High Order Differential
Equations
National Chiao Tung University
Chun-Jen Tsai
10/5/2011
Second-Order Linear Equations
A 2nd-order DE of y(x) can be expressed as
G(x, y(x), y'(x), y"(x)) = 0, ∀x ∈ I,
where I is the (open) interval of definition.
The DE is linear if the equation can be re-written as:
A(x)y" + B(x)y' + C(x)y = F(x), ∀x ∈ I,
In general, we assume that A(x), B(x), C(x), and F(x)
are continuous on the open interval I.
Recall that if F(x) = 0, the DE is a homogeneous DE.
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Modeling with Linear 2nd-Order DE
Linear 2nd-order DE are often used to model
mechanical systems and electrical systems.
dashpot (damper)
mass
spring
m
x(t)
x=0
Spring force:
FS = –kx, k > 0.
Dashpot force:
FR = –c(dx/dt), c > 0.
x>0
Since F = mx" = FS + FR, we have
d 2x
dx
m 2 + c + kx = 0.
dt
dt
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Superposition Principle
Theorem: Let y1 and y2 be two solutions of the 2nd-
order homogeneous linear DE y" + p(x)y' + q(x)y = 0
on the interval I. If c1 and c2 are arbitrary constants,
then the linear combination
y = c1y1(x) + c2y2(x)
is also a solution on I.
Proof:
The proof is as trivial as simply substituting y into the
DE.
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Existence of a Unique Solution
Theorem: Assume that p(x), q(x), and f(x) are
continuous on the open interval I containing the point
a. Then, given any two numbers b0 and b1, the
equation
y" + p(x)y' + q(x)y = f(x)
has a unique solution on I that satisfies the initial
conditions
y(a) = b0, y'(a) = b1.
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Example: y" – 2y' + y = 0, y(0) = 3, y'(0) = 1
It is easy to verify that y1(x) = ex and y2(x) = xex are
solutions of the DE. Therefore, the general solution
can be expressed as
y(x) = c1ex + c2xex.
and
y'(x) = (c1 + c2)ex + c2xex.
The initial conditions tell us
c1 = 3 and c1 + c2 = 1. Hence,
the particular solution is
y(x) = 3ex – 2xex.
C2 = –2
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Linear Independence of Functions
Two functions f and g defined on an open interval I
are said to be linearly independent on I if neither one
is a constant multiple of the other. That is, f ≠ kg for
any constant number k.
Examples:
sin x and cos x
ex and e–2x
x+1 and x2
…
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General Solutions (1/2)
Given a DE y" + p(x)y' + q(x)y = 0, can we always find
at least two linearly independent solutions y1 and y2?
From existence and uniqueness of the IVP solution,
we only need to find y1 and y2 such that
y1(a) = 1, y1'(a) = 0 and y2(a) = 0, y2'(a) = 1
because y1(a) ≠ ky2(a) for any constant k.
Given an IVP, can the solution always be written as
y = c1y1(x) + c2y2(x) for some constants c1 and c2?
If so, we have the general solution of the DE.
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General Solutions (2/2)
Let the initial condition be y(a) = b0, and y'(a) = b1
Since c1 and c2 are solutions to the linear systems
y(a) = c1y1(a) + c2y2(a) = b0
y'(a) = c1y1'(a) + c2y2'(a) = b1.
The solution exists and is unique if the matrix
 y1 (a)

 y1′ (a)
y2 (a ) 
.
y2′ (a) 
is non-singular.
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Wronskian of Two Functions
Definition: Given two functions f(x) and g(x), the
Wronskian of f(x) and g(x) is the determinant
f ( x) g ( x)
W ( f ( x), g ( x) ) =
.
f ′( x) g ′( x)
For example, f = ex and g = xex, the Wronskian is
(
)
W e x , xe x =
ex
xe x
ex
e x + xe x
= e2 x .
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Checking Independency of Solutions
Theorem: Suppose that y1 and y2 are two solutions of
the homogeneous 2nd-order linear DE
y" + p(x)y' + q(x)y = 0
on an open interval I where p and q are continuous.
a) If y1 and y2 are linearly dependent, W(y1, y2) ≡ 0 on I.
b) If y1 and y2 are linearly independent, W(y1, y2) ≠ 0
at each point of I.
In the second case, the solution y = c1y1(x) + c2y2(x)
would be a general solution.
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Generalization to nth-Order
An nth-order linear DE has the form:
y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = f(x).
The corresponding homogeneous linear DE is
y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = 0.
All the properties discussed for the 2nd-order case
can be applied to the general n-th order case.
