Ch 3.1: Second Order Linear Homogeneous
Equations with Constant Coefficients
A second order ordinary differential equation has the
general form
y ′′ = f (t , y , y ′)
where f is some given function.
This equation is said to be linear if f is linear in y and y':
y ′′ = g (t ) − p (t ) y ′ − q(t ) y
Otherwise the equation is said to be nonlinear.
A second order linear equation often appears as
P (t ) y′′ + Q(t ) y′ + R (t ) y = G (t )
If G(t) = 0 for all t, then the equation is called homogeneous.
Otherwise the equation is nonhomogeneous.
Homogeneous Equations, Initial Values
In Sections 3.6 and 3.7, we will see that once a solution to a
homogeneous equation is found, then it is possible to solve
the corresponding nonhomogeneous equation, or at least
express the solution in terms of an integral.
The focus of this chapter is thus on homogeneous equations;
and in particular, those with constant coefficients:
ay′′ + by′ + cy = 0
We will examine the variable coefficient case in Chapter 5.
Initial conditions typically take the form
y (t0 ) = y0 , y ′(t0 ) = y0′
Thus solution passes through (t0, y0), and slope of solution at
(t0, y0) is equal to y0'.
Example 1: Infinitely Many Solutions
(1 of 3)
Consider the second order linear differential equation
y′′ − y = 0
Two solutions of this equation are
y1 (t ) = e t , y2 (t ) = e −t
Other solutions include
y3 (t ) = 3et ,
y4 (t ) = 5e − t ,
y5 (t ) = 3et + 5e − t
Based on these observations, we see that there are infinitely
many solutions of the form
y (t ) = c1e t + c2 e − t
It will be shown in Section 3.2 that all solutions of the
differential equation above can be expressed in this form.
Example 1: Initial Conditions
(2 of 3)
Now consider the following initial value problem for our
equation:
y ′′ − y = 0, y (0) = 3, y ′(0) = 1
We have found a general solution of the form
y (t ) = c1e t + c2 e − t
Using the initial equations,
y (0) = c1 + c2 = 3⎫
⎬ ⇒ c1 = 2, c2 = 1
y′(0) = c1 − c2 = 1 ⎭
Thus
y (t ) = 2et + e −t
Example 1: Solution Graphs
(3 of 3)
Our initial value problem and solution are
y′′ − y = 0, y (0) = 3, y′(0) = 1 ⇒ y (t ) = 2e t + e − t
Graphs of this solution are given below. The graph on the
right suggests that both initial conditions are satisfied.
Characteristic Equation
To solve the 2nd order equation with constant coefficients,
ay ′′ + by ′ + cy = 0,
we begin by assuming a solution of the form y = ert.
Substituting this into the differential equation, we obtain
ar 2 e rt + bre rt + ce rt = 0
Simplifying,
e rt (ar 2 + br + c) = 0
and hence
ar 2 + br + c = 0
This last equation is called the characteristic equation of
the differential equation.
We then solve for r by factoring or using quadratic formula.
General Solution
Using the quadratic formula on the characteristic equation
ar 2 + br + c = 0,
we obtain two solutions, r1 and r2.
There are three possible results:
The roots r1, r2 are real and r1 ≠ r2.
The roots r1, r2 are real and r1 = r2.
The roots r1, r2 are complex.
− b ± b 2 − 4ac
r=
2a
In this section, we will assume r1, r2 are real and r1 ≠ r2.
In this case, the general solution has the form
y (t ) = c1e r1 t + c2 e r2 t
Initial Conditions
For the initial value problem
ay ′′ + by ′ + cy = 0, y (t0 ) = y0 , y′(t0 ) = y0′ ,
we use the general solution
y (t ) = c1e r1 t + c2 e r2 t
together with the initial conditions to find c1 and c2. That is,
c1e r t + c2 e r t = y0 ⎫⎪
y0′ − y0 r2 − r t
y0 r1 − y0′ − r t
e , c2 =
e
⎬ ⇒ c1 =
rt
r t
r1 − r2
r1 − r2
c1r1e + c2 r2 e = y0′ ⎪⎭
1 0
2 0
1 0
1 0
2 0
2 0
Since we are assuming r1 ≠ r2, it follows that a solution of the
form y = ert to the above initial value problem will always
exist, for any set of initial conditions.
Example 2
Consider the initial value problem
y′′ + y′ − 12 y = 0, y (0) = 0, y′(0) = 1
Assuming exponential soln leads to characteristic equation:
y (t ) = e rt ⇒ r 2 + r − 12 = 0 ⇔ (r + 4)(r − 3) = 0
Factoring yields two solutions, r1 = -4 and r2 = 3
The general solution has the form
y (t ) = c1e −4 t + c2 e3t
Using the initial conditions:
c1 + c2 = 0⎫
−1
1
c
c
⇒
=
=
,
⎬
1
2
− 4c1 + 3c2 = 1 ⎭
7
7
− 1 − 4 t 1 3t
Thus
y (t ) =
e + e
7
7
Example 3
Consider the initial value problem
2 y′′ + 3 y′ = 0, y (0 ) = 1, y′(0 ) = 3
Then
y (t ) = e rt ⇒ 2r 2 + 3r = 0 ⇔ r (2r + 3) = 0
Factoring yields two solutions, r1 = 0 and r2 = -3/2
The general solution has the form
y (t ) = c1e 0 t + c2 e −3t / 2 = c1 + c2 e −3t / 2
Using the initial conditions:
c1 + c2 = 1 ⎫
⎪
⎬ ⇒ c1 = 3, c2 = −2
3c2
−
= 3⎪
2
⎭
Thus
y (t ) = 3 − 2e −3t / 2
Example 4: Initial Value Problem
(1 of 2)
Consider the initial value problem
y′′ + 5 y′ + 6 y = 0, y (0 ) = 2, y′(0 ) = 3
Then
y (t ) = e rt ⇒ r 2 + 5r + 6 = 0 ⇔
(r + 2)(r + 3) = 0
Factoring yields two solutions, r1 = -2 and r2 = -3
The general solution has the form
y (t ) = c1e −2 t + c2 e −3t
Using initial conditions:
c1 + c2 = 2⎫
⎬ ⇒ c1 = 9, c2 = −7
− 2c1 − 3c2 = 3 ⎭
Thus
y (t ) = 9e −2 t − 7e −3t
Example 4: Find Maximum Value
(2 of 2)
Find the maximum value attained by the solution.
y (t ) = 9e −2 t − 7e −3t
y′(t ) = −18e
−2t
6e − 2 t = 7e −3t
et = 7/6
t = ln(7 / 6)
t ≈ 0.1542
y ≈ 2.204
+ 21e
−3 t
set
=0
Ch 3.2: Fundamental Solutions of Linear
Homogeneous Equations
Let p, q be continuous functions on an interval I = (α, β),
which could be infinite. For any function y that is twice
differentiable on I, define the differential operator L by
L[ y ] = y′′ + p y′ + q y
Note that L[y] is a function on I, with output value
L[ y ](t ) = y′′(t ) + p (t ) y′(t ) + q (t ) y (t )
For example,
p (t ) = t 2 , q (t ) = e 2t , y (t ) = sin(t ), I = (0, 2π )
L[ y ](t ) = − sin(t ) + t 2 cos(t ) + 2e 2t sin(t )
Differential Operator Notation
In this section we will discuss the second order linear
homogeneous equation L[y](t) = 0, along with initial
conditions as indicated below:
L[ y ] = y′′ + p (t ) y′ + q (t ) y = 0
y (t0 ) = y0 , y′(t0 ) = y1
We would like to know if there are solutions to this initial
value problem, and if so, are they unique.
Also, we would like to know what can be said about the form
and structure of solutions that might be helpful in finding
solutions to particular problems.
These questions are addressed in the theorems of this section.
Theorem 3.2.1
Consider the initial value problem
y′′ + p (t ) y′ + q (t ) y = g (t )
y (t0 ) = y0 , y′(t0 ) = y0′
where p, q, and g are continuous on an open interval I that
contains t0. Then there exists a unique solution y = φ(t) on I.
Note: While this theorem says that a solution to the initial
value problem above exists, it is often not possible to write
down a useful expression for the solution. This is a major
difference between first and second order linear equations.
