Order Differential Equations 2.1.0 Introduction In Chapter
1, we saw how to solve few first-order differential equations by
recognizing them as separable, homogeneous and exact. We now turn
to the solution of differential equations of order two or higher.
We will now examine various techniques for solving linear
homogeneous and non-homogeneous differential equations. First we
will concentrate on second-order equations for two main reasons.
First, their theory is typical of that of linear differential
equations on any order n and second they have important
applications in many fields of science and engineering. Numerical
methods for higher-order differential equations will also be
presented. Definition: For a linear differential equation, an
nth-order initial value problem (IVP), is defined as:
Solve: )()(....)()( 011
1 xgyxadxydxa
dxydxa n
n
nn
n
n =+++
Subject: to : 10
11000 )(,...,)(',)(
=== nn yxyyxyyxy Observe that n initial conditions for y(x) and
its derivatives are specified. Theorem 1: Let an(x), an-1(x),,a0(x)
and g(x) be continuous on an interval I and let an(x) 0 for all x
in this interval. If x = x0 is any point in this interval then the
solution y(x) of the initial-value problem exists on the interval
and is unique. The requirements in the above theorem that ai(x)
,i=0,1,2,,n be continuous and an(x)0 for every x in I are both
important. For example you can easily verify that y = cx2 + x + 3
is a solution of the initial value problem: x2 y 2 x y + 2 y = 6,
y(0) = 3, y (0) = 1 on the interval ),( for any choice of the
parameter c. So the solution is not unique. The continuity
conditions of the Theorem 1, are satisfied, but a2(x) = x2 is zero
at x =0. Thus Theorem 1 does not apply.
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Definition: For a linear differential equation, a 2nd-order
boundary value problem (BVP), is defined as:
Solve: )()()()( 0122
2 xgyxadxdyxa
dxydxa =++
Subject to: y(a) = y0 , y(b) = y1 The values y(a) = y0 , y(b) =
y1 are called boundary conditions. For a second order differential
equation other pairs of boundary conditions could be: y (a) = y0,
y(b) = y1
y(a) = y0, y (b) = y1 y (a) = y0, y (b) = y1 where y0 and y1 are
arbitrary constants. Note that even when the conditions of Theorem
1 are fulfilled a boundary-value problem can have several solutions
, a unique solution or no solution at all. (problem # 2).
Definition: A linear nth-order differential equation of the
form:
0)()(...)()(011
1
1 =++++
yxadxdyxa
dxydxa
dxydxa n
n
nn
n
n (1)
is called a homogeneneous, whereas an equation of the form:
)()()(...)()(011
1
1 xgyxadxdyxa
dxydxa
dxydxa n
n
nn
n
n =++++
(2)
is called non-homogeneous. As we shall see in subsequent
sections, in order to solve a non-homogeneous linear equation (2),
we must first be able to solve the associated homogeneous equation
(1). Note: To avoid needless repetition we will assume throughout
the reminder of the chapter that on some interval I:
the coefficients ai(x), i=0,1,2,,n are continuous; the function
g(x) is continuous; an(x)0 for every x in the interval.
The following theorems play an important role in obtaining the
general solution of a linear differential equation:
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Theorem 2: Let y1, y2, yk be solutions of the homogeneous
differential equation (1), on the interval I. Then the linear
combination: y= c1 y1(x) + c2 y2(x) + . . . + ck yk(x) where the
ci, i = 1,2,,k are arbitrary constants, is also a solution on that
interval. Corollary: If y1(x) is a solution of a linear homogeneous
differential equation, any constant multiple: y(x) = c1 y1(x) is
also a solution. Definition: A set of functions f1(x), f2(x),,fn(x)
is called linear independent on an interval I if there exists
constants ci, i = 1,2,,n (not all zero), such that: c1 f1(x) + c2
f2(x) + . . . + cn fn(x) = 0 (3) for every x in the interval. If
the only constants that satisfy (3), for all x in the interval, are
c1 =c2 =. . . = cn = 0, then the set is called linear independent.
Corollary: A set of functions f1(x), f2(x),,fn(x) is linear
dependent if at least one of the function can be expressed as a
linear combination of remaining functions. (Note that if ci0, then
we can divide both sides of (3) by ci and express fi(x) as a linear
combination of the remaining functions). Example: The set of
functions: f1(x)= cos2x, f2(x) = sin2x, f3(x)= sec2x, f4(x) = tan2x
are linear dependent on the interval (- /2 , /2). Solution: From
trigonometry, we know that: cos2x + sin2x =1 and 1+ tan2x =sec2x
thus for the choice of constants c1 = c2 = c4=1, c3 = -1 the
following is satisfied on the given interval: c1 cos2x + c2 sin2x +
c3 sec2x+ c4 tan2x =0. We are primarily interested in linear
independent solutions of a differential equation. It turns out that
there a rather mechanical way to check in a set of solutions: y1, .
. . ,yn of an nth-order differential equation are linear
independent: Definition: The Wronskian of functions f1(x),
f2(x),,fn(x) is the determinant:
.
............
'...''...
),...,,(
)1()1(2
)1(1
21
21
21
=
nn
nn
n
n
n
fff
ffffff
fffW
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Theorem 3: Let y1, y2,. . . ,yn be n solutions of the
homogeneous linear nth-order differential equation (1), on an
interval I. Then, the sent of solutions is linearly independent on
I if and only if W(y1, y2,. . . ,yn) 0 for every x in the interval.
