UNENE Math Refresher Course
Ordinary Differential Equations
1
Ordinary Differential Equations — Introduction © Wei-Chau Xie
Example – Vibration of a Mass-Spring System
Consider a mass-spring system as shown
m
x(t)
F(t)k
m
x, x, x
F(t)kxm
x(t)
F(t)k
x(t) is the displacement of mass m from its equilibrium position.
The spring force is kx.
Newton’s Second Law requires ma =∑
F :
m · x(t) = F(t) − kx =⇒ mx + kx = F(t)
This equation is a differential equation.
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Ordinary Differential Equations — Introduction © Wei-Chau Xie
Definition
A differential equation is an equation that contains derivatives of dependent
variables with respect to one or more independent variables.
A differential equation is called an ordinary differential equation (ODE)
if there is only one independent variable and only ordinary derivatives (not
partial derivatives) are involved.
Let x be an independent variable and y be a dependent variable. In general, an
ODE is of the form
F(
x, y,dy
dx, . . . ,
dny
dxn
)
= 0
The order of a differential equation is the order of the highest derivative
appearing in the differential equation.
If y and its various derivatives y′, y′′, . . . appear linearly in the equation, it
is a linear differential equation; otherwise, it is nonlinear.
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Ordinary Differential Equations — Introduction © Wei-Chau Xie
Definition – Linear Ordinary Differential Equations
The general form of an nth-order linear ordinary differential equation is
an(x)dny
dxn+ an−1(x)
dn−1y
dxn−1+ · · · + a1(x)
dy
dx+ a0(x) y = f(x)
If a0(x), a1(x), . . . , an(x) are constants, the ordinary differential equation is said
to have constant coefficients. Otherwise it is said to have variable coefficients.
Example
.d2y
dx2+ 2 y = sin 3x Second-order linear with constant coefficients
.
(dy
dx
)2+ 2 y = sin 3x First-order nonlinear
. x2 d2y
dx2+ 2x
dy
dx− y = ln x Second-order linear with variable coefficients
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Ordinary Differential Equations — Introduction © Wei-Chau Xie
Definition – Partial Differential Equations
If the dependent variable u is a function of more than one independent variable,
say x1, x2, . . . , xm, an equation involving the variables x1, x2, . . . , xm, u and
various partial derivatives of u with respect to x1, x2, . . . , xm is called a partial
differential equation (PDE).
Example
.∂2u
∂x2+
∂2u
∂y2= f(x, y) Poisson’s equation in two-dimensions
.∂2u
∂x2=
1
c
∂u
∂tHeat equation in one-dimension
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Ordinary Differential Equations — Introduction © Wei-Chau Xie
Definition – General Solution
The general solution of an ordinary differential equation contains all possible
solutions. In general, an nth-order ordinary differential equation will contain n
arbitrary constants in its general solution.
The solution of an ordinary differential equation satisfying a set of conditions is
a particular solution.
Example
Consider the differential equation y ′′ + y = 0.
It can be shown that
y = C1 sin x + C2 cos x
is a solution for any constants C1 and C2 =⇒ general solution
y1 = sin x is a particular solution (C1 = 1, C2 = 0)
y2 = cos x is a particular solution (C1 = 0, C2 = 1)
y3 = 2 sin x−3 cos x is a particular solution (C1 = 2, C2 = −3)6
Ordinary Differential Equations — Introduction © Wei-Chau Xie
Definition – Initial Value Problem
If a differential equation is required to satisfy conditions on the dependent
variable and its derivatives specified at one value of the independent variable,
these conditions are called initial conditions (IC) and the problem is called an
initial value problem (IVP).
Definition – Boundary Value Problem
If a differential equation is required to satisfy conditions on the dependent
variable and possibly its derivatives specified at two or more values of the
independent variable, these conditions are called boundary conditions (BC) and
the problem is called a boundary value problem (BVP).
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Ordinary Differential Equations — Introduction © Wei-Chau Xie
Example – Initial Value Problem
Consider the differential equation y′′ + y = 0.
If we impose the initial conditions y(0)= 2, y′(0) = −1, determine the solution
of this IVP.
We know that y(x) = C1 sin x + C2 cos x is the general solution.
. y′(x) = C1 cos x − C2 sin x
Applying the initial conditions:
. y(0) = C1 sin 0 + C2 cos 0 = C2 = 2
. y′(0) = C1 cos 0 − C2 sin 0 = C1 = −1
The solution of this IVP is
y(x) = − sin x + 2 cos x
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Ordinary Differential Equations — Introduction © Wei-Chau Xie
Example – Boundary Value Problem
Consider the differential equation y′′ + y = 0.
If we impose the boundary conditions y(0)= 2, y( π4 ) = −1, determine the
solution of this BVP.
We know that y(x) = C1 sin x + C2 cos x is the general solution.
Applying the boundary conditions:
. y(0) = C1 sin 0 + C2 cos 0 = C2 = 2
. y( π4 ) = C1 sin π
4 + C2 cos π4 = 1√
2(C1 + C2) = −1 =⇒ C1 = −2−
√2
The solution of this BVP is
y(x) = −(2+√
2) sin x + 2 cos x.
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First-Order Differential Equations — Method of Separation of Variables © Wei-Chau Xie
The Method of Separation of Variables
Consider the first-order DE:dy
dx= f(x) φ( y)
Case 1: If φ( y)=/ 0, divide the DE by φ( y):
1
φ( y)dy
︸ ︷︷ ︸
= f(x) dx︸ ︷︷ ︸
Contains y only Contains x only
General solution can be obtained by integrating both sides∫
1
φ( y)dy =
∫
f(x) dx + C C = arbitrary constant
Case 2: If φ( y) = 0 =⇒ y = y0 is a root of the equation
. =⇒ y = y0 is a solution.
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First-Order Differential Equations — Method of Separation of Variables © Wei-Chau Xie
Example
Solvedy
dx= 2x y
Case 1: y = 0 is a solution.
Case 2: y =/ 0, divide the DE by y and integrate both sides
.
∫1
ydy =
∫
2x dx + C
. ln∣∣ y
∣∣ = x2 + C =⇒
∣∣ y
∣∣ = ex2+C =⇒ y = ±eC · ex2
Let A = ±eC. The general solution is
y = Aex2or y = Cex2
C = arbitrary real constant
Case 1: y = 0 is included in the general solutionwith C = 0.
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First-Order Differential Equations — Method of Separation of Variables © Wei-Chau Xie
Example
Solve the IVPdy
dx=
6x5 + 2x + 1
cos y + e y, y(0) = π
The DE is variable separable
.
∫
(cos y + e y) dy =∫
(6x5 + 2x + 1) dx + C
. ∴ sin y + e y = x6 + x2 + x + C General solution
Using the initial condition: x = 0, y = π :
sin π + eπ = C =⇒ C = eπ
The solution of the IVP is
sin y + e y = x6 + x2 + x + eπ Particular solution
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First-Order Differential Equations — Method of Separation of Variables © Wei-Chau Xie
Example
Solvedr
dϕ=
sin ϕ + e2r sin ϕ
3er + er cos ϕ
dr
dϕ=
sin ϕ(1 + e2r)
er(3 + cos ϕ)Variable separable
∴
∫er
1 + e2rdr =
∫sin ϕ
cos ϕ + 3dϕ + C
∫1
1 + e2rd(er) = −
∫1
cos ϕ+3d(cos ϕ+3) + C
∫1
1 + x2dx = tan−1 x
∴ tan−1er = − ln(cos ϕ+3) + C General solution
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First-Order Differential Equations — Linear First-Order Equations © Wei-Chau Xie
Linear First-Order Equations
Linear first-order equations are of the formdy
dx+ P(x)· y = Q(x)
The roles of x and y may be exchanged to give linear first-order DE of the form
dx
dy+ P( y)·x = Q( y)
Linear First-Order Equations
dy
dx+ P(x)· y = Q(x) =⇒ y = e
−∫
P(x)dx[ ∫
Q(x)e
∫
P(x)dxdx + C]
dx
dy+ P( y)·x = Q( y) =⇒ x = e
−∫
P( y)dy[ ∫
Q( y)e
∫
P( y)dydy + C]
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First-Order Differential Equations — Linear First-Order Equations © Wei-Chau Xie
Example
Solvedy
dx= 5 y + 3e5x
Rewrite the DE in the standard formdy
dx+ P(x)· y = Q(x):
.dy
dx− 5 · y = 3e5x, P(x) = −5, Q(x) = 3e5x
.