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nth-Order Superposition Principle
Theorem: Let y1, y2,…,yn be n solutions of the
homogeneous nth-order DE on an interval I. If c1,
c2, …, cn are arbitrary constants, then the linear
combination
y = c1y1(x) + c2y2(x) +… + cnyn(x),
is also a solution on I.
Example: e–3x, cos 2x, sin 2x are solutions of the 3rd-
order homogeneous DE y(3) + 3y" + 4y' + 23y = 0. Is
c1cos 2x + c2 sin 2x also a solution?
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Existence of a Unique Solution
Theorem: Assume that p1(x), p2(x), …, pn(x), and f(x)
are continuous on the open interval I containing the
point a. Then, given any n numbers b0, b1, …, bn–1, the
nth-order equation
y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = f(x)
has a unique solution on I, given the initial conditions
y(a) = b0, y'(a) = b1 , …, y(n–1)(a) = bn–1.
Note: If f(x) = 0 and all the initial conditions are zero,
then y(x) = 0 is the unique particular solution.
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Example: y(3) + 3y" + 4y' + 12y = 0
y(x) = c1e–3x + c2cos 2x + c3sin 2x is a solution to the DE.
If y(0) = 0, y'(0) = 5, y"(0) = –39,
then c1=–3, c2 = 3, c3 = –2.
If only partial initial conditions are given, we have
different family of solutions
y(0) undetermined
y'(0) undetermined
y"(0) undetermined
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Linear Dependence of Functions
Definition: A set of functions f1(x), f2(x), …, fn(x) is
said to be linearly dependent on an interval I if there
exist constants c1, c2,…, cn, not all zero, such that
c1 f1(x)+ c2 f2(x)+…+cn fn(x) = 0.
Otherwise, it’s said to be linearly independent.
Example: The function f1(x) = sin 2x, f2(x) = sin x cos x,
and f3(x) = ex are linearly dependent on R since
1·f1 + (–2)·f2 + 0·f3 = 0.
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Wronskian of n Functions
We are more interested in linearly independent
solutions of DE. How can we verify this?
→ Use the Wronskian function.
Suppose each of the functions f1(x), f2(x),…, fn(x)
possesses at least n–1 derivatives. The determinant
W ( f1 , f 2 , L , f n ) =
f1
f1′
M
f1( n −1)
f2
f 2′
M
L
L
f 2( n −1) L
fn
f n′
M
f n( n −1)
is called the Wronskian of the functions.
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Example: x3y(3) – x2y" + 2xy' – 2y = 0
Given three solutions y1(x) = x, y2(x) = x ln x, and
y3(x) = x2, and the initial conditions y(1) = 3, y'(1) = 2,
y"(1) = 1, verify that y1, y2, and y3 are linearly
independent and find the particular solution.
Note that ln x is defined for x > 0, then
x
x ln x
x2
W = 1 1 + ln x 2 x = x ≠ 0.
0
1/ x
2
The particular solution is y(x) = x – 3x ln x + 2x2.
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Independency of Solutions (1/3)
Theorem: Let y1, y2,…,yn be n solutions of the
homogeneous nth-order linear DE
y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = 0
on an interval I, where each pi is continuous. Let
W = W(y1, y2,…,yn).
a) If y1, y2 , …, yn are linearly dependent, W ≡ 0 on I.
b) If y1, y2 , …, yn are linearly independent, W ≠ 0
at each point of I.
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Independency of Solutions (2/3)
Proof a):
If y1, y2 , …, yn are linearly dependent, ∃c1, c2,…, cn that
are not all zeros such that c1y1(x)+…+cn yn(x) = 0.
Differentiate both sides n times, we have the
equivalent system of linear equations
 y1 ( x)

 y1′ ( x)

M
 ( n −1)
y
 1 ( x)
yn ( x)  c1   0 
   
L
yn′ ( x)  c2   0 
=  .



M
M
M
M





( n −1)
( n −1)




y2 ( x) L yn ( x)  cn   0 
y2 ( x)
y2′ ( x)
L
Since c1 = c2 = … = cn = 0 is also a solution, W ≡ 0 on I.
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Independency of Solutions (3/3)
Proof (b):
If y1, y2 , …, yn are linearly independent, then the only
set of c1,…, cn such that c1y1(x)+…+cn yn(x) = 0 is the
set of all zeros. Then the system
 y1 ( x)

 y1′ ( x)

M
 ( n −1)
y
 1 ( x)
yn ( x)  c1   0 
   
′
yn ( x)  c2   0 
L
=  .



M
M
M
M





( n −1)
( n −1)




y2 ( x) L yn ( x)  cn   0 
y2 ( x)
y2′ ( x)
L
has only the trivial solution ∀x ∈ I → the system is
non-singular. Therefore W ≠ 0.