A Computer Algebra System, like Maple could be very useful for
evaluating Wronskians, and checking whether or not a set of
solutions forms a linear independent set. Example: Assume that the
functions : ex, xex, x2ex are solutions of a linear differential
equation. Do they form a linear independent set? Solution: We use
the Maple command Wronkian which is included in the linalg package:
> restart: > with(linalg): Warning, the protected names norm
and trace have been redefined and unprotected We define the set of
functions: > s:= [exp(x), x*exp(x), x^2*exp(x)];
:= s [ ], ,e x x e x x2 e x
Let's calculate the Wronskian. Be very careful on the spelling
of the word Wronskian: > ws:= Wronskian(s,x);
:= ws
e x x e x x2 e x
e x + e x x e x + 2 x e x x2 e xe x + 2 e x x e x + + 2 e x 4 x
e x x2 e x
We need to calculate the determinant: > result:= det(ws);
:= result 2 ( )e x3
Since the Wronskian is not zero, the given set in linearly
independent. Definition: Any set y1, y2, . . .,yn of linearly
independent solutions of an nth-order homogeneous differential
equation on interval I is called a fundamental set of solutions on
that interval. Note: To determine if a given set of functions form
a fundamental set of a differential equation, we must show that: i)
each function is indeed a solution of the differential equation;
ii) the Wronskian of the given set is not-zero for each value of
x.
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Theorem 4: If { y1, y2, . . .,yn } is a fundamental set (i.e.,
linearly independent set) of solutions to an nth-order differential
equation (1), on an interval I, then the general solution of (1)
is: y(x) = c1 y1(x) + c2 y2(x) + . . . + cn yn(x) where c1, . .
.,cn are arbitrary constants. Lets now examine the overall solution
methodology for linear non-homogeneous differential equations of
the form (2). Definition: Any function yp free or arbitrary
parameters that statisfies (2) is called a particular solution of
the differential equation. Theorem 5: Let yp(x) be any particular
solution to the non-homogeneous differential equation (2), and let
{ y1, y2, . . .,yn } be a fundamental set of solutions of the
associated homogeneous differential equation (1) on the interval I.
Then the general solution of the non-homogeneous differential
equation is: y(x) = c1 y1(x) + c2 y2(x) + . . . + cn yn(x) + yp(x)
We call the linear combination: yc(x) = c1 y1(x) + c2 y2(x) + . . .
+ cn yn(x) the complimentary function/solution. Thus the solution
of a linear non-homogeneous differential equation can be written
as: y(x) = (complimentary solution ) + (any particular solution) =
yc(x) + yp(x) The above theorem can be generalized for linear
non-homogeneous differential equations where the right-hand side
can be decomposed as a sum of simpler functions. The following
theorem is also known as the Superposition Principle:
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Theorem 6: Consider the linear non-homogeneous differential
equation of the form:
)(...)()()()(...)()( 21111
1 0xgxgxgyxa
dxdyxa
dxydxa
dxydxa kn
n
nn
n
n +++=++++
(4)
where g(x) = g1(x) + g2(x) + . . . + gk(x). If ypi(x) is a
particular solution of the differential equation:
)()()(...)()(011
1
1 xgyxadxdyxa
dxydxa
dxydxa in
n
nn
n
n =++++
for i = 1,2,,k
then the particular solution for (4) will be: yp(x) = yp1(x) + .
. . + ypk(x) Thus the general solution of (4) will be: y(x) =
(complimentary solution ) + (particular solution). Example: If
yp1(x) = 3e2x and yp2(x) = x2 + 3x are respectively particular
solutions of: y 6 y + 5 y = - 9e2x and y 6y + 5y = 5x2 + 3x 16,
find the particular solution of: y 6y + 5y = 5x2 + 3x 16 - 9e2x
Solution: Observing that the right-hand side can be decomposed into
two functions: g1(x) = - 9e2x and g2(x) = 5x2 + 3x 16 , by the
superposition principle, the particular solution will be: yp(x) =
x2 + 3x + 3e2x 2.1.1 Homogenenous Linear Differential Equations
with Constant
Coefficients
We now turn our attention to solving linear homogeneous
differential equations with constant coefficients. Solutions of any
nth-order homogeneous liner differential equations with constant
coefficients are determined by the solutions of the characteristic
equation: Definition: The equation an mn + an 1 mn 1 + . . . + a1 m
+ a0 = 0 is called the characteristic equation of the nth-order
linear homogeneous differential equation with constant
coefficients:
0...011
1
1 =++++
yadxdya
dxyda
dxyda n
n
nn
n
n
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2nd Order Equations: We begin our investigation by considering
the second-order homogeneous linear differential equation with
constant coefficients a y + b y + c y = 0 (5) The characteristic
equation is: a m2 + b m + c = 0. The following theorem gives the
form of the general solution of (5), which depends on the nature of
the 1solutions of the characteristic equation: Theorem 7: Let a y +
b y + c y = 0 be a homogeneous second-order equation with constants
(real) coefficients , and let m1 and m2 be the solutions of the
characteristic equation a m2 + b m + c = 0. Then:
If m1 m2 and both m1 and m2 are real, the set },{ 21 xmxm eeS =
is a fundamental set and the general solution is: xmxm ececy 21 21
+=
If m1 = m2, the set },{ 11 xmxm xeeS = is a fundamental set and
the general
solution is: xmxm xececy 11 21 +=
If m1 = a + i and m2 = a - i , the set )}sin(),cos({ xexeS axax
= is a fundamental set and the general solution is: ( ))sin()cos(
21 xcxcey ax +=
Example: Solve the initial value problem: y + y 2y = 0, y(0) =
4, y (0)=-5 Solution: The characteristic equation is: m2 + m 2 = 0.
Solving the quadratic equation we obtain: m1 = 1 and m2 =-2.
According the Theorem 7, the general solution is: y = c1 ex + c2
e-2x . To find the values of the coefficients we use the initial
conditions. Differentiating we get: y = c1 ex 2c2 e-2x Therefore:
y(0) = c1 + c2 = 4 y(0)= c1 - 2 c2 = -5 Solving the above system,
we get: c1 = 1 and c2 = 3. Thus the general solution is: y(x) = ex
+ 3 e-2 x
Higher Order Equations: As with second-order equations, a
general solution of the nth-order linear homogeneous differential
equation with constant coefficients is determined by the solutions
of the characteristic equation.