∫
P(x) dx =∫
−5 dx = −5x, e
∫
P(x)dx = e−5x
.
∫
Q(x)e
∫
P(x)dx dx =∫
3e5x · e−5x dx = 3x
. y = e−∫
P(x)dx[ ∫
Q(x)e
∫
P(x)dx dx + C]
= e5x(3x + C)
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First-Order Differential Equations — Linear First-Order Equations © Wei-Chau Xie
Example
Solve y′ =1
x − 3 y
.dy
dx=
1
x − 3 y=⇒
dx
dy= x − 3 y
.dx
dy− 1 · x = −3 y, P( y) = −1, Q( y) = −3 y
.
∫
P( y) dy =∫
−1 · dy = − y, e
∫
P( y)dy = e−y
.
∫
Q( y)e
∫
P( y)dydy =∫
−3 y · e−y dy = −3− y − 1
(−1)2e−y = 3( y + 1)e−y
. x = e−∫
P( y)dy[ ∫
Q( y)e
∫
P( y)dydy + C]
. = e y[3( y + 1)e−y + C] = 3( y + 1) + Ce y
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Applications of First-Order Differential Equations — Heating and Cooling © Wei-Chau Xie
Newton’s Law of Cooling
The rate of change in the temperature T(t), dT/dt, of a body in a medium of
temperature Tm is proportional to the temperature difference between the body
and the medium, i.e.,
dT
dt= −k(T − Tm),
where k>0 is a constant of proportionality.
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Applications of First-Order Differential Equations — Heating and Cooling © Wei-Chau Xie
Example – Newton’s Law of Cooling
A thermometer is removed from a room where the temperature is 20◦C to the
outside where the temperature is −10◦C. After 1 minute, the thermometer
reads 10◦C.
1. What is the reading of the thermometer at t = 2 minutes?
2. How long will it take for the thermometer to read −5◦C?
By Newton’s Law of CoolingdT
dt= −k(T − Tm)
T: Temperature of the thermometer at time t
Tm: Outside temperature, Tm = −10◦C
The DE can be written as
.dT
dt+ k·T = kTm, P(t) = k, Q(t) = kTm Linear first-order
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Applications of First-Order Differential Equations — Heating and Cooling © Wei-Chau Xie
.
∫
P(t) dt = kt, e
∫
P(t)dt = ekt,
∫
Q(t)e
∫
P(t)dtdt =∫
kTm · ekt dt = Tm ekt
. T = e−∫
P(t)dt[ ∫
Q(t) e
∫
P(t)dtdx + C]
= e−kt(Tm ekt + C)
. ∴ T = −10 + C e−kt
At t = 0, T = 20◦C: 20 = −10 + Ce−k·0 =⇒ C = 30
At t = 1, T = 10◦C: 10 = −10 + 30e−k·1 =⇒ e−k = 23
=⇒ k = − ln23
. ∴ T(t) = −10 + 30·et· ln 23 = −10 + 30·
(23
)t
At t = 2: T(2) = −10 + 30·(2
3
)2 = 103
= 3.33◦C
When T = −5: −5 = −10 + 30·(2
3
)t=⇒
(23
)t = 16
=⇒ t· ln23
= ln16
t =ln
16
ln23
= 4.4 min
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Applications of First-Order Differential Equations — Radioactive Decay Law © Wei-Chau Xie
Example – Radioactive Decay Law
Suppose that, in a sample of radioactive materials, there are N(t) nuclei which
have not decayed at time t.
During the time interval from t to t+1t, the number of radioactive nuclei
which will decay is proportional to N(t) and 1t, i.e.
N(t) − N(t+1t) = λ·N(t)·1t
where λ is the constant of proportionality called the decay constant.
Divide by 1t and take the limit 1t→0
−N(t+1t) − N(t)
1t= λN(t) =⇒ −
dN
dt= λN
The differential equation is variable separable∫
1
NdN =
∫
−λ dt + C =⇒ ln N = −λt + C =⇒ N(t) = C e−λt
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Applications of First-Order Differential Equations — Radioactive Decay Law © Wei-Chau Xie
Initial condition: t = 0, N(0) = N0 =⇒ C = N0
N(t) = N0 e−λt Radioactive Decay Law
Half-Life t1/2 is the length of time it takes for the radioactivity of a
radioisotope to decrease by a factor of two, i.e.
N(t1/2) = N0 e−λt1/2 = 1
2N0
e−λt1/2 = 1
2=⇒ − λt1/2 = ln
12
= − ln 2
∴ t1/2 =ln 2
λor λ =
ln 2
t1/2
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Applications of First-Order Differential Equations — Motion of a Particle © Wei-Chau Xie
Newton’s Second Law
The product of the mass of an object and its acceleration is equal to the sum of
forces applied on the object, i.e., ma =∑
F.
Example
A motor boat is travelling at a speed v0 and the motor is turned off at time
t = 0. The resistance from air is proportional to the velocity, i.e., Rair = −αv,
and the resistance from water is proportional to the square of the velocity, i.e.,
Rwater = −βv2, in which α and β are positive constants. Determine the velocity
of the boat for t>0.
αv
x,v,a
βv2
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Applications of First-Order Differential Equations — Motion of a Particle © Wei-Chau Xie
Apply Newton’s Second Law to the boat:
ma =∑
F : mdv
dt= −α v − β v2 = −β v (v + k), k =
α
β
.
∫dv
v (v + k)= −
β
m
∫
dt + C Variable separable
.1
k
∫ ( 1
v−
1
v + k
)
dv = −β
mt + C =⇒
1
k[ ln v − ln(v + k)] = −
β
mt + C
.1
kln
v
v + k= −
β
mt + C =⇒
v
v + k= C e
−kβm t
At t = 0, v = v0 =⇒v0
v0 + k= C · e0
∴ C =v0
v0 + k=
v0
v0 +α
β
=β v0
α + β v0
v(t) =C k e
−αtm
1 − C e−αt
m
=
β v0
α + β v0
·α
β· e
−αtm
1 −β v0
α + β v0
· e−αt
m
=α v0 e
−αtm
(α+βv0) − βv0 e−αt
m
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Applications of First-Order Differential Equations — Mixture Problem © Wei-Chau Xie
Example – Mixture Problem
Initially 50 kg of salt is dissolved in a large tank holding 1000 L of water. A brine
solution with a concentration of 0.2 kg/L is pumped into the tank at the rate of
10 L/min, and the well-stirred solution is pumped out at the same rate.
1. Determine the amount of salt in the tank at any time.
2. How much salt is present after 50 minutes?
3. How much salt is present after a long long time?
OutflowQout , cout
InflowQin , cin
Volume = 1000
Amount of salt = s(t)
Concentration c(t)=s(t)
1000
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Applications of First-Order Differential Equations — Mixture Problem © Wei-Chau Xie
s(t)=Amount of salt in the tank at time t =⇒ Salt concentration c(t) =s(t)
1000
Inflow
Flow rate Qin = 10 (L/min)
Salt concentration cin = 0.2 (kg/L)
Outflow
Flow rate Qout = 10 (L/min)
Salt concentration cout = c(t) =s(t)
1000(kg/L)
Consider time interval 1t:
. 1s = Change of amount of salt in 1t
. = (Amount of salt in inflow in 1t) − (Amount of salt in outflow in 1t)
. = cin Qin 1t − cout Qout 1t
. = 0.2 × 10 × 1t −s
1000× 10 × 1t
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Applications of First-Order Differential Equations — Mixture Problem © Wei-Chau Xie
Divide by 1t and take the limit 1→0 =⇒ds
dt= 2 −
s
100
.ds
dt+
1
100·s = 2, P(t) =
1
100, Q(t) = 2 Linear first-order
.