Example 1
y′′ + p (t ) y′ + q (t ) y = g (t )
y (t0 ) = y0 , y′(t0 ) = y1
Consider the second order linear initial value problem
y′′ − y = 0, y (0 ) = 3, y′(0 ) = 1
In Section 3.1, we showed that this initial value problem had
the following solution:
y (t ) = 2et + e −t
Note that p(t) = 0, q(t) = -1, g(t) = 0 are each continuous on
(-∞, ∞), and the solution y is defined and twice differentiable
on (-∞, ∞).
Example 2
Consider the second order linear initial value problem
y′′ + p (t ) y′ + q (t ) y = 0, y (0 ) = 0, y′(0 ) = 0
where p, q are continuous on an open interval I containing t0.
In light of the initial conditions, note that y = 0 is a solution
to this homogeneous initial value problem.
Since the hypotheses of Theorem 3.2.1 are satisfied, it
follows that y = 0 is the only solution of this problem.
Example 3
Determine the longest interval on which the given initial
value problem is certain to have a unique twice differentiable
solution. Do not attempt to find the solution.
(t + 1) y′′ − (cos t ) y′ + 3 y = 1, y (0) = 1, y′(0) = 0
First put differential equation into standard form:
cos t
3
1
y′′ −
y′ +
y=
, y (0 ) = 1, y′(0 ) = 0
t +1
t +1
t +1
The longest interval containing the point t = 0 on which the
coefficient functions are continuous is (-1, ∞).
It follows from Theorem 3.2.1 that the longest interval on
which this initial value problem is certain to have a twice
differentiable solution is also (-1, ∞).
Theorem 3.2.2 (Principle of Superposition)
If y1and y2 are solutions to the equation
L[ y ] = y′′ + p (t ) y′ + q (t ) y = 0
then the linear combination c1y1 + y2c2 is also a solution, for
all constants c1 and c2.
To prove this theorem, substitute c1y1 + y2c2 in for y in the
equation above, and use the fact that y1 and y2 are solutions.
Thus for any two solutions y1 and y2, we can construct an
infinite family of solutions, each of the form y = c1y1 + c2 y2.
Can all solutions can be written this way, or do some
solutions have a different form altogether? To answer this
question, we use the Wronskian determinant.
The Wronskian Determinant
(1 of 3)
Suppose y1 and y2 are solutions to the equation
L[ y ] = y′′ + p (t ) y′ + q (t ) y = 0
From Theorem 3.2.2, we know that y = c1y1 + c2 y2 is a
solution to this equation.
Next, find coefficients such that y = c1y1 + c2 y2 satisfies the
initial conditions
y (t0 ) = y0 , y′(t0 ) = y0′
To do so, we need to solve the following equations:
c1 y1 (t0 ) + c2 y2 (t0 ) = y0
c1 y1′ (t0 ) + c2 y2′ (t0 ) = y0′
c1 y1 (t0 ) + c2 y2 (t0 ) = y0
c1 y1′ (t0 ) + c2 y2′ (t0 ) = y0′
The Wronskian Determinant
(2 of 3)
Solving the equations, we obtain
y0 y2′ (t0 ) − y0′ y2 (t0 )
c1 =
y1 (t0 ) y2′ (t0 ) − y1′ (t0 ) y2 (t0 )
− y0 y1′ (t0 ) + y0′ y1 (t0 )
c2 =
y1 (t0 ) y2′ (t0 ) − y1′ (t0 ) y2 (t0 )
In terms of determinants:
y0 y2 (t0 )
y1 (t0 ) y0
y0′ y2′ (t0 )
y1′ (t0 ) y0′
c1 =
, c2 =
y1 (t0 ) y2 (t0 )
y1 (t0 ) y2 (t0 )
y1′ (t0 ) y2′ (t0 )
y1′ (t0 ) y2′ (t0 )
The Wronskian Determinant
(3 of 3)
In order for these formulas to be valid, the determinant W in
the denominator cannot be zero:
c1 =
y0
y0′
y1 (t0 )
W=
y1′ (t0 )
y2 (t0 )
y2′ (t0 )
, c2 =
W
y1 (t0 ) y0
y1′ (t0 ) y0′
W
y2 (t0 )
= y1 (t0 ) y2′ (t0 ) − y1′ (t0 ) y2 (t0 )
y2′ (t0 )
W is called the Wronskian determinant, or more simply,
the Wronskian of the solutions y1and y2. We will sometimes
use the notation
W ( y1 , y2 )(t0 )
Theorem 3.2.3
Suppose y1 and y2 are solutions to the equation
L[ y ] = y′′ + p (t ) y′ + q (t ) y = 0
(1)
and that the Wronskian
W = y1 y2′ − y1′ y2
is not zero at the point t0 where the initial conditions
y (t0 ) = y0 , y′(t0 ) = y0′
( 2)
are assigned. Then there is a choice of constants c1, c2 for
which y = c1y1 + c2 y2 is a solution to the differential
equation (1) and initial conditions (2).
Example 4
Recall the following initial value problem and its solution:
y′′ − y = 0, y (0 ) = 3, y′(0 ) = 1 ⇒ y (t ) = 2e t + e − t
Note that the two functions below are solutions to the
differential equation:
y1 = et , y2 = e −t
The Wronskian of y1 and y2 is
y1 y2
W=
= y1 y2′ − y1′ y2 = −et e −t − et e −t = −2e 0 = −2
y1′ y2′
Since W ≠ 0 for all t, linear combinations of y1 and y2 can be
used to construct solutions of the IVP for any initial value t0.
y = c1 y1 + c2 y2
Theorem 3.2.4 (Fundamental Solutions)
Suppose y1 and y2 are solutions to the equation
L[ y ] = y′′ + p (t ) y′ + q (t ) y = 0.
If there is a point t0 such that W(y1,y2)(t0) ≠ 0, then the family
of solutions y = c1y1 + c2 y2 with arbitrary coefficients c1, c2
includes every solution to the differential equation.
The expression y = c1y1 + c2 y2 is called the general solution
of the differential equation above, and in this case y1 and y2
are said to form a fundamental set of solutions to the
differential equation.
Example 5
Recall the equation below, with the two solutions indicated:
y′′ − y = 0, y1 = e t , y2 = e − t
The Wronskian of y1 and y2 is
y1
W=
y1′
y2
= −et e −t − et e −t = −2e 0 = −2 ≠ 0 for all t.
y2′
Thus y1 and y2 form a fundamental set of solutions to the
differential equation above, and can be used to construct all
of its solutions.
The general solution is
y = c1e t + c2 e − t
Example 6
Consider the general second order linear equation below,
with the two solutions indicated:
y′′ + p (t ) y′ + q (t ) y = 0
Suppose the functions below are solutions to this equation:
y1 = e r1t , y2 = e r2t , r1 ≠ r2
The Wronskian of y1and y2 is
y1 y2
e r1t
e r2t
( r1 + r2 )t
(
)
= rt
=
r
−
r
e
≠ 0 for all t.
W=
2
1
r2t
1
y1′ y2′ r1e
r2 e
Thus y1and y2 form a fundamental set of solutions to the
equation, and can be used to construct all of its solutions.
The general solution is
y = c1e r1t + c2 e r2t
Example 7: Solutions
(1 of 2)
Consider the following differential equation:
2t 2 y′′ + 3t y′ − y = 0, t > 0
Show that the functions below are fundamental solutions:
y1 = t 1/ 2 , y2 = t −1
To show this, first substitute y1 into the equation:
⎛ − t −3 / 2 ⎞
⎛ t −1/ 2 ⎞ 1/ 2 ⎛ 1 3 ⎞ 1/ 2
⎟⎟ + 3t ⎜⎜
⎟⎟ − t = ⎜ − + − 1⎟t = 0
2t ⎜⎜
⎝ 2 2 ⎠
⎝ 4 ⎠
⎝ 2 ⎠
2
Thus y1 is a indeed a solution of the differential equation.