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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 DE on an interval I is said to be a
fundamental set of solutions on the interval I.
Let y1, y2,…,yn be a fundamental set of solutions of the
homogeneous linear nth-order DE on an interval I.
Then, the general solution of the equation on I is
y = c1y1(x)＋c2y2(x)＋…＋cnyn(x),
where c1, c2,…, cn are arbitrary constants.
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Non-homogeneous Solutions (1/2)
Theorem: Let yp be any particular solution of the non-
homogeneous linear nth-order DE
y(n) + p1(x)y(n–1) + … + pn–1(x)y' + pn(x)y = f(x)
on an interval I, and let y1,y2,…, yn be a fundamental
set of solutions. Then the general solution of the
equation on I is:
y = c1y1(x)＋c2y2(x)＋…＋cnyn(x)＋yp,
where c1, c2,…, cn are arbitrary constants.
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Non-homogeneous Solutions (2/2)
Proof:
Let yp(x) be a particular solution of the DE and Y(x) be
any possible particular solution of the DE.
Let yc(x) = Y(x)－yp(x), we have
yc(n) + p1yc(n–1) + … + pn–1yc' + pnyc
= [Y(n) + p1Y(n–1) + … + pn–1Y' + pnY]
– [yp(n) + p1yp(n–1) + … + pn–1yp' + pnyp]
= f(x) – f(x) = 0.
→ yc(x) is a solution to the relating homogeneous DE.
→ Y(x) = yc(x) + yp(x).
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Complementary Function
Definition: The general solution of a homogeneous
linear nth-order DE is called the complementary
function for the associated non-homogeneous DE.
Let yc(x) = c1y1(x)＋c2y2(x)＋…＋cnyn(x), the general
solution of a non-homogeneous linear nth-order DE
has the form:
y(x) = yc(x)＋yp(x).
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Example: y" + 4y = 12x
It is easy to verify that for the DE, yp = 3x is a
particular solution and yc(x) = c1cos 2x + c2sin 2x is its
complementary solution.
If the initial condition is y(0) = 5, y'(0) = 7, since
y(x) = c1cos 2x + c2sin 2x + 3x,
y'(x) = –2c1sin 2x + 2c2cos 2x + 3.
Hence, c1 = 5 and c2 = 2.
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Reduction of Order†
For a 2nd order linear DE, one can construct a 2nd
solution y2 from a known nontrivial solution y1. If y1
and y2 are linearly independent, we must have
y2/y1 ≠ constant,
Therefore, y2(x) = u(x)y1(x). Substitute this y2(x) into
the DE and solve for u(x) is called reduction of order.
†See problem 36 in section 2.2, page 124 in the textbook
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Example: y" – y = 0
Solution:
Given y1(x) = ex, let y2(x) = u(x) ex,
→ y' = uex + exu', y" = uex + 2exu' + exu"
→ y" – y = ex(u"+ 2u') = 0
→ u"+ 2u' = 0
Let w = u', the DE becomes w' + 2w = 0. Multiplying
by the integrating factor e2x, we have d[e2xw]/dx = 0.
Therefore, w = c1e–2x or u' = c1e–2x.
→ u = (–1/2) c1e–2x + c2.
→ y2(x) = u(x) ex = (–c1/2) e–x + c2ex, let c1 = –2, c2 = 0.
→ Check W(ex, e–x)≠0
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General Case
Put the 2nd order DE into standard form:
y" + P(x)y' + Q(x)y = 0,
where P(x) and Q(x) are continuous on some interval
I. If y1 is a known solution on I and that y1(x)≠0 for
all x ∈ I, we have:
e ∫
y2 = y1 ( x) ∫ 2
dx.
y1 ( x)
− P ( x ) dx
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Example: x2y" – 3xy' + 4y = 0
Solution:
y1 = x2 is a known solution.
3
4
′
′
′
→ y − y + 2 y=0
x
x
3 ∫ dx / x
3 ∫ dx / x
2 e
ln x 3
3
y2 = x ∫
dx
←
e
=
e
=
x
x4
dx
= x2 ∫
x
= x 2 ln x
The general solution is y = c1x2＋c2x2lnx.
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Linear Constant Coefficients DEs
A homogeneous linear nth-order constant coefficient
DE has the form:
any(n) + an–1y(n–1) + … + a2y" + a1y' + a0y = 0,
where a0, a1, …, an are constant coefficients and an ≠ 0.
Recall that for a1y' + a0y = 0, the non-singular family of
solution is y = cerx on (–∞, ∞), where r = –a0/a1.
→ Is it possible that for higher order linear constant
coefficient DE also has solution in exponential form?