∫
P(t) dt =t
100,
∫
Q(t)e
∫
P(t)dtdt =∫
2 · et
100 dt = 200 et
100
. s(t) = e−∫
P(t)dt[ ∫
Q(t) e
∫
P(t)dtdx + C]
. = e− t
100(
200 et
100 + C)
= 200 + C e− t
100
C is determine from the initial condition: t = 0, s = 50
50 = 200 + C e0 =⇒ C = −150 =⇒ s(t) = 200 − 150 e− t
100
At t = 50 min: s(50) = 200 − 150 e− 50
100 = 109.02 kg
When t → ∞: limt→∞
s(t) = limt→∞
(
200 − 150 e− t
100)
= 200 kg
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Applications of First-Order Differential Equations — Tank Problem © Wei-Chau Xie
Example – Tank Problem
The moderator in a CANDU unit has been poisoned with gadolinium nitrate
Gd(NO3)3 due to activation of shut-down system two (SDS-2). Referring to
the schematic diagram, it is given that the volume of heavy water D2O in
the calandria is V = 220, 000 L, with the initial concentration of gadolinium
nitrate being C0 = 12 mg/L, the flow rate through the ion exchange columns is
Q = 2, 300 L/min, and the removal efficiency is α = 95%. Determine how long
it will take to get the gadolinium nitrate concentration below 0.01 mg/L in order
to begin plant startup activities.
Initial Gd(NO3)3
concentration
C0=12 mg/L
Flow rate
Q=2,300 L/min
Calandria
V=220,000 LIon Exchanger
Removal efficience α
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Applications of First-Order Differential Equations — Tank Problem © Wei-Chau Xie
At time t, the Gd(NO3)3 concentration is C(t).
Amount of Gd(NO3)3 in calandria is C(t)·V.
During time interval 1t,
Outflow = Q1t
Amount of Gd(NO3)3 removed = C(t)·Q1t·α
At time t+1t, the Gd(NO3)3 concentration is C(t+1t).
Amount of Gd(NO3)3 in calandria is C(t+1t)·V.
∴ (Amount of Gd(NO3)3 at t) − (Amount of Gd(NO3)3 at t+1t)
= Amount of Gd(NO3)3 removed
∴ C(t)·V − C(t+1t)·V = C(t)·Q1t·α
Divide by 1t and take the limit 1t→0
−VC(t+1t) − C(t)
1t= αQ C(t) =⇒ V
dC(t)
dt= −αQ C(t)
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Applications of First-Order Differential Equations — Tank Problem © Wei-Chau Xie
∫1
CdC = −
∫1
τdt + D, τ =
V
αQVariable separable
ln C = −t
τ+ D =⇒ C(t) = D e−t/τ
When t = 0, C(0) = C0 =⇒ D = C0
∴ C(t) = C0 e−t/τ =⇒ t = −τ lnC(t)
C0
= −V
αQln
C(t)
C0
When C(t) = 0.01 mg/L
t = −220, 000
0.95×2, 300ln
0.01
12= 713.87 min = 11 hr 54 min
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Applications of First-Order Differential Equations — Chemical Reaction © Wei-Chau Xie
Law of Mass Action
The Law of Mass Action states that the rate of reaction is directly proportional
to the product of the masses of reactants.
Example – Chemical Reaction
A chemical reaction may be expressed as
A + B → C
in which M grams of chemical A reacts with N grams of chemical B to
generate M+N grams of compound C.
Suppose we add a grams of chemical A to b grams of chemical B.
At time t, x(t) grams of compound C is produced, in whichM
M+Nx(t) grams
of chemical A andN
M+Nx(t) grams of chemical B are used.
The amounts of chemicals A and B left are, respectively,
a −M
M+Nx(t), b −
N
M+Nx(t)
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Applications of First-Order Differential Equations — Chemical Reaction © Wei-Chau Xie
By the Law of Mass Action, the rate of reaction satisfies
.dx
dt= k
(
a −M
M+Nx)(
b −N
M+Nx)
. = k ·M
M+N
[a (M+N)
M− x
]
·N
M+N
[b (M+N)
N− x
]
= k (x − α) (x − β)
where k = k ·M
M+N·
N
M+N, α =
a (M+N)
M, β =
b (M+N)
N
. ∴
∫dx
(x−α)(x−β)= k dt + C =⇒
1
α−β
∫ ( 1
x−α−
1
x−β
)
dx = k t + C
.1
α−βln
∣∣∣∣
x−α
x−β
∣∣∣∣= k t + C =⇒
x−α
x−β= C ek t k = k (α−β)
C is determined by the initial condition t = 0, x(0) = 0:
0−α
0−β= C e0 =⇒ C =
α
β=⇒
x−α
x−β=
α
βek t
∴ x(t) =αβ (1 − e−k t)
α − β e−k t
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Applications of First-Order Differential Equations — Chemical Reaction © Wei-Chau Xie
Numerical Example – Chemical Reaction
Suppose M = 1, N = 4, a = 50, b = 32. If 30 grams of compound C are
formed in 10 minutes
1. Determine the amount of compound C at any time.
2. How much of compound C is produced in 15 minutes?
3. What happens after a long long time?
. x(10) = 30, α =a (M+N)
M=
50 (1+4)
1= 250
. β =b (M+N)
N=
32 (1+4)
4= 40
. x(t) =αβ (1 − e−k t)
α − β e−k t=
250·40 (1 − e−k t)
250 − 40 e−k t=
1000 (1 − e−k t)
25 − 4 e−k t
At t = 10, x(10) =1000 (1 − e−10 k)
25 − 4 e−10 k= 30 =⇒ k = 1
10ln
8825
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Applications of First-Order Differential Equations — Chemical Reaction © Wei-Chau Xie
x(t) =1000
(
1 − e− t
10 ln8825
)
25 − 4 e− t
10 ln8825
At time t = 15: x(15) =1000
(
1 − e−3
2 ln8825
)
25 − 4 e−3
2 ln8825
= 34.79
When t → ∞: limt→∞
x(t) =1000
(
1 − e− t
10 ln8825
)
25 − 4 e− t
10 ln8825
=1000
25= 40
After a long long time, 40 grams of compound C is produced, which uses 8
grams of chemical A and all 32 grams of chemical B. Since there is no chemical
B left, the chemical reaction stops.
33
Linear Ordinary Differential Equations © Wei-Chau Xie
Linear Ordinary Differential Equations
In general, an nth-order linear ODE is of the form
an(x)dny
dxn+ an−1(x)
dn−1y
dxn−1+ · · · + a1(x)
dy
dx+ a0(x) y = F(x)
If a0, a1, . . . , an are constants, the ODE has constant coefficients.
Using the D-operator, the general nth-order linear ODE can be rewritten as
[
an(x)Dn + an−1(x)Dn−1 + · · · + a1(x)D + a0(x)]
︸ ︷︷ ︸y = F(x), Dy ≡
d y
dx
φ(D)
. ∴ φ(D) y = F(x)
34
Linear Ordinary Differential Equations © Wei-Chau Xie
Procedure for Finding the General Solution
φ(D) y = F(x), φ(D) = an Dn + an−1 Dn−1 + · · · + a1 D + a0
1. Find a particular solution yP(x) of the differential equation:
φ(D) yP = F(x) Particular solution
where the subscript P stands for particular solution.
2. Find the general solution yC(x) of the complementary differential equation:
φ(D) yC = 0 Complementary solution
where the subscript C stands for complementary solution.
3. Add yP(x) and yC(x) to obtain the general solution y(x):
y(x) = yC(x) + yP(x) General solution
35
Linear Ordinary Differential Equations — Complementary Solutions © Wei-Chau Xie
Procedure for Finding the Complementary Solution
For the nth-order linear differential equation
φ(D) y = F(x), φ(D) = anDn+an−1Dn−1+ · · · +a1D+a0
set the right-hand side to zero to obtain the complementary DE
φ(D) y = 0
Replace D by λ to obtain the characteristic equation
φ(λ) = 0 or an λn + an−1 λn−1 + · · · + a1 λ + a0 = 0
Solve the characteristic equation, which is an algebraic equation, to find the
characteristic numbers (roots) λ1, λ2, . . . , λn.
Write the complementary solution yC using the characteristic numbers.
36
Linear Ordinary Differential Equations — Complementary Solutions © Wei-Chau Xie
Forms of Complementary Solution
Having obtained the characteristic numbers, the complementary solution can be
easily written.