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Theorem 8: Let an mn + an-1 mn-1 + . . . + a1 m + a0 = 0 be the
characteristic equation of an nth-order linear homogeneous
differential equations with constant coefficients. Then:
If m is a root of multiplicity 1, then emx is a solution
associated with the root m.
If m is a real root of multiplicity k >1, then the k linear
independent solutions associated with m are: {emx, xemx, x2emx,. .
. , xk 1emx}
If m is complex root ( )ia of multiplicity k > 1, then the 2k
linear
independent solutions associated with m are:
{eax cos(x), xeax cos(x), . . . , xk 1 eax cos(x) } and {eax
sin(x), xeax sin(x), . . . , xk 1 eax sin(x) }
A Computer Algebra System, like Maple is used to find the roots
of a characteristic equation of order greater than two. Example:
Find the general solution of the differential equation: y + 3 y + 3
y + y = 0 Solution: The characteristic equation is: m3 + 3m2 + 3m +
1 =0 Lets use Maple to find its roots: > solve(m^3 + 3*m^2 + 3*m
+ 1=0,m);
, ,-1 -1 -1
We observe that the root is of multiplicity three. Thus
according Theorem 8, the fundamental set of solutions associated
with the root is: {e-x , x e-x , x2 e-x }. So the general solution
is: y(x) = c1 e-x + c2 x e-x + c3 x2 e-x
Using Maples dsolve command we can also obtain the solution as:
> gensol:=dsolve(diff(y(x), x$3) + 3*diff(y(x), x$2) +
3*diff(y(x), x) + y(x) =0, y(x));
:= gensol = ( )y x + + _C1 e( )x _C2 e( )x x _C3 e( )x x2
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Example: Find the general solution of the differential equation:
2 y(5) - 7 y(4) + 12 y + 8 y =0 Solution: The characteristic
equation is: 2 m5 7m4 + 12 m3 + 8 m2 =0 Lets use Maple to find the
roots: > solve(2*m^5- 7*m^4 + 12*m^3 + 8*m^2=0, m);
, , , ,0 0-12 + 2 2 I 2 2 I
Observe that one root is of multiplicity two. Also we got a pair
of complex conjugates. According to Theorem 8, the general solution
is: y(x) = c1 + c2 x + c3 e-x / 2 + e2 x (c4 cos(2x) + c5 sin(2x) )
Modeling Free Oscillations of Mass-Spring System: Assume a spring
that resists compression as well extension is suspended vertically
from a fixed support. At the lower end of the spring we attach a
body of mass m . We assume m to be large, so we disregard the mass
of the spring. If we pull the body down a certain distance and then
release it, it undergoes a motion. Analyze the free undamped as
well as the free damped motion. Solution: We will examine
separately each of the cases, by setting up the differential
equation that models the motion of the spring and analyze the
nature of the solutions. Free Undamped Motion: Hookes law states
that a spring exerts a restoring force F opposite to the direction
of elongation, which is proportional to the amount of elongation s.
Thus after a mass m is attached to the spring, it stretches the
spring by an amount s and attains a position of equilibrium at
which the weight is balanced by the restoring force k s. Thus m g =
k s. We refer to k as the constant of the spring. If the mass is
displaced by an amount x from the equilibrium the restoring force
of the spring is: k (x + s). Assuming there are no other forces
acting on the system, Newtons second law describes the free
undeamped motion:
Fnet = m a Thus: kxksmgkxmgxskdt
xdm =+=++= )(22
Rearranging the above, and defining 2 = k / m we obtain the
equation of motion:
0222
=+ xdt
xd (6)
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Equation (6) describes the free undampled motion (simple
harmonic motion). The initial conditions associated with (6) are :
the amount of initial displacement x(0) = x0 and the initial
velocity x (0) = x1. We assume that if x0 > 0 the mass is
released form a point below the equilibrium (if x0 < 0 is
released above). Also if x1 > 0 the mass is released with a
downward velocity (if x1 > 0 , is released with upward velocity,
if x1 = 0 is released from rest). To solve (6), we observe that the
characteristic equation is: m2 + 2 =0 which has complex roots: m1 =
i and m2 = - i. Thus solution will be: x(t) = c1 cos ( t) + c2 sin(
t) (6a) The above can also be written as: x(t) = A sin( t + ) where
( ) 2/12221 ccA += and tan = c1 / c2 The above is an oscillatory
motion of period: T = 2 / and frequency f = 1 / T and amplitude A.
Free Damped Motion: The concept of free harmonic motion is somewhat
unrealistic, since it assumes that no other forces act on the
system. Unless the mass is suspended in a perfect vacuum there will
be at least a resisting force due to the surrounding medium. In the
study of mechanics damping forces acting on a body are proportional
to the instantaneous velocity dx / dt of the body. Thus assuming
that no other external forces are impressed on the system, Newtons
second law states:
dtdxckx
dtxdm =2
2
Rearranging the above and defining 2 = k / m and 2 a = c / m, we
obtain the equation of motion:
02 222
=++ xdtdxa
dtxd (7)
To (fully) obtain a solution for (7), we need two initial
conditions. The adopt the same conventions for the initial
conditions as described in the case of free harmonic motion. The
characteristic equation for (7) is: m2 + 2 a m + 2 =0, and the
corresponding roots are: m1 = -a + ( a2 - 2)1 / 2 and m2 = = -a - (
a2 - 2)1 / 2 Thus the nature of the solutions and thus the nature
of the motion depends on the relationship between a and . Note that
in each case the solution contains a damping factor e- a t which
causes the displacements of the mass to become negligible as t
increases. We examine three cases:
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Overdamping: The damping constant a is large and a2 - 2 > 0.
