High Order Differential Equations3/82 Modeling with Linear 2nd-Order DE Linear 2nd-order DE are...
Transcript of High Order Differential Equations3/82 Modeling with Linear 2nd-Order DE Linear 2nd-order DE are...
High Order Differential
Equations
National Chiao Tung University
Chun-Jen Tsai
10/5/2011
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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.
Spring force:FS = –kx, k > 0.
Dashpot force:FR = –c(dx/dt), c > 0.
Since F = mx" = FS + FR, we have
m
spring
x(t)
x = 0 x > 0
mass
dashpot (damper)
.02
2
=++ kxdt
dxc
dt
xdm
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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
equationy" + p(x)y' + q(x)y = f(x)
has a unique solution on I that satisfies the initial
conditionsy(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.
andy'(x) = (c1 + c2)e
x + c2xex.
The initial conditions tell usc1 = 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 Iare 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 asy = 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
is non-singular.
.)()(
)()(
21
21
′′ ayay
ayay
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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
� For example, f = ex and g = xex, the Wronskian is
( ) .)()(
)()()(),(
xgxf
xgxfxgxfW
′′=
( ) ., 2x
xxx
xx
xxe
xeee
xeexeeW =
+=
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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
is called the Wronskian of the functions.
( )
)1()1(
2
)1(
1
21
21
21 ,,,
−−−
′′′=
n
n
nn
n
n
n
fff
fff
fff
fffW
L
MMM
L
L
L
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Example: x3y(3) – x2y" + 2xy' – 2y = 0
� Given three solutions y1(x) = x, y2(x) = x ln x, andy3(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
The particular solution is y(x) = x – 3x ln x + 2x2.
.0
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2ln11
ln 2
≠=+= x
x
xx
xxxx
W
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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
Since c1 = c2 = … = cn = 0 is also a solution, W ≡ 0 on I.
.
0
0
0
)()()(
)()()(
)()()(
2
1
)1()1(
2
)1(
1
21
21
=
′′′
−−−
MM
L
MMM
L
L
n
n
n
nn
n
n
c
c
c
xyxyxy
xyxyxy
xyxyxy
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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
has only the trivial solution ∀x ∈ I → the system is non-singular. Therefore W ≠ 0.
.
0
0
0
)()()(
)()()(
)()()(
2
1
)1()1(
2
)1(
1
21
21
=
′′′
−−−
MM
L
MMM
L
L
n
n
n
nn
n
n
c
c
c
xyxyxy
xyxyxy
xyxyxy
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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:
.)(
)(2
1
)(
12 dxxy
exyy
dxxP
∫∫
=
−
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Example: x2y" – 3xy' + 4y = 0
� Solution:y1 = x2 is a known solution.
→
The general solution is y = c1x2+c2x
2lnx.
xx
x
dxx
xeedxx
exy
yx
yx
y
xxdxxdx
ln
043
2
2
3ln/3
4
/3
2
2
2
3
=
=
==∫←∫
=
=+′−′′
∫
∫
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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–1r
n–1erx + … + a2r2erx + a1rerx + a0e
rx = 0.
Therefore, y = erx is a solution if
anrn + an–1r
n–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 = er1x and
y2 = er2x form a fundamental set of solutions.
The general solutions is
� Case II, b2 – 4ac = 0:
r has one real root r1 and y1 = er1x. By variation of
parameters (or reduction of order), the 2nd solution of the DE is y2 = xer1x. The general solution is
.21
21
xrxrececy +=
.11
21
xrxrxececy +=
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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:
� 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
.)(
2
)(
1
xixiececy
βαβα −+ +=
).sincos( 21 xcxceyx ββα +=
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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), andy' = (–c1/2 + 2c2)e
–x/2cos 2x + (–c2/2 – 2c1)e–x/2sin 2x
→ y = e–x/2 (– cos 2x + ¾ sin 2x)
y → 0, as x → ∞.x1
y
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–1r
n–1 + … + a2r2 + a1r + a0 = 0.
The general solution of the DE is:
Case I (no repeated roots):
Case II (with repeated roots):
....21
21
xr
n
xrxr necececy +++=
....... 2111
1
1
21
xr
n
xr
k
xrk
k
xrxr knececexcexcecy −++++++= +−
solution form ofrepeated roots
solution form ofdistinct 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–1D
n–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 asD2y+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 ofany
(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)er0x.
Note that(D – r0)[u(x)er0x] = (Du(x))er0x.
Applying the operator k times on y(x), we have
(D – r0)k[u(x)er0x] = (Dku(x))er0x for any u(x).
u(x)er0x is a solution of the DE ↔ Dku(x) = 0.
Possible u(x) that meets this condition is a polynomial
with degree less than k.
→ y(x) = (c1+c2x+ … +ckxk–1)er0x is a family of solutions.
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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
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, ∞).
,0011
11
1 =++++−
−−
− yadx
dyxa
dx
ydxa
dx
ydxa
n
nn
nn
nn
n L
†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:
with independent variable v. If r1 and r2 are distinct
roots of the characteristic equation, we have a general solution y(x) = c1x
r1 + c2xr2.
� Alternatively, we can substitute y = xm into an Euler
equation and find an alternative way to solve it.
,0)(2
2
=+−+ cydv
dyab
dv
yda
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Alternative Solution to Euler Equation
� Assume that y = xm is a solution, we have
.)1)...(2)(1(
)1)...(2)(1(
...
)1()1( 2
22
22
22
2
1
1
11
00
m
k
kmk
k
k
kk
k
mm
mm
m
xkmmmma
xkmmmmxa
dx
ydxa
xmmaxmmxadx
ydxa
mxamxxadx
dyxa
xaya
+−−−=
+−−−⋅=
−=−⋅=
=⋅=
=
−
−
−
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Example:
� 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.