λ = a =⇒ yC = C eax
λ = α± iβ =⇒ yC = eαx(A cos βx + B sin βx)
If the characteristic equation φ(λ)=0 has a p-fold root λ = a
λ = a, a, . . . , a =⇒ yC =(
C0 + C1x + · · · + Cp−1x p−1)
eax
︸ ︷︷ ︸ ︸ ︷︷ ︸
p times polynomial of degree p−1
If φ(λ)=0 has a pair of complex roots λ = α±iβ of p-fold
λ = α±iβ, . . . , α±iβ =⇒ yC = eαx[(
A0 + A1x + · · · + Ap−1x p−1)
cos βx
+(
B0 + B1x + · · · + Bp−1x p−1)
sin βx ]︸ ︷︷ ︸ ︸ ︷︷ ︸
p times polynomials of degree p−1
37
Linear Ordinary Differential Equations — Complementary Solutions © Wei-Chau Xie
Example
Solve y′′ − 9 y = 0
Using the D-operator, (D2 − 9) y = 0 =⇒ φ(D) = D2 − 9
The characteristic equation is φ(λ) = 0 =⇒ λ2 − 9 = 0 =⇒ λ = ±3
The complementary solution is yC = C1 e−3x + C2 e3x
Example
Solve (D2 + 2 D + 2) y = 0
The characteristic equation is λ2 + 4λ + 20 = 0
λ =−4 ±
√
42 − 4×1×20
2×1=
−4 ±√
−64
2= −2 ± i 4 i =
√−1
The complementary solution is yC = e−2x(A cos 4x + B sin 4x)
38
Linear Ordinary Differential Equations — Complementary Solutions © Wei-Chau Xie
Example
Solve (D2 − 4D + 4) y = 0.
The characteristic equation is λ2−4λ+4 = 0 =⇒ (λ−2)2 = 0 =⇒ λ = 2, 2
The complementary solution is y(x) = (C0 + C1x) e2x
Example
Given the characteristic numbers of a DE, write the complementary solution.
1. λ = 2, 2, 2, 0, 0
yC = (C0 + C1 x + C2 x2) e2x + (D0 + D1 x)
2. λ = 1 ± i√
3, −2, −2, 0, −1
yC = ex(A cos√
3x + B sin√
3x) + (C0 + C1 x) e−2x + D0 + D1 e−x
3. λ = 1, 1, −1, ± i, ± i
yC = (C0 + C1 x) ex + D e−x + [(A0 + A1 x) cos x + (B0 + B1 x) sin x]
39
Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie
Method of Undetermined Coefficients
The method of undetermined coefficients assumes a form for a particular
solution, with coefficients to be determined, according to the form of the
right-hand side of the DE.
Corresponding to RHS F(x) Assumed Form of yP
(1) Polynomial of degree k Polynomial of degree k
(2) eαx C eαx
(3) sin βx, cos βx A cos βx + B sin βx
(4) eαx(
a0+a1x+ · · · +akxk)
cos βx, eαx[(
A0+A1x+ · · · +Akxk)
cos βx
eαx(
b0+b1x+ · · · +bkxk)
sin βx +(
B0+B1x+ · · · +Bkxk)
sin βx]︸ ︷︷ ︸ ︸ ︷︷ ︸
Polynomial of degree k Polynomial of degree k
The coefficients are determined by substituting the assumed particular
solution into the DE.
40
Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie
Example
Find yP for y′′ + 4 y′ − 2 y = 2x2 − 3x + 6.
RHS function F(x) = 2x2 − 3x + 6 Polynomial of degree 2
Assume yP = C2 x2 + C1 x + C0 Polynomial of degree 2
Substitute into the DE
yP′′ + 4 yP
′ − 2 yP = 2C2 + 4(2C2 x + C1) − 2(C2 x2 + C1 x + C0)
= −2C2 x2 + (8C2 − 2C1)x + (2C2 + 4C1 − 2C0)
= 2x2 − 3x + 6
Equate coefficients of corresponding terms
−2C2 = 2 =⇒ C2 = −1
8C2 − 2C1 = −3 =⇒ C1 = 12(8C2 + 3) = − 5
2
2C2 + 4C1 − 2C0 = 6 =⇒ C0 = C2 + 2C1 − 3 = −9
∴ yP = −x2 − 52
x − 9 Particular solution
41
Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie
Example
Solve (D2 − 3D + 2) y = 4 e3x.
Characteristic equation λ2−3λ+2 = 0 =⇒ (λ−1)(λ−2) = 0 =⇒ λ = 1, 2
yC = C1 ex + C2 e2x
Corresponding to F(x) = 4 e3x, assume yP = C e3x
Substitute into the DE
(D2 − 3D + 2) yP = 9C e3x − 3 · 3C e3x + 2 · C e3x = 2C e3x = 4 e3x
∴ 2C = 4 =⇒ C = 2 =⇒ yP = 2 e3x
∴ y = yC + yP = C1 ex + C2 e2x + 2 e3x General solution
42
Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie
Example
Find yP for y′′ − y′ + y = 2 sin 3x.
Corresponding to F(x) = 2 sin 3x, assume yP = A cos 3x + B sin 3x
Substitute into the DE
yP′′ − yP
′ + yP = (−9A cos 3x − 9B sin 3x) − (−3A sin 3x + 3B cos 3x)
+ (A cos 3x + B sin 3x)
= (−8A − 3B) cos 3x + (3A − 8B) sin 3x
= 2 sin 3x
Equate coefficients of corresponding terms
−8A − 3B = 0
3A − 8B = 2
=⇒
A = − 1673
B = 673
∴ yP = − 1673
cos 3x + 673
sin 3x Particular solution
43
Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie
Example
Solve (D2 − 2D − 3) y = 4x − 5 + 6x e2x.
Characteristic equation λ2−2λ−3 = 0 =⇒ (λ+1)(λ−3) = 0 =⇒ λ = −1, 3
yC = C1 e−x + C2 e3x
Corresponding to F1(x) = 4x − 5, assume yP1 = A1 x + A0
Corresponding to F2(x) = 6x e2x, assume yP2 = (B1 x + B0) e2x
yP = yP1 + yP2 = A1 x + A0 + (B1 x + B0) e2x
yP′ = A1 + B1 e2x + 2(B1 x + B0) e2x, yP
′′ = 4B1 e2x + 4(B1 x + B0) e2x
Substitute into the DE
(D2−2D−3) yP = [4B1e2x+4(B1 x+B0)e2x]−2[A1+B1e2x+2(B1 x+B0)e2x]
−3[A1 x+A0+(B1 x+B0)e2x]
= −3A1x − (2A1+3A0) + [(2B1−3B0) − 3B1 x]e2x
= 4x − 5 + 6x e2x
44
Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie
Equate coefficients of corresponding terms
−3A1 = 4
2A1+3A0 = 5
2B1−3B0 = 0
−3B1 = 6
=⇒
A1 = −43
A0 = 13
(5 − 2A1) = 239
B0 = 23
B1 = −43
B1 = −2
∴ y = yC + yP
= C1 e−x + C2 e3x − 43
x + 239
−(
2x + 43
)
e2x General solution
45
Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie
Exceptions in the Method of Undetermined Coefficients
In general, if any of the normally assumed terms of a particular solution occurs
in the complementary solution, one must multiply these assumed terms by a
power of x which is sufficiently high, but not higher, so that none of these
assumed terms occur in the complementary solution.
Example
Given the complementary solution yC and the right-hand side F(x) of the DE, specify
the form of yP using the method of undetermined coefficients.