Therefore the characteristic equation has real roots. The solution
is: ( )tataat ececetx 2/1222/122 )(2)(1)( += (7a) The above
equation represent a nonoscillatory motion . Critical damping: In
this case, a2 - 2 = 0. The characteristic equation will have a real
root of multiplicity two (double real root). The solution is: (
)tccetecectx atatat 2121)( +=+= (7b) The motion is very similar to
that of an overdamped system. Note that since the exponential term
is never zero, and the term (c1 + c2 t) can have at most one
positive zero (root), it follows that the motion can have at most
one passage through the equilibrium position ( x(t) = 0 ).
Underdamping: The damping coefficient is small compared to the
spring constant and a2 - 2 < 0. Therefore the characteristic
equation has complex roots. The solution is: ( ))sin()cos()( 222221
atcatcetx at += The above equation can also be written as: x(t) = A
e-a t sin((2 a2)1/2 t + ) ( ) 2/12221 ccA += and tan = c1 / c2.
This is an oscillatory motion, but because the coefficient e-a t
the amplitudes of vibrations tends to zero as t tends to
infinity.
The quantity A e-a t is called the damped amplitude and the
quantity 22
2a
the quasi period of the motion.
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Example: A 16-pound weight is attached to a 5-foot long string.
At equilibrium the spring measures 8.2 feet. If the weight is
pushed up and released from rest at a point 2ft above the
equilibrium position, find an expression for the displacement
function x(t) and also graph it. Assume that the surrounding medium
offers a resistance equal to the instantaneous velocity. Solution:
From the given data, the elongation of the spring after the weight
is attached is: 8.2 5 = 3.2 ft. Using Hookes law we can calculate
the constant k of the spring: 16 = k (3.2) or k = 5 lb/ft. Also the
mass is: m = 16 / 32 =1/2. Thus the equation of motion is:
dtdxx
dtxd = 5
21
2
2
or 010222
=++ xdtdx
dtxd
The characteristic equation is: m2 + 2m + 10 = 0 with roots: im
312,1 = Thus the system is underdamped. Thus the displacement
function x(t) is given by: x(t) = e-t ( c1 cos3t + c2 sin3t ) The
constants can be found from the initial conditions: x(0) = -2
(above the equilibrium) and x(0) = 0 (released from rest). Using
the initial conditions we obtain: c1 = -2 and c2= - 2/3. Thus the
equation of motion becomes:
= ttetx t 3sin323cos2)(
Using the plotting capabilities of Maple, we can plot and
visualize the underdamped motion of the system: > plot(exp(-t)
*(-2*cos(3*t) - (2/3)*sin(3*t)), t=0..2*Pi);
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Example: A box with rectangular cross-section a x b (feet) and
height H, and weight w floats in fluid whose weight per volume
(density) is . If h is the height of the submerged box at
equilibrium point, describe the motion subsequent to its being
pushed down into the fluid. Create a mathematical model describing
the motion. Is the motion oscillatory? What is the period?
Solution: At equilibrium point, the weight of the box is a force
balanced by the buoyant force exerted by the fluid. According to
Archimedes Principle, the buoyant force is equal to the weight of
the displaced fluid. Thus: Weight of the box = Buoyant Force =
Weight of fluid displaced w = (a b) h (Note: The quantity (a b )
represent the submerged volume of the box which is equal to the
volume of the displaced fluid) We now consider the non-equilibrium
configuration. For this, mark a line around the box at equilibrium.
Let y(t) denote the amount by which this line is displayed above or
below the surface of the fluid. Lets assume that y(t) is positive
is the box rises ( line is visible) and negative if the box
descends. Then, the buoyancy force will be: a b ( h y ) p (Note: y
= y(t) ) According to Newtons second law Fnet = m a, where m is the
mass of the box and a the acceleration. The model assumes that no
other forces act on the box. Therefore:
22
)()(dt
ydgwwyhab =+
But, w = (a b) h so substituting to the previous equation and
simplifying, we obtain:
0)(22
=
+ tywgab
dtyd
Denoting
=wgab2 the equation of motion is:
0222
=+ ydt
yd The above denotes a free undamped motion. Lets assume that
the initial conditions are: y(0) = y0 and y (0) = u0 ( initial
velocity). Solving the equation of motion and using the initial
conditions to find the appropriate constants, we obtain:
-
)sin()cos()( 00 tuttyty
+=
Thus the box, bobs with periodic motion. Its period is 2=T
2.1.2 Non-Homegeneous Linear Differential Equations: Method
of
Undetermined Coefficients: In the previous section we learned
how to solve nth-order linear homogeneous equations with constant
(real) coefficients. These techniques are also useful in solving
some non-homogeneous equations of the form:
)(...011
1
1 xgyadxdya
dxyda
dxyda n
n
nn
n
n =++++
(8)
Recall from Theorem 5, that the general solution for the above
type of differential equations is: y(x) = (complimentary solution )
+ (any particular solution) = yc(x) + yp(x) The underlying idea in
the method of undetermined coefficients, it to make a conjecture,
or an educated guess about the form of particular solution yp(x)
according to the type of function described by g(x). The method
applies only linear non-homogeneous equations of the form (8),
where:
the coefficients are constants; the function g(x) is a constant,
a polynomial, and exponential of the form
eax, a since or cosine function sinx or cosx of a finite sume or
products of these functions.
Method: 1) Solve the corresponding homogeneous differential
equation and obtain the complimentary solution yc(x). 2) Determine
the form of the particular solution yp(x) (see below). 3) Determine
the unknown coefficients in yp(x) by substituting the yp(x) into
the given differential equation and equate like terms. 4) Form the
general solution.