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Characteristic Equations
Note that if you assume y = erx as a possible solution,
and substitute it into the DE, we have
anrnerx + an–1rn–1erx + … + a2r2erx + a1rerx + a0erx = 0.
Therefore, y = erx is a solution if
anrn + an–1rn–1 + … + a2r2 + a1r + a0 = 0.
The equation is called the characteristic equation or
auxiliary equation of the homogeneous constant
coefficient linear DE.
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Example: 2nd-Order Cases
Consider a 2nd-order DE, ay" + by' + cy = 0.
Let y = erx, and substituting y' = rerx and y" = r2erx into
the DE, we have: ar2erx + brerx + cerx = 0.
erx > 0 for x ∈ R → ar2 + br + c = 0.
The roots of the characteristic equation gives us the
general solution of the DE.
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Classify 2nd-Order Roots (1/2)
Case I, b2 – 4ac > 0:
r has two real roots r1 and r2, and y1 = er x and
y2 = er x form a fundamental set of solutions.
The general solutions is
1
2
y = c1e r1x + c2 e r2 x .
Case II, b2 – 4ac = 0:
r has one real root r1 and y1 = er x. By variation of
parameters (or reduction of order), the 2nd solution of
the DE is y2 = xer x. The general solution is
1
1
y = c1e r1x + c2 xe r1x .
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Classify 2nd-Order Roots (2/2)
Case III, b2 – 4ac < 0:
r has two complex roots r1 = α + iβ and r2 = α – iβ.
Similar to Case I, the general solution is:
y = c1e (α +iβ ) x + c2 e (α −iβ ) x .
By proper selection of c1 and c2, and using Euler’s
formula, it can be shown that a general solution can
also be represented by
y = eαx (c1 cos βx + c2 sin βx).
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Example: 4y"+4y'+17y = 0
Solve the IVP: y(0) = –1, y'(0) = 2.
Solution:
The roots of the auxiliary equation 4m2+4m+17 = 0 are
m1 = –½ + 2i and m2 = –½ – 2i
→ y = e–x/2 (c1cos 2x + c2sin 2x), and
y' = (–c1/2 + 2c2)e–x/2cos 2x + (–c2/2 – 2c1)e–x/2sin 2x
→ y = e–x/2 (– cos 2x + ¾ sin 2x)
y
1
x
y → 0, as x → ∞.
1
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Higher-Order Auxiliary Equations
In general, to solve
any(n) + an–1y(n–1) + … + a2y" + a1y' + a0y = 0,
where ai ∈ R and an ≠ 0, we must solve
anrn + an–1rn–1 + … + a2r2 + a1r + a0 = 0.
The general solution of the DE is:
Case I (no repeated roots):
y = c1e r1x + c2 er2 x + ... + cn e rn x .
Case II (with repeated roots):
y = c1e r1 x + c2 xer1 x + ... + ck x k −1e r1x + ck +1e r2 x + ... + cn e rn−k x .
solution form of
repeated roots
solution form of
distinct roots
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Differential Operators
The symbol D, defined by Dy = dy/dx, is called a
differential operator. D transforms a function into
another function. For example:
D(cos 4x) = –4sin 4x, D(5x3 – 6x2) = 15x2 – 12x
In general, we define Dny = dny/dxn.
Polynomial expressions involving D, such as D + 3
and D2 + 3D – 4 are also differential operators. In
particular, they are called polynomial differential
operators.
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nth-Order Differential Operator
Definition: An nth-order differential operator is:
L = anDn＋an–1Dn–1＋…＋a1D＋a0.
Note that a DE can be expressed in L. In particular,
Ly = 0 stands for
Ly = any(n) + an–1y(n–1) + … + a1y' + a0y = 0.
Example: y" + 5y' + 6y = 5x – 3 can be written as
D2y＋5Dy＋6y = 5x－3, or (D2＋5D＋6)y = 5x－3.
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Properties of L
L is a linear operator. That is,
L{α f(x)＋β g(x)}＝α L(f(x))＋β L(g(x)).
L is also commutative. For example,
(D – a)(D – b)y = (D – b)(D – a)y.
Proof:
(D – a)(D – b)y = (D – a)(y' – by)
= D(y' – by) – a(y' – by)
= y" – by' – ay' + aby
= (D – b)Dy – (D – b)ay
= (D – b)(Dy – ay) = (D – b)(D – a)y.
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Solution of Repeated Roots (1/2)
For a 2nd-order DE, the general solution when the
auxiliary equation has repeated roots can be
obtained using variation of parameters.
For the nth-order case, assuming that the auxiliary
equation of
any(n) + an–1y(n–1) + … + a1y' + a0y = 0
has k repeated roots r0. This means that the DE can
be expressed as (D – r0)k(D – r1) … (D – rn–k)y = 0.