(1) yC = c1ex + c2e2x + (d0 + d1x)e−x, F(x) = 3ex + 2e3x − x e−x
F(x) Normally Assumed Form for yP Contained in yC Modification
3ex C1ex c1ex x·C1ex
2e3x C2e3x ---------- ----------
−x e−x (D0 + D1 x)e−x (d0 + d1x)e−x x2 ·(D0 + D1 x)e−x
∴ yP = x·C1ex + C2e3x + x2 ·(D0 + D1 x)e−x
46
Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie
(2) yC = c0 + c1 x + c2 x2 + a cos 2x + b sin 2x, F(x) = 2 − x + sin x − cos 2x
F(x) Normally Assumed Form for yP Contained in yC Modification
2 − x C0 + C1 x c0 + c1 x + c2 x2 x3 ·(C0 + C1 x)
sin x A1 cos x + B1 sin x ---------- ----------
− cos 2x A2 cos 2x+B2 sin 2x a cos 2x + b sin 2x x·(A2 cos 2x+B2 sin 2x)
∴ yP = x3 ·(C0 + C1 x) + A1 cos x + B1 sin x + x·(A2 cos 2x + B2 sin 2x)
(3) yC = (c0 + c1 x)ex + a cos x + b sin x, F(x) = ex + e−x − x sin x
F(x) Normally Assumed Form for yP Contained in yC Modification
ex C1 ex (c0 + c1 x)ex x2 ·C1 ex
e−x C2 e−x ---------- ----------
−x sin x (A0+A1x) cos x a cos x + b sin x x·[(A0+A1x) cos x
+(B0+B1x) sin x +(B0+B1x) sin x]
∴ yP = x2 ·C1 ex + C2 e−x + x·[(A0+A1x) cos x + (B0+B1x) sin x]
47
Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie
Example
Solve y′′ − 2 y′ + y = ex.
Characteristic equation λ2 − 2λ + 1 = 0 =⇒ (λ − 1)2 = 0 =⇒ λ = 1, 1
yC = (C0 + C1 x)ex
Corresponding to F(x) = ex, the normally assumed form for yP is Cex.
Since Cex is contained in yC, assume yP = x2 · Cex.
yP′ = 2Cxex + Cx2 ex, yP
′′ = 2Cex + 4Cxex + Cx2 ex.
Substitute into the DE
yP′′ − 2 yP
′ + yP = (2Cex + 4Cxex + Cx2 ex) − 2(2Cxex + Cx2 ex) + Cx2 ex
= 2Cex = ex
2C = 1 =⇒ C = 12
=⇒ yP = 12
x2 ex
∴ y = yC + yP = (C0 + C1 x)ex + 12
x2 ex General solution
48
Applications of Linear Differential Equations © Wei-Chau Xie
Vibration of a Single Degree-of-Freedom System
F(t)m
x(t)
k
c m
x0(t) x(t)
k
c
Consider a single degree-of-freedom system consisting of a mass m, a spring
of stiffness k, and a viscous damper with damping coefficient c.
The motion is described by the horizontal displacement x(t) of mass m from
its equilibrium position.
The system is called single degree-of-freedom (DOF) because it needs one
variable to describe its motion.
49
Applications of Linear Differential Equations © Wei-Chau Xie
1. Single DOF System under Externally Applied Force
F(t)
F(t)F(t)
m
x(t)
k
c
m
x, x, x
cx
kx
m
x(t)
k
c
Consider mass m. It moves towards the right by a displacement x(t).
The spring is stretched by x(t) =⇒ the spring pulls the mass back towards
the left and the spring force is kx.
The damping force is similar to the spring force but is proportional to the
velocity =⇒ the damping force is cx towards the left.
Newton’s Second Law requires ma =∑
F :
m · x(t) = F(t) − kx − cx =⇒ mx + cx + kx = F(t)
50
Applications of Linear Differential Equations © Wei-Chau Xie
2. Single DOF System under Base Excitation
m
x0(t) x(t)
k
c
m
x0(t)x(t)
k
c m
x, x, x
c(x−x0)
k(x−x0)
The support moves towards the right by a displacement x0(t).
Applications of this model include a structure under earthquake excitation or
a robotic arm under base movement.
Consider mass m. It moves towards the right by a displacement x(t).
Mass moves towards the right =⇒ the spring pulls it back towards the left.
The stretch of the spring is x(t)−x0(t) =⇒ the spring force is k(x−x0).
The damping force is similar to the spring force but is proportional to the
velocity =⇒ the damping force is c(x−x0) towards the left.
51
Applications of Linear Differential Equations © Wei-Chau Xie
m
x0(t) x(t)
k
c
m
x0(t)x(t)
k
c m
x, x, x
c(x−x0)
k(x−x0)
Newton’s Second Law requires ma =∑
F :
m · x(t) = −k(x−x0) − c(x−x0)
Let y(t) = x(t)−x0(t) be the relative displacement of the mass.
The equation of motion becomes
my + cy + ky = −mx0
where x0(t) is the base acceleration.
The loading on the mass created from base excitation is F(t) = −mx0(t),
which is proportional to the mass and the base acceleration.
52
Applications of Linear Differential Equations © Wei-Chau Xie
Equation of motion of a single degree-of-freedom system
mx + cx + kx = F(t)
Divide the equation by m
x +c
mx +
k
mx =
F(t)
m, ω2
0 =k
m, 2ζω0 =
c
m
ω0 natural circular frequency of the system
ζ nondimensional damping coefficient
Equation of motion in standard form x + 2ζω0 x + ω20 x =
F(t)
m
Solution x(t) has two parts: complementary solution and particular solution.
Complementary solution xC(t) corresponds to F(t) = 0. The system is in
free (not forced) vibration, and the solution is the response of free vibration.
Particular solution xP(t) corresponds to the right-hand side of the DE or
the forcing term F(t) and is the response of forced vibration.
53
Applications of Linear Differential Equations © Wei-Chau Xie
Free Vibration—Complementary Solution
Complementary DE x + 2ζω0 x + ω20 x = 0
Characteristic equation λ2 + 2ζω0λ + ω20 = 0 =⇒ λ = ω0
(
−ζ ±√
ζ 2−1)
Underdamped System 06ζ <1
Most engineering structures fall in this category with ζ usually less than 10%.
λ = ω0
(
−ζ ± i√
1−ζ 2)
= −ζω0 ± iωd
where ωd = ω0
√
1−ζ 2 is the damped natural circular frequency.
xC(t) = e−ζω0t(A cos ωd t + B sin ωd t)
A and B are determined from the initial conditions x(0) = x0 and x(0) = v0:
A = x0, B =v0+ζω0x0
ωd
∴ xC(t) = e−ζω0t(
x0 cos ωd t +v0+ζω0x0
ωd
sin ωd t)
, 06ζ <1.
54
Applications of Linear Differential Equations © Wei-Chau Xie
Special Case: Undamped System ζ = 0, ωd = ω0
The response becomes
. xC(t) = x0 cos ω0t +v0
ω0
sin ω0t
. =√
x20+
( v0
ω0
)2{
x0√
x20+(v0/ω0)
2cos ω0t +
v0/ω0√
x20+(v0/ω0)
2sin ω0t
}
. ︸ ︷︷ ︸ ︸ ︷︷ ︸ ︸ ︷︷ ︸
. a cos ϕ sin ϕ
. = a cos(ω0t−ϕ) Harmonic function: amplitude a, phase angle ϕ
x0
acos(ω0t−ϕ)
xC(t)
t
Slope=v0 at t=0
Amplitude a
−a
a
Period T= 2πω0
ϕω0
55
Applications of Linear Differential Equations © Wei-Chau Xie
Harmonic solution xC(t) = a cos(ω0t−ϕ)
Period T =2π
ω0
ω0 is called the natural circular frequency of the system, unit rad/sec.
The natural frequency is f =ω0
2π=
1
T, unit
cycles
secor Hz.
The maximum displacement is
max xC(t) = a =√
x20+
( v0
ω0
)2= amplitude of the motion
56
Applications of Linear Differential Equations © Wei-Chau Xie
Underdamped Free Vibration
xC(t) = ae−ζω0t cos(ωd t−ϕ) a =√
x20 +
(v0+ζω0 x0ωd
)2, ϕ = tan−1
(v0+ζω0 x0ωd x0
)
cos(ωdt−ϕ)
ae−ζω0tcos(ωdt−ϕ)
xC(t)
t
t
−a
−1
1
−ae−ζω0t (Envelope)
Harmonic oscillatory response with exponentially decaying amplitude
ae−ζω0t
a
Td= 2πωd
ϕωd
57
Applications of Linear Differential Equations © Wei-Chau Xie
Forced Vibration—Particular Solution
For an underdamped system with 0<ζ <1, the complementary solution or
the response of free vibration is
xC(t) = e−ζω0t(
x0 cos ωd t +v0+ζω0x0
ωdsin ωd t
)
The response decays exponentially and approaches zero as t→∞.