Similarly, y2 is also a solution:
( )
(
)
2t 2 2t −3 + 3t − t −2 − t −1 = (4 − 3 − 1)t −1 = 0
Example 7: Fundamental Solutions
(2 of 2)
Recall that
y1 = t 1/ 2 , y2 = t −1
To show that y1 and y2 form a fundamental set of solutions,
we evaluate the Wronskian of y1 and y2:
y1
W=
y1′
t 1/ 2
y2
= 1 −1/ 2
t
y2′
2
t −1
− t −2
1
3
3
= −t −3 / 2 − t −3 / 2 = − t −3 / 2 = −
2
2
2 t3
Since W ≠ 0 for t > 0, y1, y2 form a fundamental set of
solutions for the differential equation
2t 2 y′′ + 3t y′ − y = 0, t > 0
Theorem 3.2.5: Existence of Fundamental Set
of Solutions
Consider the differential equation below, whose coefficients
p and q are continuous on some open interval I:
L[ y ] = y′′ + p (t ) y′ + q (t ) y = 0
Let t0 be a point in I, and y1 and y2 solutions of the equation
with y1 satisfying initial conditions
y1 (t0 ) = 1, y1′ (t0 ) = 0
and y2 satisfying initial conditions
y2 (t0 ) = 0, y2′ (t0 ) = 1
Then y1, y2 form a fundamental set of solutions to the given
differential equation.
Example 7: Theorem 3.2.5
(1 of 3)
Find the fundamental set specified by Theorem 3.2.5 for the
differential equation and initial point
y′′ − y = 0, t0 = 0
We showed previously that
y1 = et , y2 = e − t
were fundamental solutions, since W(y1, y2)(t0) = -2 ≠ 0.
But these two solutions don’t satisfy the initial conditions
stated in Theorem 3.2.5, and thus they do not form the
fundamental set of solutions mentioned in that theorem.
Let y3 and y4 be the fundamental solutions of Thm 3.2.5.
y3 (0) = 1, y3′ (0) = 0; y4 (0) = 0, y4′ (0) = 1
Example 7: General Solution
(2 of 3)
Since y1 and y2 form a fundamental set of solutions,
y3 = c1et + c2 e − t , y3 (0) = 1, y3′ (0) = 0
y4 = d1et + d 2 e −t , y4 (0) = 0, y4′ (0) = 1
Solving each equation, we obtain
y3 (t ) =
1 t 1 −t
1
1
e + e = cosh(t ), y4 (t ) = e t − e −t = sinh(t )
2
2
2
2
The Wronskian of y3 and y4 is
y1
W=
y1′
y2 cosh t sinh t
=
= cosh 2 t − sinh 2 t = 1 ≠ 0
y2′ sinh t cosh t
Thus y3, y4 forms the fundamental set of solutions indicated
in Theorem 3.2.5, with general solution in this case
y (t ) = k1 cosh(t ) + k 2 sinh(t )
Example 7:
Many Fundamental Solution Sets
Thus
{
(3 of 3)
}
S1 = et , e − t , S 2 = { cosh t , sinh t}
both form fundamental solution sets to the differential
equation and initial point
y′′ − y = 0, t0 = 0
In general, a differential equation will have infinitely many
different fundamental solution sets. Typically, we pick the
one that is most convenient or useful.
Summary
To find a general solution of the differential equation
y′′ + p (t ) y′ + q (t ) y = 0, α < t < β
we first find two solutions y1 and y2.
Then make sure there is a point t0 in the interval such that
W(y1, y2)(t0) ≠ 0.
It follows that y1 and y2 form a fundamental set of solutions
to the equation, with general solution y = c1y1 + c2 y2.
If initial conditions are prescribed at a point t0 in the interval
where W ≠ 0, then c1 and c2 can be chosen to satisfy those
conditions.
Ch 3.3:
Linear Independence and the Wronskian
Two functions f and g are linearly dependent if there
exist constants c1 and c2, not both zero, such that
c1 f (t ) + c2 g (t ) = 0
for all t in I. Note that this reduces to determining whether
f and g are multiples of each other.
If the only solution to this equation is c1 = c2 = 0, then f
and g are linearly independent.
For example, let f(x) = sin2x and g(x) = sinxcosx, and
consider the linear combination
c1 sin 2 x + c2 sin x cos x = 0
This equation is satisfied if we choose c1 = 1, c2 = -2, and
hence f and g are linearly dependent.
Solutions of 2 x 2 Systems of Equations
When solving
c1 x1 + c2 x2 = a
c1 y1 + c2 y2 = b
for c1 and c2, it can be shown that
ay2 − bx2
ay2 − bx2
=
,
c1 =
x1 y2 − y1 x2
D
x1
− ay1 + bx1 − ay1 + bx1
c2 =
=
, where D =
y1
x1 y2 − y1 x2
D
x2
y2
Note that if a = b = 0, then the only solution to this system of
equations is c1 = c2 = 0, provided D ≠ 0.
Example 1: Linear Independence
(1 of 2)
Show that the following two functions are linearly
independent on any interval:
f (t ) = e t , g (t ) = e − t
Let c1 and c2 be scalars, and suppose
c1 f (t ) + c2 g (t ) = 0
for all t in an arbitrary interval (α, β ).
We want to show c1 = c2 = 0. Since the equation holds for
all t in (α, β ), choose t0 and t1 in (α, β ), where t0 ≠ t1. Then
c1et0 + c2 e − t0 = 0
c1et1 + c2 e −t1 = 0
Example 1: Linear Independence
(2 of 2)
The solution to our system of equations
c1e t0 + c2 e − t0 = 0
c1e t1 + c2 e −t1 = 0
will be c1 = c2 = 0, provided the determinant D is nonzero:
e t0
D= t
e1
e − t0
t 0 −t1
−t 0 t1
t 0 −t1
t1 −t 0
=
e
e
−
e
e
=
e
−
e
e −t1
Then
1
t 0 −t1
t1 − t 0
t 0 − t1
D=0 ⇔ e
=e
⇔ e
= t0 −t1 ⇔ e t0 −t1
e
⇔ e t0 −t1 = 1 ⇔ t0 = t1
(
)
2
=1
Since t0 ≠ t1, it follows that D ≠ 0, and therefore f and g are
linearly independent.
Theorem 3.3.1
If f and g are differentiable functions on an open interval I
and if W(f, g)(t0) ≠ 0 for some point t0 in I, then f and g are
linearly independent on I. Moreover, if f and g are
linearly dependent on I, then W(f, g)(t) = 0 for all t in I.
Proof (outline): Let c1 and c2 be scalars, and suppose
c1 f (t ) + c2 g (t ) = 0
for all t in I. In particular, when t = t0 we have
c1 f (t0 ) + c2 g (t0 ) = 0
c1 f ′(t0 ) + c2 g ′(t0 ) = 0
Since W(f, g)(t0) ≠ 0, it follows that c1 = c2 = 0, and hence
f and g are linearly independent.
Theorem 3.3.2 (Abel’s Theorem)
Suppose y1 and y2 are solutions to the equation
L[ y ] = y′′ + p (t ) y′ + q (t ) y = 0
where p and q are continuous on some open interval I. Then
W(y1,y2)(t) is given by
− ∫ p ( t ) dt
W ( y , y )(t ) = ce
1
2
where c is a constant that depends on y1 and y2 but not on t.
Note that W(y1,y2)(t) is either zero for all t in I (if c = 0) or
else is never zero in I (if c ≠ 0).
Example 2: Wronskian and Abel’s Theorem
Recall the following equation and two of its solutions:
y′′ − y = 0, y1 = e t , y2 = e − t
The Wronskian of y1and y2 is
y1 y2
W=
= −et e −t − et e −t = −2e 0 = −2 ≠ 0 for all t.
y1′ y2′
Thus y1 and y2 are linearly independent on any interval I, by
Theorem 3.3.1. Now compare W with Abel’s Theorem:
W ( y1 , y2 )(t ) = ce ∫
− p ( t ) dt
= ce ∫
− 0 dt
=c
Choosing c = -2, we get the same W as above.
Theorem 3.3.3
Suppose y1 and y2 are solutions to equation below, whose
coefficients p and q are continuous on some open interval I:
L[ y ] = y′′ + p (t ) y′ + q (t ) y = 0
Then y1 and y2 are linearly dependent on I iff W(y1, y2)(t) = 0
for all t in I. Also, y1 and y2 are linearly independent on I iff
W(y1, y2)(t) ≠ 0 for all t in I.