Hence, the solution of (D – r0)ky = 0 will also be a
solution of the nth-order DE.
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Solution of Repeated Roots (2/2)
Since y1 = er0x is a solution of (D – r0)ky = 0, let
y(x) = u(x)er x.
0
Note that
(D – r0)[u(x)er x] = (Du(x))er x.
0
0
Applying the operator k times on y(x), we have
(D – r0)k[u(x)er x] = (Dku(x))er x for any u(x).
0
0
u(x)er x is a solution of the DE ↔ Dku(x) = 0.
0
Possible u(x) that meets this condition is a polynomial
with degree less than k.
→ y(x) = (c1+c2x+ … +ckxk–1)er x is a family of solutions.
0
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Example: (D2 + 6D + 13)2y = 0
The characteristic equation is
(r2 + 6r + 13)2 = 0.
By completing the square, we have
[(r + 3)2 + 4]2 = 0.
The roots are –3±2i of multiplicity 2. Hence the
general solution is
y(x) = e–3x(c1cos 2x+d1sin 2x) + xe–3x(c2cos 2x+d2sin 2x).
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Euler Equation†
Any linear differential equation of the form
n
n −1
d
y
d
y
dy
an x n n + an −1 x n −1 n −1 + L + a1 x + a0 y = 0,
dx
dx
dx
where the coefficients ai are constants, is called an
Euler equation.
Note that anxn = 0 at x = 0. Therefore, we focus on
solving the equation on (0, ∞).
†See problem 51 in section 2.1, page 113 in the textbook
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Example: ax2y″ + bxy′ + cy = 0
If x > 0, the substitution v = ln x transforms Euler
equation into a constant-coefficient linear equation:
d2y
dy
a 2 + (b − a) + cy = 0,
dv
dv
with independent variable v. If r1 and r2 are distinct
roots of the characteristic equation, we have a
general solution y(x) = c1xr1 + c2xr2.
Alternatively, we can substitute y = xm into an Euler
equation and find an alternative way to solve it.
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Alternative Solution to Euler Equation
Assume that y = xm is a solution, we have
a0 y = a0 x m
dy
= a1 x ⋅ mx m −1 = a1mx m
dx
2
d
y
a2 x 2 2 = a2 x 2 ⋅ m(m − 1) x m − 2 = a2 m(m − 1) x m
dx
...
a1 x
k
d
y
ak x k k
dx
= ak x k ⋅ m(m − 1)(m − 2)...(m − k + 1) x m − k
= ak m(m − 1)(m − 2)...(m − k + 1) x m .
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Example:
3
2
d
y
d
y
dy
x3 3 + 5x 2 2 + 7 x + 8 y = 0
dx
dx
dx
In this example, n = 3, we have the auxiliary equation
am(m–1)(m–2) + bm(m–1) + cm + d = 0,
where a = 1, b = 5, c = 7, and d = 8.
The auxiliary equation then becomes:
m3 + 2m2 + 4m + 8 = 0.
Therefore, m = –2, ±2i are distinct roots. Again,
complex exponentials can be removed by Euler’s
formula.
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Mechanical Vibrations
Many dynamic systems can be approximated using a
2nd-order linear DE with constant coefficients:
mx" + cx' + kx = F(t),
where F(t) is the external input force to the system,
and cx' is the damping force of the system.
The behavior of the system can be predicted by the
solution x(t) to an IVP x(t0) = x0, x'(t0) = x1 on an
interval containing t0.
mass
spring
dashpot (damper)
m
x(t)
x=0
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x>0
Free Undamped Motion
Hooke’s law gives the restoring force of a spring as
F = ks.
Newton’s 2nd law describes the motion of a mass as
m(d2x/dt2) = –k(s + x) + mg
= –kx + (mg – ks) = –kx.
l
l+ s
System of free undamped motion is
s
d 2x
2
m
+ ω0 x = 0, x(0) = x0 , x′(0) = x1 ,
x
2
equilibrium
dt
position
where ω0 =
the solution is
x(t) = A cos ω0t + B sin ω0t.
(k/m)½,
mg - ks = 0
m
motion
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Alternative Form of Solution
By applying trigonometric formula, we have:
B
x(t ) = C cos(ω0t − α ), C = A2 + B 2 , tan α =
A
ω0: natural frequency of the system.
x negative
x =0
t
x positive
The motion is called a simple harmonic motion.
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Aging Spring
In real world, the spring constant k usually varies as
the spring gets old. Replace k with k(t) = ke–αt, k > 0,
α > 0, we have a more realistic system model:
mx" + ke–αtx = 0.