The complementary solution is called the transient solution.
Because its value becomes negligible after some time, its effect is small and is
not important in practice.
The particular solution xP(t) is associated with the right-hand side of the DE
and hence corresponds to forced vibration.
The particular solution is called the steady-state solution, because it is the
solution that persists when time is large.
58
Applications of Linear Differential Equations © Wei-Chau Xie
Suppose F(t) = F0 sin �t. The particular solution is
.xP(t) =F0
m
1√
(ω20−�2)2+(2ζω0�)2
sin(�t−ϕ), ϕ = tan−1 2ζω0�
ω20−�2
Let r =�
ω0
=Excitation frequency
Undamped natural frequency= Frequency ratio
The amplitude of the response of forced vibration is
.∣∣xP(t)
∣∣
max=
F0
m
1√
(ω20−�2)2+(2ζω0�)2
=F0
k
1√
(1−r2)2+(2ζ r)2, ω2
0 =k
m
If dynamic effect is not considered one obtains
mx + cx + kx = F(t) = F0 sin �t =⇒ xstatic =F0
k
This is the static displacement of the system under the static force F0.
59
Applications of Linear Differential Equations © Wei-Chau Xie
∣∣xP(t)
∣∣
max
xstatic
=1
√
(1−r2)2+(2ζ r)2= Dynamic Magnification Factor
0
1
2
3
4
5
6
7
0.5 1 1.5 2 2.5
ζ=0
ζ=0.1
ζ=0.2
ζ=0.3
�ω0
r=
Dyn
amic
Mag
nif
icat
ion
Fac
tor
(DM
F)
60
Applications of Linear Differential Equations © Wei-Chau Xie
When r → 0 (excitation frequency � → 0), DMF → 1.
Dynamic excitation approaches a static force. The amplitude of dynamic
response approaches the static displacement.
When r → ∞ (excitation frequency � ≫ ω0), DMF → 0.
The dynamic response approaches zero.
Maximum value of DMF occurs whend(DMF)
dr= 0 =⇒ r ≈ 1, if ζ ≪1
Maximum value of DMF occurs when r ≈ 1 (� ≈ ω0)
DMFmax ≈ DMF∣∣
r=1=
1√
(1−r2)2+(2ζ r)2
∣∣∣∣
r=1
=1
2ζ
The smaller the damping coefficient, the larger the amplitudes of dynamic
response.
When ζ = 0 and � = ω0, a response of unbound amplitude occurs and the
system is in resonance.
61
Applications of Linear Differential Equations © Wei-Chau Xie
Resonance
When ζ = 0 and � = ω0, the equation of motion becomes
xP + ω20 xP =
F0
msin ω0t =⇒ xP(t) = −
F0
2mω0
t cos ω0t
t
xP(t)
Excitation Frequency � = Natural Frequency of the System ω0
=⇒ Resonance
Amplitude of response of the system increases linearly with time.
62
The Laplace Transform © Wei-Chau Xie
The Laplace Transform
The Laplace transform is one of the most important integral transforms.
Applying the Laplace transform to a linear ODE with constant coefficients
converts it into a linear algebraic equation.
The solution of the DE can then be obtained by determining the inverse Laplace
transform.
The method of Laplace transform is particularly advantageous in solving linear
ODE with the RHS functions involving discontinuous and impulse functions.
Definition — Laplace Transform
Let f(t), t>0, be a given function. The Laplace transform F(s) of function f(t)
is defined by
F(s) = L{
f(t)}
=∫ ∞
0e−st f(t)dt, s>0
63
The Laplace Transform © Wei-Chau Xie
Example
Determine the Laplace transform of f(t) = 1.
F(s) = L{
1}
=∫ ∞
0e−st · 1 dt = −
1
se−st
∣∣∣
∞
t=0=
1
s, s>0
Example
Determine the Laplace transforms of f(t) = sin ωt and f(t) = cos ωt.
. L{
eiωt}
=∫ ∞
0e−st · eiωtdt =
∫ ∞
0e−(s−iω)tdt = −
1
s−iωe−(s−iω)t
∣∣∣
∞
t=0
. =1
s−iω=
s + iω
(s−iω)(s+iω)=
s + iω
s2 + ω2, s>0
. ∵ cos ωt = Re(eiωt), sin ωt = Im(eiωt) eiωt = cos ωt+ i sin ωt
. ∴ L{
cos ωt}
= Re[L{
eiωt}
] =s
s2 + ω2
. L{
sin ωt}
= Im[L{
eiωt}
] =ω
s2 + ω2
64
The Laplace Transform © Wei-Chau Xie
Properties of the Laplace Transform
1.. Laplace transform L { · } is a linear operator
2.. Laplace Transform of Derivatives
L{
f (n)(t)}
= snF(s) − sn−1 f(0) − sn−2 f ′(0) − · · · − s f (n−2)(0) − f (n−1)(0)
3.. Property of Shifting
L{
eat f(t)}
= F(s−a) = L{
f(t)}∣∣∣
s→(s−a)
4.. Property of Differentiation
L{
tn f(t)}
= (−1)nF(n)(s) = (−1)n dnF(s)
dsn, n = 1, 2, . . .
65
The Laplace Transform © Wei-Chau Xie
Example
Determine the Laplace transform of f(t) = eat.
Solution 1: F(s) = L{
eat}
=∫ ∞
0e−st · eatdt Definition
. =∫ ∞
0e−(s−a)tdt = −
1
s−ae−(s−a)t
∣∣∣
∞
t=0=
1
s−a, s>a
Solution 2: F(s) = L{
eat · 1}
= L{
1}∣∣
s→(s−a)Property of Shifting
. =1
s
∣∣∣
s→(s−a)=
1
s−a
66
The Laplace Transform © Wei-Chau Xie
Example
Determine the Laplace transform of f(t) = tn, n=0, 1, . . . .
Using the formula of Laplace transform of derivative
. L{
f ′(t)}
= s L{
f(t)}
− f(0)f(t) = tn
==⇒ L{
n tn−1}
= s L{
tn}
− 0n
L{
tn}
=n
sL
{
tn−1}
. n = 1 : L{
t}
=1
sL
{
1}
=1
s2, L
{
1}
=1
s
. n = 2 : L{
t2}
=2
sL
{
t}
=2
s·
1
s2=
2·1s3
. n = 3 : L{
t3}
=3
sL
{
t2}
=3
s·
2·1s3
=3·2·1
s4
. · · · · · ·
In general, L{
tn}
=n!
sn+1, n = 0, 1, . . . 0!=1
67
The Heaviside Step Function © Wei-Chau Xie
The Heaviside Step Function
The Heaviside step function is defined by
H(t−a) ={
0, t<a
1, t>aa is a real number
t
a
H(t−a)
1
0
. L{
H(t−a)}
=1
se−as, s>0
. L{
f(t−a) H(t−a)}
= e−asF(s), s > a > 0
68
The Heaviside Step Function © Wei-Chau Xie
Example
Evaluate (a) L{
(t−2)3 H(t−2)}
(b) L{
sin t H(t−2π)}
(a) L{
(t−2)3 H(t−2)}
= e−2sL
{
t3}
= e−2s 3!s4
=6
s4e−2s
(b) L{
sin t H(t−2π)}
= L{
sin(t−2π) H(t−2π)}
. = e−2πsL
{
sin t}
=e−2πs
s2 + 1
69
The Heaviside Step Function © Wei-Chau Xie
The Heaviside step function is very useful in dealing with functions with
discontinuities or piecewise smooth functions.
f(t) ={
0, t< t0
f1(t), t>t0
= f1(t)H(t−t0)
tt0
f (t)
f1(t)
f(t) ={
f2(t), t< t0
0, t>t0
= f2(t)[1−H(t−t0)]
tt0
f (t)
f2(t)
f(t) =
0, t<a
g(t), a< t<b
0, t>b
= g(t)[H(t−a) − H(t−b)]
ta b
f (t)
g(t)
70
Impulse Functions © Wei-Chau Xie
Impulse Functions
Consider an impulse function f(t) over a time interval t0< t< t0+ε with constant
amplitude I/ε as shown in Figure (a).
tt0 t0+ε1ε1
f (t)
ε1
I
(a) (b) (c)
Area=I
tt0 t0+ε2ε2
f (t)
ε2
IArea=I
tt0
f (t)
Area=I
ε→0
The area under the function f(t), or the impulse, is I.
f(t) =I
ε
{
H(t−t0) − H[t−(t0+ε)]}
When ε decreases as in Figures (a) and (b)
The width of the time interval over which the impulse is defined decreases.