08752
22
3
33 =+++ y
dx
dyx
dx
ydx
dx
ydx
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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.
m
spring
x(t)
x = 0 x > 0
mass dashpot (damper)
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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.
System of free undamped motion is
where ω0 = (k/m)½, the solution is
x(t) = A cos ω0t + B sin ω0t.
,)0(,)0(,0 10
2
02
2
xxxxxdt
xd=′==+ ω
equilibrium
position
s
l
m
mmg - ks = 0
motion
x
l + s
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Alternative Form of Solution
� By applying trigonometric formula, we have:
� The motion is called a simple harmonic motion.
A
BBACtCtx =+=−= ααω tan,),cos()( 22
0
x negative
x = 0
x positive
t
ω0: natural frequency of the system.
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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:
→
→ The roots of the auxiliary eq.:
dt
dxkx
dt
xdm β−−=
2
2
02 2
02
2
=++ xdt
dx
dt
xdωλ
2
0
2 ωλλ −±−=mm
m
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Three Cases of Damped Motion
� Case I: Over-damped
� Case II: Critically damped
� Case III: Under-damped
)()(20
220
2
21
tttececetx
ωλωλλ −−−− +=
)()( 21 tccetxt += −λ
)sin
cos()(
22
02
22
01
tc
tcetxt
λω
λωλ
−+
−= −
t
x
x
t
x
t
undampedunderdamped
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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, and2. f(x) is a linear combination of functions of the
following types:
P(x) = pnxn + pn–1x
n–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 haveyp' = 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 haveyp' = – 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 + c2e
4x), 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)”.
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
1. 1 (any constant)
2. x3-x+1
3. sin4x, or cos4x
4. e5x
5. x2e5x
6. e3xsin4x
7. 5x2sin4x
8. xe3xcos4x
yp
f(x)
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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
formyp = 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
[ ] [ ][ ]
[ ] [ ] [ ]
[ ] [ ] ).()(
)(
)(
)()()()(
)()(
221122112211
221122112211
2211221122221111
22221111
xfuyuyuyuyxPuyuydx
d
uyuyuyuyxPuydx
duy
dx
d
uyuyuyuyxPyuuyyuuy
yxQyxPyuyxQyxPyu
yxQyxPy ppp
=′′+′′+′+′+′+′=
′′+′′+′+′+′+′=
′′+′′+′+′+′+′′+′+′′+
+′+′′++′+′′=
+′+′′
We want this to be zero!
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Variation of Parameters (3/3)
� The solution of the system
can be expressed in terms of determinants:
and
where
W
xfy
W
Wu
)(211 −==′
=′′+′′
=′+′
)(
0
2211
2211
xfuyuy
uyuy
,)(12
2W
xfy
W
Wu ==′
.)(
0,
)(
0,
1
1
2
2
2
1
21
21
xfy
yW
yxf
yW
yy
yyW
′=
′=
′′=
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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 externalforce f(t) on the damped motion system:
→
Note: ω0 is the natural frequency of the system.
)(2
2
tfdt
dxkx
dt
xdm +−−= β
)(2 2
02
2
tFxdt
dx
dt
xd=++ ωλ
m
moves theceiling ofthe platform
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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.
1
-1
transient
steady-state
x
t
1
-1
x( t) = transient
+ steady-state
x
t
π/2 π/2
xp(t)
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Example: Transient/Steady State
� The solution of the IVP
isx(t) = (x1 – 2)e–t sin t + 2 sin t,
12
2
)0(,0)0(,sin2cos422 xxxttxdt
dx
dt
xd=′=+=++
transient steady-state
4
- 4
x1
= 7
x
t
- 2
2
x1
= 3
x1
= 0
x1
= -3
2ππ
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Undamped Forced Motion
� The solution of
is
where c1 = 0, c2 = –γF0 /ω0(ω02 –γ2).
→
→ There is no transient term.
0)0(,0)0(,sin0
2
02
2
=′==+ xxtFxdt
xdγω
,sinsincos)(22
0
00201 t
Ftctctx γ
γωωω
−++=
.,)sinsin()(
)( 00022
00
0 ωγγωωγγωω
≠+−−
= ttF
tx
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Pure Resonance
� In the previous example, when γ → ω0, the displacement of the system become large as t → ∞.
)(
)sinsin(
lim
)(
sinsinlim)(
2
0
3
0
00
0
22
00
000
0
0
γωωγ
γωωγγ
γωω
γωωγ
ωγ
ωγ
−
+−
=
−
+−=
→
→
d
d
ttd
d
F
ttFtx
.cos2
sin2
0
0
002
0
0 ttF
tF
ωω
ωω
−=
t
x
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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.
75/82
Damping System of Taipei 101
� Taipei 101 uses a 730-ton damping ball to stabilize the building
76/82
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 toy(a) = y0, y(b) = y1.
Such problems are called end-pointor boundary-value problems (BVP).
y
x
I
(a, y0)
(b, y1)
solutions of the DE
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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
x
( /2, 0)(0, 0)
1
–1
π
02 =c
21
2 −=c
12 =c
41
2 =c
21
2 =c
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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 non-
trivial 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:
...,3,2,1,2
22
== nL
nn
πλ
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The Deflection of a Uniform Beam
� The deflection of a curve can be represented by a 4th-order DE:
)(4
4
xFdx
ydEI =
load per unit lengthflexural rigidity
A flexible beam The deflection curve
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Boundary Conditions
� Endpoint conditions of a flexible beam:
y″′ = 0y″ = 0Free end
y″ = 0y = 0Simply supported
y′ = 0y = 0Embedded
Endpoint conditionsEnd of beam
x = 0 x = L x = 0 x = L x = 0 x = L
Embedded at both ends Free at the right end Supported at both ends