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Determining the form of yp(x):
If g(x) contain a term of the form xm , then the associated
particular solution contains a linear combination of elements form
the set: S = { xm, xm-1,. . . , x, 1}
If g(x) contains a term of the form xm ekx then the associated
particular solution has a linear combination of elements from the
set:
S = { xmekx, xm 1 ekx, . . ., x0 ekx} If g(x) contains a term of
the form xm eax cos(x) or of the form xm eax sin(x)
then the associated particular solution has a linear combination
of elements form the set: S = { xm eax cos(x), xm -1 eax cos(x) ,.
. . ., xeax cos(x), eax cos(x),
{ xm eax sin(x), xm -1 eax sin(x) ,. . . ., xeaxsins(x), eax
sins(x)}
If any of the functions is the set S appears in the general
solution yc(x) of the corresponding homogeneous equation, multiply
each function in the set S by xr to obtain a set S where r is the
smallest integer so that each function in S is not a function in
S
Trials for the particular solution yp(x):
g(x) Form of yp(x) 3 x 10 A x + B 2x3 + 12x 3 Ax3 + Bx2 + Cx + D
sin5x A sin5x + B cos 5x ( 5x2 + 4) e3x (Ax2 + Bx + C) e3x x3 e2x
sin5x (Ax3 + Bx2 + Cx + D) e2x cos5x + (Ex3 + Fx2 + Gx + H) e2x
sin5x Example: Solve y 2 y - 3y = 4x 5 + 6xe2x Solution: First we
solve the associated homogeneous differential equation: y 2y 3y = 0
The characteristic equation is: m2 2m -3 =0, with roots m1,2 = -1,
3 Thus the complimentary solution is: yc(x) = c1 e-x + c2 e3x
Observe that g(x) consists of a polynomial and an exponential type
function. Thus we seek a particular solution of the form: yp(x) =
Ax + B + C x e2x + E e2x Differentiate the above, substitute to the
given non-homogeneous differential equation and group like terms:
-3Ax 2A 3B 3Cx e2x + ( 2C - 3E) e2x = 4x 5 + 6x e2x Equating like
terms, we obtain a linear system of algebraic equations (to be
solved by hand or by using a Computer Algebra System): 3 A = 4, -2A
3B = -5, -3C = 6, 2C 3E = 0
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Solving the above we have: A = - 4/3, B= 23/9, C = -2, E = -4/3
Thus the general solution is:
xxxxpc exexececxyxyxy223
21 342
923
34)()()( +=+=
Note the values of c1 and c2 could be determined if initial or
boundary conditions were available. Example: Solve y 6 y + 9y = 6x2
+ 2 12 e3x Solution: It is easy to determine that the complimentary
solution is: yc(x) = c1 e3x + c2 x e3x (The characteristic equation
has a double root) From the form of g(x) we seek (at first), a
particular solution of the form: yp(x) = A x2 + Bx + C + D e3x
Inspection of this solution shows that one term is duplicated in
the complimentary solution. To avoid duplication, we need to
multiply the exponential part of the particular solution by x2
(smallest integer is r = 2). Thus the correct form of the
particular solution is: yp(x) = A x2 + Bx + C + D x2 e3x
Differentiating, substituting into the differential equation and
collecting like terms, gives: -9A x2 + (-12 A + 9 B ) x + 2A 6B +
9C + 2 D e3x = 6x2 + 2 12 e3x Equating like terms, we obtain the
values for the parameters. Thus the general solution is: y(x) = c1
e3x + c2 x e3x + 2/3 x2 + 8/9 x + 2/3 6 x2 e3x
Example: Determine the correct form of the particular solution
of: y(4) + y + 1 x2 e-x Solution: The characteristic equation of
the associated homogeneous differential equation is: m4 + m3 = 0.
The roots are: 0, 0, 0, -1 Thus the complimentary solution is:
yc(x) = c1 + c2 x + c3 x2 + c4 e-x
Our initial guess for a particular solution is: yp(x) = A + (B
x2 + C x + D) e-x Observe that the first part of the particular
solution has a term (constant term) which also exists in the
complimentary solution. To eliminate duplications, we must multiply
the first part of the particular solution by x2 Thus the final form
of the particular solution is: yp(x) = Ax2 + (B x2 + C x + D)
e-x
-
Lets explore the use of Maple in the solution process on linear
non-homogeneous differential equations, with the method of
undetermined coefficients: Example: Solve : y + 4 y + 14 y + 20 y =
10 e-2x e-x cos(3x) Solution: First we find the solution of the
associated homogeneous differential equation: > restart: >
homog_solut:= dsolve(diff( y(x), x$3) + 4*diff(y(x), x$2) +
14*diff(y(x), x) + 20*y(x) =0, y(x));
:= homog_solut = ( )y x + + _C1 e( )2 x _C2 e( )x ( )sin 3 x _C3
e( )x ( )cos 3 x We consider at first the following form for:
yp(x)=A e-2x + e-x (B cos3x + C sin3x ) Since there re duplications
with terms of the complimentary solution, we multiply the above by
x and the revised version for the particular solution (avoiding
duplications) is: yp(x) = A x e-2x + x e-x ( B cos3x + C sin3x )
Maple can be very helpful in obtaining the values for the
parameters A,B, C, D. This can be achieved as follows: >
restart: > yp:=(x)-> A*x*exp(-2*x) + x*exp(-x)*( B* cos(3*x)
+ C*sin(3*x));
:= yp x + A x e( )2 x x e( )x ( ) + B ( )cos 3 x C ( )sin 3 x
> > left_side:= simplify(diff(yp(x), x$3) + 4* diff(yp(x),
x$2) + 14*diff(yp(x), x) + 20 * yp(x)); left_side 10 A e
( )2 x18 e
( )xB ( )cos 3 x 18 e
( )xC ( )sin 3 x 6 e
( )xB ( )sin 3 x :=
6 e( )x
C ( )cos 3 x + The following command mathes the like terms from
both sides and produces a set which contain the numeric values for
the parameters A, B, C: > match(10*exp(-2*x) - exp(-x)*cos(3*x)
= left_side,x, 'val');
true
> val; { }, , = C -160 = B 3 C = A 1
-
> Therefore the general solution is:
++++= )3sin(601)3cos(3)3cos()3sin()( 232
21 xxxexexecxececxy
xxxxx
2.1.3 Non-Homegeneous Linear Differential Equations: Method
of
Variation of Parameters : The method of variations of parameters
is more general than the previous method since it applies for any
type of function g(x) in a non-homogeneous linear differential
equation of order two or higher. First we present the method for
2nd order equations, followed by its generalization for nth order
equations: Method for second order equations: Given a second order
equation: a2(x) y + a1(x) y + a0(x) y = g(x)
1) Divide by a2(x) to rewrite the equation in the standard form:
y + p(x) y + q(x) y = f(x)
2) Find the complimentary solution yc(x) = c1 y1(x) + c2 y2(x)
of the associated homogeneous equation.