Summary
Let y1 and y2 be solutions of
y′′ + p (t ) y′ + q (t ) y = 0
where p and q are continuous on an open interval I.
Then the following statements are equivalent:
The functions y1 and y2 form a fundamental set of solutions on I.
The functions y1 and y2 are linearly independent on I.
W(y1,y2)(t0) ≠ 0 for some t0 in I.
W(y1,y2)(t) ≠ 0 for all t in I.
Linear Algebra Note
Let V be the set
V = {y : y′′ + p(t ) y′ + q (t ) y = 0, t ∈ (α , β ) }
Then V is a vector space of dimension two, whose bases are
given by any fundamental set of solutions y1 and y2.
For example, the solution space V to the differential equation
y′′ − y = 0
has bases
{
}
S1 = e t , e − t , S 2 = { cosh t , sinh t}
with
V = Span S1 = Span S 2
Ch 3.4:
Complex Roots of Characteristic Equation
Recall our discussion of the equation
ay′′ + by′ + cy = 0
where a, b and c are constants.
Assuming an exponential soln leads to characteristic equation:
y (t ) = e rt ⇒ ar 2 + br + c = 0
Quadratic formula (or factoring) yields two solutions, r1 & r2:
− b ± b 2 − 4ac
r=
2a
If b2 – 4ac < 0, then complex roots: r1 = λ + iμ, r2 = λ - iμ
Thus
y1 (t ) = e (λ +iμ )t , y2 (t ) = e (λ −iμ )t
Euler’s Formula; Complex Valued Solutions
Substituting it into Taylor series for et, we obtain Euler’s
formula:
n
n 2n
n −1 2 n −1
∞
∞
(
it
)
(
−
1
)
t
(
−
1
)
t
eit = ∑
=∑
+ i∑
= cos t + i sin t
(2n )! n=1 (2n − 1)!
n =0 n !
n =0
∞
Generalizing Euler’s formula, we obtain
eiμt = cos μt + i sin μt
Then
e (λ +iμ )t = e λt eiμt = e λt [cos μt + i sin μt ] = e λt cos μt + ie λt sin μt
Therefore
y1 (t ) = e (λ +iμ )t = e λ t cos μt + ie λ t sin μt
y2 (t ) = e (λ −iμ )t = e λ t cos μt − ie λ t sin μt
Real Valued Solutions
Our two solutions thus far are complex-valued functions:
y1 (t ) = e λ t cos μt + ie λ t sin μt
y2 (t ) = e λ t cos μt − ie λ t sin μt
We would prefer to have real-valued solutions, since our
differential equation has real coefficients.
To achieve this, recall that linear combinations of solutions
are themselves solutions:
y1 (t ) + y2 (t ) = 2e λ t cos μ t
y1 (t ) − y2 (t ) = 2ie λ t sin μ t
Ignoring constants, we obtain the two solutions
y3 (t ) = e λ t cos μ t , y4 (t ) = e λ t sin μ t
Real Valued Solutions: The Wronskian
Thus we have the following real-valued functions:
y3 (t ) = e λ t cos μt , y4 (t ) = e λ t sin μt
Checking the Wronskian, we obtain
e λt cos μt
e λt sin μt
W = λt
e (λ cos μt − μ sin μt ) e λt (λ sin μt + μ cos μt )
= μ e 2 λt ≠ 0
Thus y3 and y4 form a fundamental solution set for our ODE,
and the general solution can be expressed as
y (t ) = c1e λ t cos μt + c2 e λ t sin μt
Example 1
Consider the equation
y′′ + y′ + y = 0
Then
y (t ) = e rt ⇒ r 2 + r + 1 = 0 ⇔ r =
−1± 1− 4 −1 ± 3 i
1
3
=
=− ±
i
2
2
2 2
Therefore
λ = −1 / 2, μ = 3 / 2
and thus the general solution is
(
)
(
y (t ) = c1e − t / 2 cos 3t / 2 + c2 e − t / 2 sin 3t / 2
)
Example 2
Consider the equation
y′′ + 4 y = 0
Then
y (t ) = e rt ⇒ r 2 + 4 = 0 ⇔ r = ±2 i
Therefore
λ = 0, μ = 2
and thus the general solution is
y (t ) = c1 cos(2t ) + c2 sin (2t )
Example 3
Consider the equation
3 y′′ − 2 y′ + y = 0
Then
y (t ) = e rt ⇒ 3 r 2 − 2r + 1 = 0 ⇔ r =
Therefore the general solution is
(
)
(
2 ± 4 − 12 1
2
= ±
i
6
3 3
)
y (t ) = c1e t / 3 cos 2t / 3 + c2 e t / 3 sin 2t / 3
Example 4: Part (a)
(1 of 2)
For the initial value problem below, find (a) the solution u(t)
and (b) the smallest time T for which |u(t)| ≤ 0.1
y′′ + y′ + y = 0, y (0) = 1, y′(0) = 1
We know from Example 1 that the general solution is
(
)
(
u (t ) = c1e − t / 2 cos 3t / 2 + c2 e − t / 2 sin 3t / 2
Using the initial conditions, we obtain
= 1⎫
3
⎪
= 3
⎬ ⇒ c1 = 1, c2 =
1
3
3
− c1 +
c2 = 1 ⎪
2
2
⎭
c1
Thus
(
)
(
u (t ) = e − t / 2 cos 3t / 2 + 3 e − t / 2 sin 3t / 2
)
)
Example 4: Part (b)
(2 of 2)
Find the smallest time T for which |u(t)| ≤ 0.1
Our solution is
(
)
(
u (t ) = e − t / 2 cos 3t / 2 + 3 e − t / 2 sin 3t / 2
)
With the help of graphing calculator or computer algebra
system, we find that T ≅ 2.79. See graph below.
Ch 3.5:
Repeated Roots; Reduction of Order
Recall our 2nd order linear homogeneous ODE
ay′′ + by′ + cy = 0
where a, b and c are constants.
Assuming an exponential soln leads to characteristic equation:
y (t ) = e rt ⇒ ar 2 + br + c = 0
Quadratic formula (or factoring) yields two solutions, r1 & r2:
− b ± b 2 − 4ac
r=
2a
When b2 – 4ac = 0, r1 = r2 = -b/2a, since method only gives
one solution:
y1 (t ) = ce −b t / 2 a
Second Solution: Multiplying Factor v(t)
We know that
y1 (t ) a solution ⇒ y2 (t ) = cy1 (t ) a solution
Since y1 and y2 are linearly dependent, we generalize this
approach and multiply by a function v, and determine
conditions for which y2 is a solution:
y1 (t ) = e −b t / 2 a a solution ⇒ try y2 (t ) = v(t )e −b t / 2 a
Then
y2 (t ) = v(t )e − b t / 2 a
b
v(t )e −b t / 2 a
2a
b2
b
b
−b t / 2 a
−b t / 2 a
−b t / 2 a
− v′(t )e
− v′(t )e
+ 2 v(t )e −b t / 2 a
y2′′(t ) = v′′(t )e
2a
2a
4a
y2′ (t ) = v′(t )e −b t / 2 a −
ay′′ + by′ + cy = 0
Finding Multiplying Factor v(t)
Substituting derivatives into ODE, we seek a formula for v:
⎧ ⎡
⎫
⎤ ⎡
b
b2
b
⎤
e
⎨a ⎢v′′(t ) − v′(t ) + 2 v(t )⎥ + b ⎢v′(t ) − v(t )⎥ + cv(t )⎬ = 0
4a
2a
a
⎦
⎦ ⎣
⎩ ⎣
⎭
b2
b2
av′′(t ) − bv′(t ) +
v(t ) + bv′(t ) − v(t ) + cv(t ) = 0
4a
2a
⎛ b2 b2
⎞
+ c ⎟⎟v(t ) = 0
av′′(t ) + ⎜⎜ −
⎝ 4a 2a
⎠
−b t / 2 a
⎛ b 2 2b 2 4ac ⎞
⎛ − b 2 4ac ⎞
⎟⎟v(t ) = 0 ⇔ av′′(t ) + ⎜⎜
⎟⎟v(t ) = 0
+
+
av′′(t ) + ⎜⎜ −
4a ⎠
⎝ 4a 4a 4a ⎠
⎝ 4a
⎛ b 2 − 4ac ⎞
⎟⎟v(t ) = 0
av′′(t ) − ⎜⎜
⎝ 4a ⎠
v′′(t ) = 0 ⇒ v(t ) = k3t + k 4
General Solution
To find our general solution, we have:
y (t ) = k1e −bt / 2 a + k 2 v(t )e −bt / 2 a
= k1e −bt / 2 a + (k3t + k 4 )e −bt / 2 a
= c1e −bt / 2 a + c2te −bt / 2 a
Thus the general solution for repeated roots is
y (t ) = c1e −bt / 2 a + c2te −bt / 2 a
Wronskian
The general solution is
y (t ) = c1e − bt / 2 a + c2te −bt / 2 a
Thus every solution is a linear combination of
y1 (t ) = e − bt / 2 a , y2 (t ) = te −bt / 2 a
The Wronskian of the two solutions is
e − bt / 2 a
W ( y1 , y2 )(t ) = b −bt / 2 a
− e
2a
te −bt / 2 a
⎛ bt ⎞ −bt / 2 a
⎜1 − ⎟e
⎝ 2a ⎠
⎛ bt ⎞ −bt / a ⎛ bt ⎞
=e
⎜ ⎟
⎜1 − ⎟ + e
⎝ 2a ⎠
⎝ 2a ⎠
= e −bt / a ≠ 0 for all t
−bt / a
Thus y1 and y2 form a fundamental solution set for equation.