The amplitude I/ε of the function increases.
The area under the function remains constant.
71
The Dirac Delta Function © Wei-Chau Xie
Dirac Delta Function
In the case of I = 1, the limiting function as ε→0,
limε→0
f(t) = limε→0
H(t−t0) − H[t−(t0+ε)]ε
≡ δ(t−t0)
is called the unit impulse function or the Dirac delta function.
Properties of the Dirac Delta Function
ta
δ(t−a)
Area=1
1. δ(t−a) = 0, if t 6=a.
2. δ(t−a) → +∞, as t→a.
3.
∫ a+α
a−α
δ(t−a)dt = 1, α>0.
4. Shifting Property If g(t) is any function,∫ a+α
a−α
g(t) δ(t−a)dt = g(a), α>0
72
The Dirac Delta Function © Wei-Chau Xie
5.
∫ t
−∞δ(t−a)dt = H(t−a) =
{
0, t<a
1, t>a=⇒
dH(t−a)
dt= δ(t−a)
6. L{
δ(t−a)}
=∫ ∞
0e−st · δ(t−a)dt = e−as, a>0 Shifting property
7. L{
f(t) δ(t−a)}
=∫ ∞
0e−st· f(t)δ(t−a)dt = e−as f(a), a>0
δ(t−a) is not a function in the usual sense.
It is known as a generalized function.
73
The Inverse Laplace Transform © Wei-Chau Xie
The Inverse Laplace Transform
Given the Laplace transform F(s) of function f(t), F(s)=L{
f(t)}
, the inverse
Laplace transform is f(t)=L−1
{
F(s)}
.
Properties of the Inverse Laplace Transform
1. L{
·}
is a linear operator =⇒ L−1
{
·}
is a linear operator
2. L{
eat f(t)}
= F(s−a) =⇒ L−1
{
F(s−a)}
= eatL
−1{
F(s)}
= eat f(t)
3. L{
tn f(t)}
= (−1)nF(n)(s) =⇒ L−1
{
F(n)(s)}
= (−1)n tn f(t)
4. L{
f(t−a)H(t−a)}
= e−as F(s) =⇒ L−1
{
e−as F(s)}
= f(t−a) H(t−a)
74
The Inverse Laplace Transform © Wei-Chau Xie
Example
Evaluate L−1
{
F(s)}
, where F(s) =2s
s2+4+
1
s3−
4
s−1.
. L−1
{
F(s)}
= L−1
{2s
s2+4
}
+ L−1
{1
s3
}
− L−1
{4
s−1
}
. = 2 L−1
{s
s2+22
}
+ 12
L−1
{2!s3
}
− 4 L−1
{1
s−1
}
. = 2 cos 2t + 12
t2 − 4 et
75
The Inverse Laplace Transform © Wei-Chau Xie
Example
Evaluate L−1
{s
s2 + 6s + 11+
1
(s−3)3
}
.
.L −1
{s
s2 + 6s + 11+
1
(s−3)3
}
= L−1
{(s+3) − 3
(s+3)2 + 2
}
+ 12
L−1
{2
(s−3)3
}
. = L−1
{s+3
(s+3)2 + 2
}
− 3 L−1
{1
(s+3)2 + 2
}
+ 12
L−1
{2
(s−3)3
}
L−1
{
F(s−a)}
= eatL
−1{
F(s)}
. = e−3tL
−1
{s
s2 + (√
2)2
}
−3e−3t
√2
L−1
{ √2
s2 + (√
2)2
}
+ 12
e3tL
−1
{2!s3
}
. = e−3t cos√
2t − 3√2
e−3t sin√
2t + 12
e3t t2
76
The Inverse Laplace Transform © Wei-Chau Xie
Example
Evaluate L−1
{e−πs
s2+9
}
.
. ∵ L−1
{1
s2+9
}
= 13
L−1
{3
s2+32
}
= 13
sin 3t
. ∴ L−1
{e−πs
s2+9
}
= 13
sin 3(t−π) H(t−π)
L−1
{
e−as F(s)}
= f(t−a) H(t−a)
. = − 13
sin 3t H(t−π)
77
The Inverse Laplace Transform © Wei-Chau Xie
Theorem — Convolution Integral
If L−1
{
F(s)}
= f(t) and L−1
{
G(s)}
= g(t), then
L−1
{
F(s)G(s)}
=(
f ∗ g)
(t) =∫ t
0f(u)g(t−u)du =
∫ t
0g(u) f(t−u)du
in which the integral is known as a Convolution Integral.
Example
Evaluate L−1
{1
s2(s2+1)
}
.
. ∵ L−1
{1
s2
}
= t, L−1
{1
s2+1
}
= sin t
. ∴ L−1
{1
s2(s2+1)
}
= L−1
{1
s2·
1
s2+1
}
= t ∗ sin t =∫ t
0(t−u) sin u du
. = −∫ t
0(t−u) d(cos u) = −(t−u) cos u
∣∣∣
t
u=0+
∫ t
0cos u du = t − sin t
78
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Solving Differential Equations Using the Laplace Transform
Consider an nth-order linear ODE with constant coefficients
an y (n)(t) + an−1 y (n−1)(t) + · · · + a1 y ′(t) + a0 y(t) = f(t)
L{
y(t)}
= Y(s)
L{
y ′(t)}
= sY(s) − y(0)
L{
y ′′(t)}
= s2 Y(s)) − s y(0) − y ′(0)
· · · · · ·L
{
y(n)(t)}
= sn Y(s) − sn−1 y(0) − sn−2 y ′(0) − · · · − s y(n−2)(0) − y(n−1)(0)
Taking the Laplace transform on both sides of the equation leads to an
algebraic equation for Y(s)
an
[
snY(s) −n∑
i=1
sn−i y(i−1)(0)
]
+ an−1
[
sn−1Y(s) −n−1∑
i=1
sn−1−i y(i−1)(0)
]
+ · · ·
+ a1[sY(s) − y(0)] + a0Y(s) = L{
f(t)}
79
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Solving for Y(s) yields
Y(s) =L
{
f(t)}
+n∑
k=1
k∑
i=1
ak y(i−1)(0)sk−i
n∑
i=0
ai si
Taking the inverse transform y(t)=L−1
{
Y(s)}
gives the solution of the DE.
80
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Example
Solve the initial value problem (IVP) y ′′ + 4 y = 4t, y(0) = 1, y ′(0) = 5.