3) Let W = '' 21
21
yyyy
4) Let W
xfxyu )()(' 21= and
Wxfxyu )()(' 12 =
5) Integrate to obtain u1(x) and u2(x) 6) A particular solution
is: yp(x) = u1(x) y1(x) + u2(x) y2(x) 7) The general solution is:
y(x) = yc(x) + yp(x)
-
Method for higher order equation: The method we have just
examined can be generalized to linear nth-order equations that can
be put in the standard for: y(n) + Pn 1(x) y(n 1) + . . . + P1(x) y
+ P0(x) y = f(x) If yc(x) = c1 y1 + . . . + cn yn is the
complimentary solution of the associated homogeneous equation, then
the particular solution has the form: yp(x) = u1(x) y1(x) + u2(x)
y2(x)+ . . . + un(x) yn(x) where the uk k = 1,2,, n are determined
by solving the linear system: y1 u1 + y2 u2 + . . . + yn un = 0 y1
u1 + y2 u2 + . . . + y un = 0 . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . y1(n 1) u1 + y2(n 1) u1 + . . . + yn(n 1)
u1 = f(x) Using Cramers rule the above system can be solved
yielding:
WW
xu kk =)(' k = 1,2,3,,n where W is the Wronskian of y1, y2, . .
., yn and Wk is the determinant obtained by replacing the kth
column of the Wronskian by the column consisting of the elements (
0, 0, 0, . . . , f(x) ) Example: Solve 4 y + 36 y = csc(3x)
Solution: Note that the method of variation of undetermined
coefficients cannot be applied because of the special nature of the
function g(x). Thus we will use the method of variation of
parameters. We first bring the differential equation into the
standard form: y + 9 y = 1/4 csc(3x) We can easily obtain the
solution of the associated homogeneous equation to be: yc(x) = c1
cos3x + c2 sin3x Thus y1(x) = cos3x and y2(x) = sin3x. Also f(x) =
1/4 csc(3x)
The 33cos33sin3
3sin3cos)3sin,3(cos == xx
xxxxW
-
The 41
3cos33csc)4/1(3sin0
1 == xxx
W and xx
xxx
W3sin43cos
3csc)4/1(3sin303cos
2 ==
Thus, 121' 11 == W
Wu and x
xWWu
3sin123cos' 22 ==
Integrating the above, we obtain:
xxu121)(1 = and 36
)3ln(sin)(2xxu =
Thus the particular solution is:
( ) ( ))3sin(ln)3sin(361)3cos(
121)( xxxxxy p +=
The general solution is: y(x) = yc(x) + yp(x) Example: A mass m
= 1/5 Kg is attached to a spring with constant k = 2 Nt/m. An
external force f(t) = 5 cos(4t) is acting on the vibrating mass.
The motion is damped with a coefficient c = 1.2 Assume that the
mass is released from rest at 1/2 ft below the equilibrium.
Construct a model for the equation of motion. Solve the model,
characterize and visualize the components of the solution.
Solution: The external force f(t) acting on the vibrating mass
represents a driving force causing an oscillatory motion. Using
Newtons second law Fnet = m a we obtain:
)(22
tfdtdxckx
dtxdm +=
which can be written (standard form): )(2 222
tFxdtdx
dtxd =++
where F(t) = f(t) /m 2 = c/ m and 2 = k / m For the given
situation, the equation of motion can be modeled as an initial
value problem:
)4cos(522.151
2
2
txdtdx
dtxd =++ x(0) = 1/2 and x (0) =0
In standard for the above can be written as: )4cos(2510622
txdtdx
dtxd =++
-
The associated homogeneous differential equation has a
characteristic equation with roots: m1 = =3 + i and m2 = -3 i Thus
the solution is: ( ))sin()cos()( 213 tctcetx tc += We call the
above the transient tem or transient solution. Observe that xc(t)
thends to zero as time t tends to infinity. Using the method of
undetermined coefficients we assume a particular solution of the
form: xp(t) = A cos(4t) + B sin(4t). This describes an oscillatory
(perpetual) motion. It is called the steady-state term or
steady-state solution. Differentiating and substituting into the
non-homogeneous differential equation gives: (-6A + 24B) cos4t +
(-24A 6B) sin4t = 25 cos4t Equating like terms and solving the
resulting system we obtain: A = -25/102, B = 50/51 Thus the
solution is:
( ) )4sin(5150)4cos(
10225)sin()cos()( 21
3 tttctcetx t ++= Using the initial conditions x(0) = 1/2 and x
(0) =0 and substituting in the above equation and its derivatives,
we obtain the values for c1 and c2 Therefore:
)4sin(5150)4cos(
10225)sin(
5186)cos(
5138)( 3 ttttetx t +
=
We can use Maple to visualize the components of the motion: >
restart: > with(plots): Warning, the name changecoords has been
redefined We first graph the general solution x(t): >
p1:=plot(exp(-3*t)*(38/51*cos(t) - 86/51*sin(t)) - 25/102* cos(4*t)
+ 50/51*sin(4*t), t = 0..Pi,color = red): > display(p1);
-
The steady-state solution can be visualized as: >