Example 1
Consider the initial value problem
y′′ + 2 y′ + y = 0, y (0 ) = 1, y ′(0 ) = 1
Assuming exponential soln leads to characteristic equation:
y (t ) = e rt ⇒ r 2 + 2r + 1 = 0 ⇔ (r + 1) 2 = 0 ⇔ r = −1
Thus the general solution is
y (t ) = c1e − t + c2te − t
Using the initial conditions:
c1
− c1
+ c2
= 1⎫
⎬ ⇒ c1 = 1, c2 = 2
= 1⎭
Thus
y (t ) = e − t + 2te − t
Example 2
Consider the initial value problem
y ′′ − y ′ + 0.25 y = 0, y (0 ) = 2, y ′(0 ) = 1 / 2
Assuming exponential soln leads to characteristic equation:
y (t ) = e rt ⇒ r 2 − r + 0.25 = 0 ⇔ (r − 1 / 2) 2 = 0 ⇔ r = 1 / 2
Thus the general solution is
y (t ) = c1et / 2 + c2te t / 2
Using the initial conditions:
=
c1
1
c1
2
Thus
+ c2
=
2⎫
1
⎪
1 ⎬ ⇒ c1 = 2, c2 = −
2
2 ⎪⎭
1
y (t ) = 2et / 2 − te t / 2
2
Example 3
Consider the initial value problem
y′′ − y′ + 0.25 y = 0, y (0 ) = 2, y′(0 ) = 3 / 2
Assuming exponential soln leads to characteristic equation:
y (t ) = e rt ⇒ r 2 − r + 0.25 = 0 ⇔ (r − 1 / 2) 2 = 0 ⇔ r = 1 / 2
Thus the general solution is
y (t ) = c1et / 2 + c2te t / 2
Using the initial conditions:
=
c1
1
c1
2
Thus
+ c2
=
2⎫
1
⎪
3 ⎬ ⇒ c1 = 2, c2 =
2
2 ⎪⎭
1
y (t ) = 2et / 2 + te t / 2
2
Reduction of Order
The method used so far in this section also works for
equations with nonconstant coefficients:
y ′′ + p (t ) y′ + q (t ) y = 0
That is, given that y1 is solution, try y2 = v(t)y1:
y2 (t ) = v(t ) y1 (t )
y2′ (t ) = v′(t ) y1 (t ) + v(t ) y1′ (t )
y2′′(t ) = v′′(t ) y1 (t ) + 2v′(t ) y1′ (t ) + v(t ) y1′′(t )
Substituting these into ODE and collecting terms,
y1v′′ + (2 y1′ + py1 )v′ + ( y1′′ + py1′ + qy1 )v = 0
Since y1 is a solution to the differential equation, this last
equation reduces to a first order equation in v′ :
y1v′′ + (2 y1′ + py1 ) v′ = 0
Example 4: Reduction of Order
(1 of 3)
Given the variable coefficient equation and solution y1,
t 2 y′′ + 3ty′ + y = 0, t > 0; y1 (t ) = t −1 ,
use reduction of order method to find a second solution:
y2 (t ) = v(t ) t −1
y2′ (t ) = v′(t ) t −1 − v(t ) t − 2
y2′′(t ) = v′′(t ) t −1 − 2v′(t ) t − 2 + 2v(t ) t −3
Substituting these into ODE and collecting terms,
(
) (
)
t 2 v′′t −1 − 2v′t −2 + 2vt −3 + 3t v′t −1 − vt −2 + vt −1 = 0
⇔ v′′t − 2v′ + 2vt −1 + 3v′ − 3vt −1 + vt −1 = 0
⇔ tv′′ + v′ = 0
⇔ tu′ + u = 0, where u (t ) = v′(t )
Example 4: Finding v(t)
(2 of 3)
To solve
tu′ + u = 0, u (t ) = v′(t )
for u, we can use the separation of variables method:
du
t
+u = 0 ⇔
dt
⇔
−1
u = t eC
Thus
v′ =
du
1
∫ u = −∫ t dt ⇔ ln u = − ln t + C
⇔ u = ct −1 , since t > 0.
c
t
and hence
v(t ) = c ln t + k
Example 4: General Solution
(3 of 3)
We have
v(t ) = c ln t + k
Thus
y2 (t ) = (c ln t + k )t −1 = ct −1 ln t + k t −1
Recall
y1 (t ) = t −1
and hence we can neglect the second term of y2 to obtain
y2 (t ) = t −1 ln t.
Hence the general solution to the differential equation is
y (t ) = c1t −1 + c2t −1 ln t
Ch 3.6: Nonhomogeneous Equations;
Method of Undetermined Coefficients
Recall the nonhomogeneous equation
y′′ + p (t ) y′ + q (t ) y = g (t )
where p, q, g are continuous functions on an open interval I.
The associated homogeneous equation is
y′′ + p (t ) y′ + q (t ) y = 0
In this section we will learn the method of undetermined
coefficients to solve the nonhomogeneous equation, which
relies on knowing solutions to homogeneous equation.
Theorem 3.6.1
If Y1, Y2 are solutions of nonhomogeneous equation
y′′ + p (t ) y ′ + q (t ) y = g (t )
then Y1 - Y2 is a solution of the homogeneous equation
y′′ + p (t ) y′ + q (t ) y = 0
If y1, y2 form a fundamental solution set of homogeneous
equation, then there exists constants c1, c2 such that
Y1 (t ) − Y2 (t ) = c1 y1 (t ) + c2 y2 (t )
Theorem 3.6.2 (General Solution)
The general solution of nonhomogeneous equation
y′′ + p (t ) y ′ + q (t ) y = g (t )
can be written in the form
y (t ) = c1 y1 (t ) + c2 y2 (t ) + Y (t )
where y1, y2 form a fundamental solution set of homogeneous
equation, c1, c2 are arbitrary constants and Y is a specific
solution to the nonhomogeneous equation.
Method of Undetermined Coefficients
Recall the nonhomogeneous equation
y′′ + p (t ) y′ + q (t ) y = g (t )
with general solution
y (t ) = c1 y1 (t ) + c2 y2 (t ) + Y (t )
In this section we use the method of undetermined
coefficients to find a particular solution Y to the
nonhomogeneous equation, assuming we can find solutions
y1, y2 for the homogeneous case.
The method of undetermined coefficients is usually limited
to when p and q are constant, and g(t) is a polynomial,
exponential, sine or cosine function.