→ Do you know how to solve this equation?
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Free Damped Motion
DE of free damped motion:
d 2x
dx
m 2 = −kx − β
dt
dt
→
d 2x
dx
2
+
2
λ
+
ω
0x =0
2
dt
dt
m
→ The roots of the auxiliary eq.:
2
m = −λ ± λ − ω
2
0
m
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Three Cases of Damped Motion
x
Case I: Over-damped
x (t ) = e
− λt
(c1e
λ2 −ω02 t
+ c2 e
− λ2 −ω02 t
t
)
x
Case II: Critically damped
x(t ) = e
− λt
t
(c1 + c2t )
x
Case III: Under-damped
x(t ) = e − λt (c1 cos ω02 − λ2 t
undamped
underdamped
t
+ c2 sin ω02 − λ2 t )
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Non-homogeneous Linear DE
To solve a non-homogeneous linear DE
any(n) + an-1y(n–1) + … + a2y" + a1y' + a0y = f(x),
we must do two things:
(1) Find the complementary function yc;
(2) Find any particular solution yp of the DE.
→ two methods:
Method of undetermined coefficients
Variation of parameters
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Undetermined Coefficients
The method of undetermined coefficients can be
applied under two conditions:
1. ai, i = 0, 1, …, n, are constants, and
2. f(x) is a linear combination of functions of the
following types:
P(x) = pnxn + pn–1xn–1 + … + p2x2 + p1x + p0,
P(x)eαx,
P(x)eαxsin βx,
P(x)eαxcos βx.
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Example: y" + 4y' – 2y = 2x2 – 3x + 6
By guessing, let yp = Ax2 + Bx + C, we have
yp' = 2Ax + B, and yp" = 2A.
Therefore:
yp" + 4yp' – 2yp
= 2A + 8Ax + 4B – 2Ax2 – 2Bx – 2C
= – 2Ax2 + (8A – 2B)x + (2A + 4B – 2C)
= 2x2 – 3x + 6.
→ yp = – x2 – (5/2)x – 9.
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Example: y" – y' + y = 2 sin 3x
By guessing, let yp = A cos 3x + B sin 3x, we have
yp' = – 3A sin 3x + 3B cos 3x, and
yp" = – 9A cos 3x – 9B sin 3x.
Therefore:
yp" – yp' + yp
= (– 9A – 3B + A) cos 3x + (– 9B + 3A + B) sin 3x
= 2 sin 3x.
→ yp = (6/73) cos 3x – (16/73) sin 3x.
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Example: yp by Superposition
Solve y" – 2y' – 3y = 4x – 5 + 6xe2x.
By superposition principle, we divide the problem into
two sub-problems, that is,
f(x) = f1(x) + f2(x),
where f1(x) = 4x – 5, and f2(x) = 6xe2x.
By guessing, let yp1 = Ax + B, and yp2 = Cxe2x + Ee2x.
Substitute yp = Ax + B + Cxe2x + Ee2x into the DE, we
have:
yp = –(4/3)x + (23/9) – 2xe2x – (4/3)Ee2x
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Example: A Glitch in the Method
Solve y" – 5y' + 4y = 8ex.
Simply guessing that yp = Aex and substituting yp into
the DE gives us 0 = 8ex. What went wrong?
Note, if the assumed form of yp is in the solution
space of yc (in this case yc = c1ex + c2e4x), then we
always get 0 = f(x).
Solution, let yp = Axex. Since the derivatives of yp
contains both the term Aex and Axex, it is a reasonable
guess for a particular solution.
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Summary of Two Cases (1/2)
Case I:
No functions in the assumed particular solution is a
solution of the associated homogeneous DE.
→ Substitute with yp = “the form of f(x)”.
f(x)
1. 1 (any constant)
2. x3－x＋1
3. sin4x, or cos4x
4. e5x
5. x2e5x
6. e3xsin4x
7. 5x2sin4x
8. xe3xcos4x
yp
A
Ax3＋Bx2＋Cx＋E
A cos 4x＋B sin 4x
Ae5x
(Ax2＋Bx＋C)e5x
Ae3xcos4x＋Be3xsin4x
(Ax2＋Bx＋C)cos4x＋(Ex2＋Fx＋G)sin4x
(Ax＋B)e3xcos4x＋(Cx＋E)e3xsin4x
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Summary of Two Cases (2/2)
Case II:
A function in the assumed particular solution is also a
solution of the associated homogeneous DE.
→ Substitute with yp = xn × “the form of solution for
f(x)”, where n is the smallest positive integer so that yp
is not in the solution space of yc.