p2:=plot(-25/102*cos(4*t) + 50/51*sin(4*t), t = 0..Pi, color =
blue): > display(p2);
The tansient solution can be visualized as: > p3:=
plot(exp(-3*t)*(38/51*cos(t) - 86/51* sin(t)), t= 0..Pi, color=
cyan): > display(p3);
Superimposing all the graphs : > display({p1,p2,p3});
-
Example: Consider the case of a forced undamped motion described
by the equation:
)sin(02
2
2
tFxdt
xd =+ with x(0) =0 and x (0) =0 Use a Computer Algebra System to
examine the motion. As times increases what happens to the
magnitude on x(t) as the period of free oscillations T = 2 /
approaches the period of the external force T = 2 / (i.e., when = )
? Solution: We use Maples dsolve command in which we can
incorporate the given initial conditions: > dsolve({diff(x(t),
t$2) + omega^2 * x(t) = F0*sin(gamma*t),x(0) = 0, D(x)(0)
=0},x(t));
= ( )x t ( )sin t F0 ( ) + 2 2F0 ( )sin t + 2 2
> combine(%); = ( )x t ( )sin t F0 F0 ( )sin t + 3 2
Lets take the limit as approaches the value of : >
limit(%,gamma= omega);
= ( )x t 12F0 ( ) + ( )sin t ( )cos t t
2
> expand(%); = ( )x t 12
F0 ( )sin t2
12
F0 ( )cos t t
Allowing sufficient time, or taking the limit t :
-
> limit(%, t =infinity); = lim
t ( )x t undefined
The value of x(t) becomes infinite (undefined). This phenomenon
is called pure resonance. Thus the displacements of the spring /
mass system will become bigger and bigger thus forcing the spring
beyond its elastic limit and the system will fail. Example:
Consider a long slender vertical column of length L which has
uniform cross section. The column is hinged at both ends. Find and
analyze the deflection if the column is subjected to a vertical
load P. Solution: During the eighteenth century, Leonard Euler was
one of the first mathematicians to analyze how a thin elastic
column buckles under a compressive axial force. Let y(x) represent
the deflection of the column when a constant vertical compressive
force, or load P, is applied to its top. From engineering analysis,
using the equation of bending moments we get:
Pydx
ydEI =22
or 022
=+ Pydx
ydEI
where E is the Youngs module of elasticity and I the moment of
inertia. Since the vertical column is hinged at the two ends: y(0)
= y(L) =0 Thus the model for deflection is represented by the
boundary value problem (BVL):
022
=+ Pydx
ydEI y(0) = 0 and y(L) = 0
Observe that y = 0 is a perfectly good solution for this
problem. This solution has the interpretation: If the load P is not
great enough then there is no deflection. Thus the important
question is: For what values of P will the column bend? That is,
can we find non-trivial solution. Bringing the above equation into
the standard form we obtain: y + y = 0, y(0) = 0, y(L) =0 where =
(P / EI ) Note that > 0. The characteristic equation has roots:
im =2,1 Thus the solution is: ( ) ( )xcxcxy sincos)( 21 += Using
the initial condition y(0) =0 we get c1 =0. From y(L) =0 we obtain
( ) 0sin2 =Lc . If c2 =0 then y=0 (trivial solution). Thus c2 0.
Thus the previous relation implies: ( ) 0sin =L
-
Therefore:
nL = or 222
Ln = , n= 1,2,3,. . .
The deflection curves y(x) (also called eigenfunctions) are
given by:
=L
xncxy sin)( 2 n = 1,2,3, corresponding the eigenvalues n = ( n2
2 / L2 ) Physically this implies that the column will buckle of
deflect if and only if the compressive force has one of the values:
Pn = ( n2 2 E I / L2 ) n = 1,2,3, The different forces are called
critical loads. The deflection curve corresponding to the smallest
critical load P1 is called the Euler load and the corresponding
curve y1(x) = c2 sin ( x / L) is called the first buckling mode.
2.1.4 Homework Problems and Projects Problems 1) Find an interval
centered around x = 0 for which the initial value problems have a
unique solution:
( x 5) y + 3y =x, y(0)= 0, y (0) =1
y + (tanx ) y = ex , y(0) = 1, y (0) =0
2) The differential equation: 01622
=+ ydx
yd has solution the expression
y(x) = c1 cos(4x) + c2 sin(4 x). Show that :
i) The given differential equation with boundary conditions:
y(0) = 0, y(/2)=0 has infinite solutions;
ii) The given differential equation with boundary conditions:
y(0) = 0, y(/8) =0 has only one solution;
iii) The given differential equation with boundary conditions:
y(0) = 0, y(/2) =0 has no solution.
-
3) Show that the set of functions: f1(x)= x1 / 3 + 4, f2(x) = x1
/ 3 + 4x, f3(x)= x 1, f4(x)= x2 are linear dependent.