Example 1: Exponential g(t)
Consider the nonhomogeneous equation
y′′ − 3 y′ − 4 y = 3e 2t
We seek Y satisfying this equation. Since exponentials
replicate through differentiation, a good start for Y is:
Y (t ) = Ae 2t ⇒ Y ′(t ) = 2 Ae 2t , Y ′′(t ) = 4 Ae 2t
Substituting these derivatives into differential equation,
4 Ae 2t − 6 Ae 2t − 4 Ae 2t = 3e 2t
⇔ − 6 Ae 2t = 3e 2t
⇔ A = −1 / 2
Thus a particular solution to the nonhomogeneous ODE is
1 2t
Y (t ) = − e
2
Example 2: Sine g(t), First Attempt
(1 of 2)
Consider the nonhomogeneous equation
y ′′ − 3 y ′ − 4 y = 2 sin t
We seek Y satisfying this equation. Since sines replicate
through differentiation, a good start for Y is:
Y (t ) = A sin t ⇒ Y ′(t ) = A cos t , Y ′′(t ) = − A sin t
Substituting these derivatives into differential equation,
− A sin t − 3 A cos t − 4 A sin t = 2 sin t
⇔ (2 + 5 A)sin t + 3 A cos t = 0
⇔ c1 sin t + c2 cos t = 0
Since sin(x) and cos(x) are linearly independent (they are not
multiples of each other), we must have c1= c2 = 0, and hence
2 + 5A = 3A = 0, which is impossible.
y′′ − 3 y ′ − 4 y = 2 sin t
Example 2: Sine g(t), Particular Solution
(2 of 2)
Our next attempt at finding a Y is
Y (t ) = A sin t + B cos t
⇒ Y ′(t ) = A cos t − B sin t , Y ′′(t ) = − A sin t − B cos t
Substituting these derivatives into ODE, we obtain
(− A sin t − B cos t ) − 3( A cos t − B sin t ) − 4( A sin t + B cos t ) = 2 sin t
⇔ (− 5 A + 3B )sin t + (− 3 A − 5B ) cos t = 2 sin t
⇔ − 5 A + 3B = 2, − 3 A − 5 B = 0
⇔ A = −5 / 17, B = 3 / 17
Thus a particular solution to the nonhomogeneous ODE is
3
−5
Y (t ) =
sin t + cos t
17
17
Example 3: Polynomial g(t)
Consider the nonhomogeneous equation
y′′ − 3 y′ − 4 y = 4t 2 − 1
We seek Y satisfying this equation. We begin with
Y (t ) = At 2 + Bt + C ⇒ Y ′(t ) = 2 At + B, Y ′′(t ) = 2 A
Substituting these derivatives into differential equation,
(
)
2 A − 3(2 At + B ) − 4 At 2 + Bt + C = 4t 2 − 1
⇔ − 4 At 2 − (6 A + 4 B )t + (2 A − 3B − 4C ) = 4t 2 − 1
⇔ − 4 A = 4, 6 A + 4 B = 0, 2 A − 3B − 4C = −1
⇔ A = −1, B = 3 / 2 , C = −11 / 8
Thus a particular solution to the nonhomogeneous ODE is
3 11
Y (t ) = − t 2 + t −
2
8
Example 4: Product g(t)
Consider the nonhomogeneous equation
y′′ − 3 y′ − 4 y = −8e t cos 2t
We seek Y satisfying this equation, as follows:
Y (t ) = Aet cos 2t + Bet sin 2t
Y ′(t ) = Aet cos 2t − 2 Aet sin 2t + Bet sin 2t + 2 Bet cos 2t
= ( A + 2 B )et cos 2t + (− 2 A + B )et sin 2t
Y ′′(t ) = ( A + 2 B )et cos 2t − 2( A + 2 B )et sin 2t + (− 2 A + B )et sin 2t
+ 2(− 2 A + B )et cos 2t
= (− 3 A + 4 B )et cos 2t + (− 4 A − 3 B )et sin 2t
Substituting derivatives into ODE and solving for A and B:
A=
2
10
2
10
, B=
⇒ Y (t ) = et cos 2t + et sin 2t
13
13
13
13
Discussion: Sum g(t)
Consider again our general nonhomogeneous equation
y′′ + p (t ) y′ + q (t ) y = g (t )
Suppose that g(t) is sum of functions:
g (t ) = g1 (t ) + g 2 (t )
If Y1, Y2 are solutions of
y′′ + p (t ) y′ + q (t ) y = g1 (t )
y′′ + p (t ) y′ + q (t ) y = g 2 (t )
respectively, then Y1 + Y2 is a solution of the
nonhomogeneous equation above.
Example 5: Sum g(t)
Consider the equation
y′′ − 3 y′ − 4 y = 3e 2t + 2 sin t − 8e t cos 2t
Our equations to solve individually are
y′′ − 3 y′ − 4 y = 3e 2t
y′′ − 3 y′ − 4 y = 2 sin t
y′′ − 3 y′ − 4 y = −8e t cos 2t
Our particular solution is then
1
3
5
10
2
Y (t ) = − e 2t + cos t − sin t + et cos 2t + et sin 2t
2
17
17
13
13
Example 6: First Attempt
(1 of 3)
Consider the equation
y′′ + 4 y = 3 cos 2t
We seek Y satisfying this equation. We begin with
Y (t ) = A sin 2t + B cos 2t
⇒ Y ′(t ) = 2 A cos 2t − 2 B sin 2t , Y ′′(t ) = −4 A sin 2t − 4 B cos 2t
Substituting these derivatives into ODE:
(− 4 A sin 2t − 4 B cos 2t ) + 4( A sin 2t + B cos 2t ) = 3 cos 2t
(− 4 A + 4 A)sin 2t + (− 4 B + 4 B ) cos 2t = 3 cos 2t
0 = 3 cos 2t
Thus no particular solution exists of the form
Y (t ) = A sin 2t + B cos 2t
Example 6: Homogeneous Solution
(2 of 3)
Thus no particular solution exists of the form
Y (t ) = A sin 2t + B cos 2t
To help understand why, recall that we found the
corresponding homogeneous solution in Section 3.4 notes:
y′′ + 4 y = 0 ⇒ y (t ) = c1 cos 2t + c2 sin 2t
Thus our assumed particular solution solves homogeneous
equation
y′′ + 4 y = 0
instead of the nonhomogeneous equation.
y′′ + 4 y = 3 cos 2t
y′′ + 4 y = 3 cos 2t
Example 6: Particular Solution
(3 of 3)
Our next attempt at finding a Y is:
Y (t ) = At sin 2t + Bt cos 2t
Y ′(t ) = A sin 2t + 2 At cos 2t + B cos 2t − 2 Bt sin 2t
Y ′′(t ) = 2 A cos 2t + 2 A cos 2t − 4 At sin 2t − 2 B sin 2t − 2 B sin 2t − 4 Bt cos 2t
= 4 A cos 2t − 4 B sin 2t − 4 At sin 2t − 4 Bt cos 2t
Substituting derivatives into ODE,
4 A cos 2t − 4 B sin 2t = 3 cos 2t
⇒ A = 3 / 4, B = 0
3
⇒ Y (t ) = t sin 2t
4
Ch 3.7: Variation of Parameters
Recall the nonhomogeneous equation
y′′ + p (t ) y′ + q (t ) y = g (t )
where p, q, g are continuous functions on an open interval I.
The associated homogeneous equation is
y′′ + p (t ) y′ + q (t ) y = 0
In this section we will learn the variation of parameters
method to solve the nonhomogeneous equation. As with the
method of undetermined coefficients, this procedure relies on
knowing solutions to homogeneous equation.
Variation of parameters is a general method, and requires no
detailed assumptions about solution form. However, certain
integrals need to be evaluated, and this can present difficulties.
Example: Variation of Parameters
(1 of 6)
We seek a particular solution to the equation below.
y′′ + 4 y = 3 csc t
We cannot use method of undetermined coefficients since
g(t) is a quotient of sint or cost, instead of a sum or product.