Let Y(s)=L{
y(t)}
. Taking the Laplace transform of both sides of the DE yields
. [s2 Y(s) − s y(0) − y ′(0)] + 4Y(s) = 4 ·1
s2
. (s2 + 4)Y(s) =4
s2+ s + 5
Solving for Y(s) leads to
. Y(s) =4
s2(s2+4)+
s
s2+4+
5
s2+4
. =( 1
s2−
1
s2+4
)
+s
s2+4+
5
s2+4=
1
s2+
s
s2+4+
4
s2+4
Taking the inverse Laplace transform gives
. y(t) = L−1
{
Y(s)}
= L−1
{1
s2
}
+ L−1
{s
s2+22
}
+ 2 L−1
{2
s2+22
}
. = t + cos 2t + 2 sin 2t
81
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Example
Solve the IVP y ′′ + 2 y ′ + 5 y = f(t), t > 0, y(0) = 1, y ′(0) = 3, where
f(t) ={
1, 0 6 t < 2
0, t > 2
Using the Heaviside step function, f(t) can be written as
. f(t) = 1 − H(t−2) =⇒ L{
f(t)}
= L{
1}
− L{
H(t−2)}
=1
s−
e−2s
s
Taking the Laplace transform of both sides of the DE yields
. [s2 Y(s) − s y(0) − y ′(0)] + 2[sY(s) − y(0)] + 5Y(s) =1−e−2s
s
. (s2 + 2s + 5)Y(s) =1−e−2s
s+ s + 5
Solving for Y(s) leads to
.Y(s) =1−e−2s
s(s2+2s+5)+
s+5
s2+2s+5=
1−e−2s
5
{ 1
s−
s+2
s2+2s+5
}
+(s+1) + 4
(s+1)2+4
. =1−e−2s
5
{ 1
s−
(s+1)
(s+1)2+22−
1
(s+1)2+22
}
+(s+1)
(s+1)2+22+
2 · 2
(s+1)2+22
82
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
. ∵ L−1
{1
s−
(s+1)
(s+1)2+22−
1
2·
2
(s+1)2+22
}
= t − e−t(
cos 2t − 12
sin 2t)
L−1
{
F(s−a)}
= eatL
−1{
F(s)}
. ∴ L−1
{
e−2s
[1
s−
(s+1)
(s+1)2+22−
1
2·
2
(s+1)2+22
]}
. ={
(t−2) − e−(t−2) [ cos 2(t−2) − 12
sin 2(t−2)]}
H(t−2)
L−1
{
e−as F(s)}
= f(t−a) H(t−a)
Taking the inverse Laplace transform gives
. y(t) = L−1
{
Y(s)}
. = 15
{
[t − e−t(
cos 2t − 12
sin 2t)
]
. −{
(t−2) − e−(t−2) [ cos 2(t−2) − 12
sin 2(t−2)]}
H(t−2)
}
. + e−t cos 2t + 2e−t sin 2t
83
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Example
Solve the IVP y ′′ + y = 4 δ(t−2π), y(0) = 1, y ′(0) = 0.
Let Y(s)=L{
y(t)}
. Taking the Laplace transform of both sides of the DE yields
. [s2 Y(s) − s y(0) − y ′(0)] + Y(s) = 4 L{
δ(t−2π)}
. (s2 + 1)Y(s) = 4 e−2πs + s
Solving for Y(s) leads to Y(s) =4 e−2πs
s2+1+
s
s2+1
Taking the inverse Laplace transform gives
. y(t) = L−1
{
Y(s)}
= L−1
{4 e−2πs
s2+1
}
+ L−1
{s
s2+1
}
. = 4 sin(t−2π) H(t−2π) + cos t
. L−1
{
e−as F(s)}
= f(t−a) H(t−a)
84
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Example
Solve the IVP y ′′ − 2 y ′ + 5 y = 2 δ(t−3), y(0) = 0, y ′(0) = 1.
Let Y(s)=L{
y(t)}
. Taking the Laplace transform of both sides of the DE yields
. [s2 Y(s) − s y(0) − y ′(0)] − 2[sY(s) − y(0)] + 5Y(s) = 2 L{
δ(t−3)}
. (s2 − 2s + 5)Y(s) = 2 e−3s + 1 =⇒ Y(s) =2 e−3s + 1
(s−1)2+22
Taking the inverse Laplace transform gives
. ∵ L−1
{2
(s−1)2+22
}
= et sin 2t L−1
{
F(s−a)}
= eat f(t)
. ∴ y(t) = L−1
{
Y(s)}
= L−1
{
e−3s 2
(s−1)2+22
}
+1
2L
−1
{2
(s−1)2+22
}
. = et−3 sin 2(t−3) H(t−3) + 12
et sin 2t
L−1
{
e−as F(s)}
= f(t−a) H(t−a)
85
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Example
Find y(t) where y(t) = t3 +∫ t
0sin(t−u) y(u) du.
Let Y(s)=L{
y(t)}
. Take the Laplace transform of both sides of the equation
. L{
y(t)}
= L{
t3}
+ L
{ ∫ t
0sin(t−u) y(u) du
}
Using the convolution integral
. L
{ ∫ t
0sin(t−u) y(u) du
}
= L{
sin t}
L{
y(t)}
=1
s2+1Y(s)
. ∴ Y(s) =3!s4
+1
s2+1Y(s) =⇒
(
1 −1
s2+1
)
Y(s) =6
s4
Solving for Y(s) leads to Y(s) =s2+1
s2·
6
s4=
6
s4+
6
s6
. ∴ y(t) = L−1
{
Y(s)}
= L−1
{3!s4
}
+ 120
L−1
{5!s6
}
= t3 + 120
t5
86
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Example
Find y(t) where y(t) = 3t2 − e−t −∫ t
0et−u y(u) du.
Let Y(s)=L{
y(t)}
. Take the Laplace transform of both sides of the equation
. L{
y(t)}
= 3 L{
t2}
− L{
e−t}
− L
{ ∫ t
0et−u y(u) du
}
. ∵ L
{ ∫ t
0et−u y(u) du
}
= L{
et}
L{
y(t)}
=1
s−1Y(s) Convolution integral
. ∴ Y(s) = 3·2!s3
−1
s+1−
1
s−1Y(s) =⇒
(
1 +1
s−1
)
Y(s) =6
s3−
1
s+1
. ∴ Y(s) =s−1
s
( 6
s3−
1
s+1
)
=6
s3−
6
s4−
1
s+1+
1
s(s+1)
. =6
s3−
6
s4−
1
s+1+
( 1
s−
1
s+1
)
. ∴ y(t) = L−1
{
Y(s)}
= 3L−1
{2!s3
}
− L−1
{3!s4
}
+ L−1
{1
s
}
− 2L−1
{1
s+1
}
. = 3t2 − t3 + 1 − 2e−t
87
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Example
Determine the current i(t) in the series RLC circuit shown when R = 2 �,
L = 0.1 H, C = 0.1 F, and V(t) = 120t−120t H(t−1).
V(t)
R
C
L
i V(t)
0
120
120 t
t 1
For t 6 0, V(t) = 0. The capacitor behaves as an open circuit =⇒ i(0) = 0.
For t > 0, apply Kirchhoff ’s Voltage Law
. −V(t) + Ri + Ldi
dt+
1
C
∫ t
−∞i(u)du = 0
. 0.1di
dt+ 2i + 10
∫ t
−∞i(u)du = 120t − 120t H(t−1)
88
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Let I(s)=L{
i(t)}
.
.L
{ ∫ t
−∞i(u)du
}
= L
{ ∫ t
−∞i(u) · 1du
}
. = L{
1}
L{
i(t)}
=1
sI(s) Convolution integral
.L{
t H(t−1)}
= L{
[(t−1)+1] H(t−1)}
. = L{
(t−1) H(t−1)}
+ L{
H(t−1)}
=e−s
s2+
e−s
s
Take the Laplace transform of both sides of the equation
.0.1 [s I(s) − i(0)] + 2 I(s) + 10·1
sI(s) = 120·
1
s2− 120
( e−s
s2+
e−s
s
)
.
(
0.1 s + 2 +10
s
)
I(s) = 120( 1−e−s
s2−
e−s
s
)
. ∵ 0.1 s + 2 +10
s=
0.1 s2 + 2 s + 10
s=
s2 + 20 s + 100
10 s
89
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
. ∴ I(s) =1200 s
s2 + 20 s + 100
( 1−e−s
s2−
e−s
s
)
. = 1200
{1 − e−s
s (s+10)2−
e−s
(s+10)2
}
. ∵ L−1
{1
s
}
= 1, L−1
{1
(s+10)2
}
= e−10t t
. ∴ L−1
{1
s (s+10)2
}
=∫ t
0u e−10u · 1 du Convolution integral
. =−10u − 1
(−10)2e−10u
∣∣∣
t
0= −
10t + 1
100e−10t +
1
100
. L−1
{e−s
s (s+10)2
}
={
−10(t−1) + 1
100e−10(t−1) +
1
100
}
H(t−1)
. L−1
{e−s
(s+10)2
}
= e−10(t−1)(t−1) H(t−1)
90
Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie
Taking the inverse Laplace transform gives the solution to the DE
.i(t) = L−1
{
I(s)}
. = 1200
{
−10t+1
100e−10t +
1
100−
[
−10(t−1)+1
100e−10(t−1) +
1
100
]
H(t−1)
. − (t−1) e−10(t−1) H(t−1)
}
. = 12{
1 − (10t+1) e−10t − H(t−1) − [90(t−1) + 1] e−10(t−1) H(t−1)
}
91