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Examples:
Case I
y" – 8y′ + 25y = 5x3e–x – 7e–x
y" + 4y = x cos x
y" – 9y′ + 14y = 3x2 – 5 sin 2x + 7xe6x
Case II
y" – 2y′ + y = ex
y" + y = 4x + 10 sin x, y(π) = 0, y′(π) = 2
y" – 6y′ + 9y = 6x2 + 2 – 12 e3x
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Variation of Parameters (1/3)
Given a 2nd-order DE in standard form:
y" + P(x)y' + Q(x)y = f(x).
We can seek a particular solution using variation of
parameters by assuming that the solution has the
form
yp = u1(x)y1(x) + u2(x)y2(x),
where y1 and y2 form a fundamental set of solutions
on I of the associated homogeneous DE.
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Variation of Parameters (2/3)
Take the derivatives yp' and yp", and substitute them
into the DE, we have
y′p′ + P( x) y′p + Q( x) y p
= u1 [ y1′′ + P( x) y1′ + Q( x) y1 ] + u2 [ y′2′ + P( x) y2′ + Q( x) y2 ]
+ y1u1′′ + u1′ y1 + y2u2′′ + u 2′ y2 + P( x)[ y1u1′ + y2u2′ ] + y1′u1′ + y′2u′2
d
[ y1u1′ ] + d [ y2u2′ ] + P( x)[ y1u1′ + y2u2′ ] + y1′u1′ + y′2u2′
dx
dx
d
= [ y1u1′ + y2u2′ ] + P( x)[ y1u1′ + y2u2′ ] + y1′u1′ + y2′ u2′ = f ( x).
dx
=
We want this to be zero!
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Variation of Parameters (3/3)
The solution of the system
 y1u1′ + y2u 2′ = 0

 y1′u1′ + y2′ u ′2 = f ( x)
can be expressed in terms of determinants:
W
y f ( x) and
u1′ = 1 = − 2
W
W
u2′ =
W2 y1 f ( x)
=
,
W
W
where
y1
W=
y1′
y2
0
, W1 =
y2′
f ( x)
y2
y1
, W2 =
y2′
y1′
0
f ( x)
.
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Summary of the Method
To solve y" + P(x)y' + Q(x)y = f(x):
Find yc = c1y1(x) + c2y2(x).
Find u1 and u2 by integrating u1' = W1/W and u2' = W2/W.
A particular solution is yp = u1(x)y1(x) + u2(x)y2(x).
The general solution is y = yc + yp.
Note that there is no need to introduce any constants
when integrating u1' and u2' (why?).
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Examples: y" + y = tan x
Note that the complementary function is
yc(x) = c1 cos x + c2 sin x.
Compute W, W1, W2 and obtain
W = cos2x + sin2x = 1,
W1 = – sin 2x / cos x = cos x – sec x,
W2 = sin x,
→ u1' = cos x – sec x, and u2' = sin x.
→ u1 = sin x – ln |sec x + tan x|, and u2 = –cos x.
→ yp(x) = (sin x – ln |sec x + tan x|) cos x – cos x sin x.
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High-Order Variation of Parameters
For a linear nth-order DE
y(n) + Pn–1(x)y(n–1) + …+ P1(x)y' + P0(x)y = f(x),
if yc = c1y1 + c2y2 + … + cnyn is the complementary
function of the DE, then a particular solution is
yp = u1(x)y1(x) + u2(x)y2(x) + … + un(x)yn(x),
where uk′ = Wk/W, k = 1, 2, …, n and W is the
Wronskian of y1, y2, .., yn and Wk is the determinant
obtained by replacing the kth column of the
Wronskian with the column (0, 0, …, f(x))T.
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Forced Oscillation and Resonance
Now, consider the effect of external
force f(t) on the damped motion system:
d 2x
dx
m 2 = − kx − β
+ f (t )
dt
dt
→
moves the
ceiling of
the platform
d 2x
dx
2
+
2
λ
+
ω
0 x = F (t )
2
dt
dt
Note: ω0 is the natural frequency of the system.
m
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Transient and Steady-State Terms
When F(t) is a periodic function and λ > 0, the
solution is the sum of a non-periodic function xc(t) and
a periodic function xp(t). Moreover limt→∞ xc(t) = 0.
x
x
steady-state xp(t)
1
1
x( t) = transient
+ steady-state
t
t
transient
-1
-1
π/2
π/2
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Example: Transient/Steady State
The solution of the IVP
d 2x
dx
+ 2 + 2 x = 4 cos t + 2 sin t ,
2
dt
dt
x ( 0) = 0 ,
is
x′(0) = x1
x
x(t) = (x1 – 2)e–t sin t + 2 sin t,
transient steady-state
4
x1 = 7
x1 = 3
x1 = 0
2
t
-2
-4
x 1 = -3
π
2π
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Undamped Forced Motion
The solution of
d 2x
2
+
ω
0 x = F0 sin γ t ,
2
dt
is
x ( 0) = 0 ,
x′(0) = 0
F0
x(t ) = c1 cos ω0t + c2 sin ω0t + 2
sin γ t ,
2
ω0 − γ
where c1 = 0, c2 = –γF0 /ω0(ω02 –γ2).