Recall that the solution to the homogeneous equation is
yC (t ) = c1 cos 2t + c2 sin 2t
To find a particular solution to the nonhomogeneous
equation, we begin with the form
y (t ) = u1 (t ) cos 2t + u 2 (t ) sin 2t
Then
y′(t ) = u1′ (t ) cos 2t − 2u1 (t ) sin 2t + u 2′ (t ) sin 2t + 2u 2 (t ) cos 2t
or
y′(t ) = −2u1 (t ) sin 2t + 2u 2 (t ) cos 2t + u1′ (t ) cos 2t + u 2′ (t ) sin 2t
Example: Derivatives, 2nd Equation
(2 of 6)
From the previous slide,
y′(t ) = −2u1 (t ) sin 2t + 2u 2 (t ) cos 2t + u1′ (t ) cos 2t + u 2′ (t ) sin 2t
Note that we need two equations to solve for u1 and u2. The
first equation is the differential equation. To get a second
equation, we will require
u1′ (t ) cos 2t + u 2′ (t ) sin 2t = 0
Then
y′(t ) = −2u1 (t ) sin 2t + 2u 2 (t ) cos 2t
Next,
y′′(t ) = −2u1′ (t ) sin 2t − 4u1 (t ) cos 2t + 2u 2′ (t ) cos 2t − 4u 2 (t ) sin 2t
Example: Two Equations
(3 of 6)
Recall that our differential equation is
y′′ + 4 y = 3 csc t
Substituting y'' and y into this equation, we obtain
− 2u1′ (t ) sin 2t − 4u1 (t ) cos 2t + 2u 2′ (t ) cos 2t − 4u 2 (t ) sin 2t
+ 4(u1 (t ) cos 2t + u 2 (t ) sin 2t ) = 3 csc t
This equation simplifies to
− 2u1′ (t ) sin 2t + 2u 2′ (t ) cos 2t = 3 csc t
Thus, to solve for u1 and u2, we have the two equations:
− 2u1′ (t ) sin 2t + 2u 2′ (t ) cos 2t = 3 csc t
u1′ (t ) cos 2t + u 2′ (t ) sin 2t = 0
Example: Solve for u1'
(4 of 6)
To find u1 and u2 , we need to solve the equations
− 2u1′ (t ) sin 2t + 2u 2′ (t ) cos 2t = 3 csc t
u1′ (t ) cos 2t + u 2′ (t ) sin 2t = 0
From second equation,
u 2′ (t ) = −u1′ (t )
cos 2t
sin 2t
Substituting this into the first equation,
cos 2t ⎤
⎡
cos 2t = 3 csc t
− 2u1′ (t ) sin 2t + 2 ⎢− u1′ (t )
⎥
sin 2t ⎦
⎣
− 2u1′ (t ) sin 2 (2t ) − 2u1′ (t ) cos 2 (2t ) = 3 csc t sin 2t
[
]
⎡ 2 sin t cos t ⎤
2
2
′
− 2u1 (t ) sin (2t ) + cos (2t ) = 3⎢
⎣ sin t ⎥⎦
u1′ (t ) = −3 cos t
Example : Solve for u1 and u2
(5 of 6)
From the previous slide,
u1′ (t ) = −3 cos t ,
u 2′ (t ) = −u1′ (t )
cos 2t
sin 2t
Then
⎡ 1 − 2 sin 2 t ⎤
⎡1 − 2 sin 2 t ⎤
cos 2t ⎤
⎡
= 3 cos t ⎢
u 2′ (t ) = 3 cos t ⎢
⎥ = 3⎢
⎥
⎥
2
sin
t
cos
t
2
sin
t
⎣ sin 2t ⎦
⎣
⎦
⎣
⎦
⎡ 1
2 sin 2 t ⎤ 3
= 3⎢
−
⎥ = csc t − 3 sin t
⎣ 2 sin t 2 sin t ⎦ 2
Thus
u1 (t ) = ∫ u1′ (t ) dt = ∫ − 3 cos tdt = − 3 sin t + c1
3
⎛3
⎞
u 2 (t ) = ∫ u 2′ (t )dt = ∫ ⎜ csc t − 3 sin t ⎟dt = ln csc t − cot t + 3 cos t + c2
2
⎝2
⎠
Example: General Solution
(6 of 6)
Recall our equation and homogeneous solution yC:
y′′ + 4 y = 3 csc t , yC (t ) = c1 cos 2t + c2 sin 2t
Using the expressions for u1 and u2 on the previous slide, the
general solution to the differential equation is
y (t ) = u1 (t ) cos 2t + u 2 (t ) sin 2t + yC (t )
3
= −3 sin t cos 2t + ln csc t − cot t sin 2t + 3 cos t sin 2t + yC (t )
2
3
= 3 [cos t sin 2t − sin t cos 2t ] + ln csc t − cot t sin 2t + yC (t )
2
3
= 3 2 sin t cos 2 t − sin t 2 cos 2 t − 1 + ln csc t − cot t sin 2t + yC (t )
2
3
= 3 sin t + ln csc t − cot t sin 2t + c1 cos 2t + c2 sin 2t
2
[
(
)]
y′′ + p (t ) y′ + q (t ) y = g (t )
Summary
y (t ) = u1 (t ) y1 (t ) + u 2 (t ) y2 (t )
Suppose y1, y2 are fundamental solutions to the homogeneous
equation associated with the nonhomogeneous equation
above, where we note that the coefficient on y'' is 1.
To find u1 and u2, we need to solve the equations
u1′ (t ) y1 (t ) + u 2′ (t ) y2 (t ) = 0
u1′ (t ) y1′ (t ) + u 2′ (t ) y2′ (t ) = g (t )
Doing so, and using the Wronskian, we obtain
u1′ (t ) = −
y2 (t ) g (t )
y (t ) g (t )
, u 2′ (t ) = 1
W ( y1 , y2 )(t )
W ( y1 , y2 )(t )
Thus
u1 (t ) = − ∫
y (t ) g (t )
y2 (t ) g (t )
dt + c1 , u2 (t ) = ∫ 1
dt + c2
W ( y1 , y2 )(t )
W ( y1 , y2 )(t )
Theorem 3.7.1
Consider the equations
y′′ + p (t ) y′ + q (t ) y = g (t )
y′′ + p (t ) y′ + q (t ) y = 0
(1)
( 2)
If the functions p, q and g are continuous on an open interval I,
and if y1 and y2 are fundamental solutions to Eq. (2), then a
particular solution of Eq. (1) is
y2 (t ) g (t )
y1 (t ) g (t )
Y (t ) = − y1 (t ) ∫
dt + y2 (t ) ∫
dt
W ( y1 , y2 )(t )
W ( y1 , y2 )(t )
and the general solution is
y (t ) = c1 y1 (t ) + c2 y2 (t ) + Y (t )
Ch 3.8: Mechanical & Electrical Vibrations
Two important areas of application for second order linear
equations with constant coefficients are in modeling
mechanical and electrical oscillations.
We will study the motion of a mass on a spring in detail.
An understanding of the behavior of this simple system is the
first step in investigation of more complex vibrating systems.
Spring – Mass System
Suppose a mass m hangs from vertical spring of original
length l. The mass causes an elongation L of the spring.
The force FG of gravity pulls mass down. This force has
magnitude mg, where g is acceleration due to gravity.
The force FS of spring stiffness pulls mass up. For small
elongations L, this force is proportional to L.
That is, Fs = kL (Hooke’s Law).
Since mass is in equilibrium, the forces balance each other:
mg = kL
Spring Model
We will study motion of mass when it is acted on by an
external force (forcing function) or is initially displaced.
Let u(t) denote the displacement of the mass from its
equilibrium position at time t, measured downward.
Let f be the net force acting on mass. Newton’s 2nd Law:
mu ′′(t ) = f (t )
In determining f, there are four separate forces to consider:
Weight:
w = mg
(downward force)
Spring force: Fs = - k(L+ u) (up or down force, see next slide)
Damping force: Fd(t) = - γ u′ (t) (up or down, see following slide)
External force: F (t)
(up or down force, see text)
Spring Model:
Spring Force Details
The spring force Fs acts to restore spring to natural position,
and is proportional to L + u. If L + u > 0, then spring is
extended and the spring force acts upward. In this case
Fs = − k ( L + u )
If L + u < 0, then spring is compressed a distance of |L + u|,
and the spring force acts downward. In this case
Fs = k L + u = k [− (L + u )] = − k (L + u )
In either case,
Fs = − k ( L + u )
Spring Model:
Damping Force Details
The damping or resistive force Fd acts in opposite direction as
motion of mass. Can be complicated to model.