4) Show that the given functions form a fundamental set of
solutions of the differential equation on the indicated
interval:
y - 4 y = 0; { cosh(2x), sinh(2x) }, I = ),(
x2 y + x y + y = 0; { cos(ln(x)) , sin(ln(x)) }, I = ),0(
(Be careful to insert all appropriate parentheses when using
Maple)
y(4) + y = 0; { 1, x, cos(x), sin(x) }, I = ),( 5) Solve (find
the general solution of) the following differential equations: 12 y
- 5 y 2 y =0
y + y = 0, y (0) = 0, y ( / 2) = 2
022
=+ yd
yd , y( / 3) =0, y ( / 3) = 2
6) Solve(find the general solution of) the following
differential equations. Use also the dsolve command in Maple to
check your answers :
24 y + 2y 5 y y = 0
y(4) + 4y + 24 y + 40 y + 100 y = 0
7) An 8-pound weight stretches a spring 2 feet. Assuming that
the dumping force is numerically equal to 2 times the instantaneous
velocity, determine the equation of motion if the weight is
released from the equilibrium position with an upward velocity of 3
ft/s. Characterize and graph the equation of motion. 8) A 32-pound
weight stretches a spring 2 ft. Determine the amplitude and the
period of motion if the weight is released 1 ft above the
equilibrium position with an initial upward velocity of 2 ft/s. How
many complete vibrations will the weight have completed at the end
of 4 seconds? 9) A cylinder of height H, radius R and weight f,
float with its axis vertical in a fluid whose weight per unit
volume is . Assume that h is the height at equilibrium of the
submerged part of the cylinder, describe the motion of the
cylinder
-
subsequent to its being pushed down into the fluid and released.
Create a mathematical model for describing the motion. Solve the
model. Find an expression for the period of the motion. Explain the
creation and the solution methodology of the model.
10) Two roots of the characteristic equation with real
coefficients are: m1 = -1 and m2 = 5+ i. Find the corresponding
homogeneous linear equation.
11) Solve the following differential equations (You may use
Maple ONLY to help you determine the parameters in the particular
solution). You must show all the steps of the solution process:
y + 4 y = ( x2 3) sin(2x)
y + y = 2 x sin(x)
y(4) - y = 4 x + 2 x e-x
y + 4 y + 4 y = ( 3 + x) e-2x
12) Solve the following differential equations (You may use
Maple ONLY to help you determine the parameters in the particular
solution). You must show all the steps of the solution process: y 4
y + 8 y = ( 2 x2 3 x) e2x cos(2x) + (10 x2 x 1) e2x sin(2x) 13)
Solve he following differential equations:
y + y = sec2x
3 y 6 y + 6y = ex sec x
21'2'' xeyyy
x
+=+
-
14) A 4 lb weight is attached to a spring whose constant is 2 lb
/ft. The medium offers resistance to the motion of the weight
numerically equal to the instantaneous velocity (coefficient of
damping is c = 1). Assume that the weight is released from a point
1 ft above the equilibrium position, with a downward velocity of 8
ft/s.
Construct a model for the equation of motion Solve the model
Find the time at which the weight passes the equilibrium point Find
the time at which the weight attains its extreme displacement from
the
equilibrium point 15) A weight of mass m =1 is attached from a
spring whose constant is 5 lb/ft. Initially the mass is released 1
ft below the equilibrium position with a downward velocity of 5
ft/s. The motion takes place in a medium that offers a damping
force numerically equal to 2 times the instantaneous velocity.
Assume that an external force f(t) = 12 cos(2t) + 3 sin(2t) is
acting on the system:
Find the equation of motion for the system Solve the model Graph
(separately) the transient and steady-state solutions Superimpose
the equation of motion with the above graphs
16) Assume that a thin vertical column of uniform cross section
is embedded at its base ( x =0) and thus does not move. The column
is free at its top ( x = L). A load P is applied to its free end.
The load causes a small initial deflection . The differential
equation for the deflection y(x) is:
=+ Pydx
ydEI 22
Write appropriate initial and boundary conditions and solve the
above differential
equation assuming 0 Solve the differential equation for the
deflection y(x) Show that the Euler load for this column is one
fourth of the Euler load for a
column that is hinged at the two ends.
-
Projects: 1) Consider an RLC-circuit, consisting of an Ohms
resistor of resistance R [ohms], an inductor of inductance L
[henrys], a capacitor of capacitance C [farads] and a battery of
electromotive force E(t) [volts]. The equation for the current I(t)
is obtained by considering the three voltage drops: EL = L I(t)
(voltage drop across the inductor) ER = R I(t) (voltage drop across
the resistor)
EC = dttIC )(1 (voltage drop across the capacitor) Note that the
charge = dttItQ )()( According to Kirchhoffs law, the sum of the
voltage drops is equal to the electromotive force E(t). Assume a
sinusoidal force E(t) = E0 sin( t). a) Use Kirchhoffs law to
generate a model for the current I(t) for a sinusoidal
electromotive force E(t). b) Consider R = 100 ohms, L = 0.1 henry,
C = 10-3 farad and E(t) =155 sin(377 t) and assume that the charge
and current is zero at t =0:
Write the model to this particular RLC circuit Find the
transient and steady state solution Find the general solution I(t).
Use the initial conditions I(0) = Q(0) =0 to calculate the values
of all constants Superimpose the graphs of the transient and
steady-state solution. What
happens to the current after a short time? Clearly explain the
construction and solution methodology of the models in the above
cases. 2) A bead is constrained to slide along a frictionless rod
of length L. The rod is rotating in a vertical plane with constant
angular velocity about a pivot P fixed at midpoint of the rod. The
design allows the bead to move along the entire length of the rod.
Let r(t) denote the position of the bead from the midpoint. Using
Newtons second law and taking into account the gravitational, and
inertia forces (coriolis, centrifugal), the differential equation
of motion is:
)sin(222
tmgrmdt
rdm = a) Solve the differential equation subject to the initial
conditions r(0) = r0 and
r (0) = u0 b) Determine the initial conditions for which the
bead exhibits simple harmonic
motion c) What is the minimum length L of the rod for which in
can accommodate
-
the simple harmonic motion of the bead? d) For initial
conditions other than those obtained in part b), the bead will
eventually fly off the rod. Explain why this is the case? e)
Assume that = 1 rad/s. Superimpose plots of the solution r(t) for
initial conditions r(0) = 0 and r (0) = u0 where u0 = 0, 10, 15,
16, 16.1, 17 For which case(s) the bead stays always on the rod?
Explain!