F0
(−γ sin ω0t + ω0 sin γ t ) ,
→ x(t ) =
2
2
ω0 (ω0 − γ )
γ ≠ ω0 .
→ There is no transient term.
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Pure Resonance
In the previous example, when γ → ω0, the
displacement of the system become large as t → ∞.
x(t ) = lim F0
γ →ω0
− γ sin ω0t + ω0 sin γ t
ω0 (ω02 − γ 2 )
d
(−γ sin ω0t + ω0 sin γ t )
dγ
= F0 lim
γ →ω0
d
(ω03 − ω0γ 2 )
dγ
=
x
t
F0
F0
sin
ω
t
−
t cos ω0t.
0
2
2ω0
2ω0
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Tacoma Narrow Bridge, WA, USA
Opened in July 1, 1940, collapsed in Nov. 7, 1940.
The wind-blow frequency matched the natural frequency of
the bridge, which caused a pure resonance effect that
destroyed the bridge.
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Damping System of Taipei 101
Taipei 101 uses a 730-ton damping ball to stabilize
the building
75/82
Boundary-Value Problem
Solving a linear DE with y or its derivatives specified
at different points. For example, solve
y" + P(x)y' + Q(x)y = f(x),
subject to
y(a) = y0, y(b) = y1.
solutions of the DE
y
Such problems are called end-point
or boundary-value problems (BVP).
(b, y 1 )
(a, y 0 )
x
I
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Example: y" + 16y = 0
One can verify that y = c1 cos 4x + c2 sin 4x is a family
of solutions of y" + 16y = 0. What are the solutions of
the BVPs with
y(0) = 0, y(π/2) = 0?
y(0) = 0, y(π/8) = 0?
y(0) = 0, y(π/2) = 1?
y
1
c2 = 1
c2 =
1
2
c2 =
1
4
c2 = 0
x
(0, 0)
–1
c2 = −
1
2
(π /2, 0)
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Eigenvalue Problems
An eigenvalue problem in DE is a homogeneous BVP
such that the boundary conditions evaluate to 0 and
there is a parameter λ at the coefficient of y. That is
y" + p(x)y' + λq(x)y = 0, y(a) = 0, y(b) = 0.
The eigenvalue problem tries to find a λ (eigenvalue)
such that the BVP has a nontrivial solution. The nontrivial solution is then called an eigenfunction.
In last example, p(x) = 0, λ = 16, q(x) = 1, a = 0, b = π/2.
The 2nd BVP has only trivial solution.
→ what is the eigenvalue λ?
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Example: y" + λy = 0, y(0) = y(L) = 0
The problem can be solved by enumerating different
cases when λ = 0, λ < 0, and λ > 0.
(1) λ = 0, we have y″ = 0,
→ the general solution is y(x) = Ax + B.
→ y = 0 is the only solution for the BVP
→ λ = 0 is not an eigenvalue of the BVP
(2) λ < 0, let λ = –α2, α > 0, we have y″ – α2y = 0,
→ the general solution is y(x) = c1eαx + c2e–αx.
→ y = 0 is the only solution for the BVP
→ λ = 0 is not an eigenvalue of the BVP
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Example: y" + λy = 0, y(0) = y(L) = 0
(3) λ > 0, let λ = α2, α > 0, we have y″ + α2y = 0,
→ the solution is y(x) = c1cos(αx) + c2 sin(αx).
→ y(0) = 0 implies c1 = 0
→ y(L) = 0 implies sin(αL) = 0, or αL = nπ, n∈ Z
→ The BVP has infinitely many eigenvalues:
n 2π 2
λn = 2 ,
L
n = 1, 2, 3, ...
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The Deflection of a Uniform Beam
The deflection of a curve can be represented by a 4th-
order DE:
d4y
EI 4 = F ( x)
dx
flexural rigidity
A flexible beam
load per unit length
The deflection curve
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Boundary Conditions
Endpoint conditions of a flexible beam:
End of beam
x=0
Endpoint conditions
Simply supported
y=0
y″ = 0
Embedded
y=0
y′ = 0
Free end
y″ = 0
y″′ = 0
x=L
Embedded at both ends
x=0
x=L
Free at the right end
x=0
x=L
Supported at both ends
82/82