Fd may be due to air resistance, internal energy dissipation
due to action of spring, friction between mass and guides, or a
mechanical device (dashpot) imparting resistive force to mass.
We keep it simple and assume Fd is proportional to velocity.
In particular, we find that
If u′ > 0, then u is increasing, so mass is moving downward. Thus Fd
acts upward and hence Fd = - γ u′, where γ > 0.
If u′ < 0, then u is decreasing, so mass is moving upward. Thus Fd
acts downward and hence Fd = - γ u′ , γ > 0.
In either case,
Fd (t ) = −γ u ′(t ), γ > 0
Spring Model:
Differential Equation
Taking into account these forces, Newton’s Law becomes:
mu ′′(t ) = mg + Fs (t ) + Fd (t ) + F (t )
= mg − k [L + u (t )] − γ u ′(t ) + F (t )
Recalling that mg = kL, this equation reduces to
mu ′′(t ) + γ u ′(t ) + ku (t ) = F (t )
where the constants m, γ, and k are positive.
We can prescribe initial conditions also:
u (0) = u0 , u ′(0) = v0
It follows from Theorem 3.2.1 that there is a unique solution
to this initial value problem. Physically, if mass is set in
motion with a given initial displacement and velocity, then
its position is uniquely determined at all future times.
Example 1:
Find Coefficients
(1 of 2)
A 4 lb mass stretches a spring 2". The mass is displaced an
additional 6" and then released; and is in a medium that exerts
a viscous resistance of 6 lb when velocity of mass is 3 ft/sec.
Formulate the IVP that governs motion of this mass:
mu′′(t ) + γ u ′(t ) + ku (t ) = F (t ), u (0) = u0 , u ′(0) = v0
Find m:
w
4 lb
1 lb sec 2
w = mg ⇒ m = ⇒ m =
⇒ m=
2
g
32ft / sec
8 ft
Find γ :
γ u ′ = 6 lb ⇒ γ =
6 lb
lb sec
⇒γ =2
3ft / sec
ft
Fs = − k L ⇒ k =
lb
4 lb
4 lb
⇒ k=
⇒ k = 24
ft
2 in
1 / 6 ft
Find k:
Example 1: Find IVP
(2 of 2)
Thus our differential equation becomes
1
u ′′(t ) + 2 u ′(t ) + 24u (t ) = 0
8
and hence the initial value problem can be written as
u ′′(t ) + 16 u ′(t ) + 192u (t ) = 0
1
u (0) = , u ′(0) = 0
2
This problem can be solved using
methods of Chapter 3.4. Given
on right is the graph of solution.
Spring Model:
Undamped Free Vibrations
(1 of 4)
Recall our differential equation for spring motion:
mu ′′(t ) + γ u ′(t ) + ku (t ) = F (t )
Suppose there is no external driving force and no damping.
Then F(t) = 0 and γ = 0, and our equation becomes
mu ′′(t ) + ku (t ) = 0
The general solution to this equation is
u (t ) = A cos ω0t + B sin ω0t ,
where
ω02 = k / m
Spring Model:
Undamped Free Vibrations
(2 of 4)
Using trigonometric identities, the solution
u (t ) = A cos ω0t + B sin ω0t , ω02 = k / m
can be rewritten as follows:
u (t ) = A cos ω0t + B sin ω0t ⇔ u (t ) = R cos(ω0t − δ )
⇔ u (t ) = R cos δ cos ω0t + R sin δ sin ω0t ,
where
B
A = R cos δ , B = R sin δ ⇒ R = A + B , tan δ =
A
2
2
Note that in finding δ, we must be careful to choose correct
quadrant. This is done using the signs of cos δ and sin δ.
Spring Model:
Undamped Free Vibrations
(3 of 4)
Thus our solution is
u (t ) = A cos ω0t + B sin ω0t = R cos(ω0t − δ )
where
ω0 = k / m
The solution is a shifted cosine (or sine) curve, that describes
simple harmonic motion, with period
m
2π
T=
= 2π
k
ω0
The circular frequency ω0 (radians/time) is natural frequency
of the vibration, R is the amplitude of max displacement of
mass from equilibrium, and δ is the phase (dimensionless).
Spring Model:
Undamped Free Vibrations
(4 of 4)
Note that our solution
u (t ) = A cos ω0t + B sin ω0t = R cos(ω0t − δ ), ω0 = k / m
is a shifted cosine (or sine) curve with period
m
T = 2π
k
Initial conditions determine A & B, hence also the amplitude R.
The system always vibrates with same frequency ω0 , regardless
of initial conditions.
The period T increases as m increases, so larger masses vibrate
more slowly. However, T decreases as k increases, so stiffer
springs cause system to vibrate more rapidly.
Example 2: Find IVP
(1 of 3)
A 10 lb mass stretches a spring 2". The mass is displaced an
additional 2" and then set in motion with initial upward
velocity of 1 ft/sec. Determine position of mass at any later
time. Also find period, amplitude, and phase of the motion.
mu′′(t ) + ku (t ) = 0, u (0) = u0 , u ′(0) = v0
Find m:
Find k:
10 lb
5 lb sec 2
w
⇒ m=
w = mg ⇒ m = ⇒ m =
2
32ft / sec
16 ft
g
Fs = − k L ⇒ k =
10 lb
10 lb
lb
⇒ k=
⇒ k = 60
2 in
1 / 6 ft
ft
Thus our IVP is
5 / 16 u ′′(t ) + 60u (t ) = 0, u (0) = 1 / 6, u ′(t ) = −1
Example 2: Find Solution
(2 of 3)
Simplifying, we obtain
u ′′(t ) + 192u (t ) = 0, u (0) = 1 / 6, u ′(0) = −1
To solve, use methods of Ch 3.4 to obtain
1
1
u (t ) = cos 192t −
sin 192t
6
192
or
1
1
u (t ) = cos 8 3t −
sin 8 3t
6
8 3
1
1
u (t ) = cos 8 3t −
sin 8 3t
6
8 3
Example 2:
Find Period, Amplitude, Phase
(3 of 3)
The natural frequency is
ω0 = k / m = 192 = 8 3 ≅ 13.856 rad/sec
The period is
T = 2π / ω0 ≅ 0.45345 sec
The amplitude is
R = A2 + B 2 ≅ 0.18162 ft
Next, determine the phase δ :
A = R cos δ , B = R sin δ , tan δ = B / A
⎛
⎞
B
− 3
−1 − 3
⎟ ≅ −0.40864 rad
tan δ =
⇒ tan δ =
⇒ δ = tan ⎜⎜
⎟
A
4
⎝ 4 ⎠
(
Thus u (t ) = 0.182 cos 8 3t + 0.409
)
Spring Model: Damped Free Vibrations
(1 of 8)
Suppose there is damping but no external driving force F(t):
mu ′′(t ) + γ u ′(t ) + ku (t ) = 0
What is effect of damping coefficient γ on system?
The characteristic equation is
− γ ± γ 2 − 4mk
4mk ⎤
γ ⎡
r1 , r2 =
1
1
=
−
±
−
⎢
⎥
2m
2m ⎣
γ2 ⎦
Three cases for the solution:
γ 2 − 4mk > 0 :
γ 2 − 4mk = 0 :
u (t ) = Ae r1 t + Be r2 t , where r1 < 0, r2 < 0 ;
u (t ) = ( A + Bt )e −γ t / 2 m , where γ / 2m > 0 ;
2
γ
−
4
mk
γ 2 − 4mk < 0 : u (t ) = e −γ t / 2 m ( A cos μ t + B sin μ t ), μ =
> 0.
2m
Note : In all three cases, lim u (t ) = 0, as expected from damping term.
t →∞
Damped Free Vibrations: Small Damping
(2 of 8)
Of the cases for solution form, the last is most important,
which occurs when the damping is small:
γ 2 − 4mk > 0 : u (t ) = Ae r t + Be r t , r1 < 0, r2 < 0
1
2
γ 2 − 4mk = 0 : u (t ) = ( A + Bt )e −γ t / 2 m , γ / 2m > 0
γ 2 − 4mk < 0 : u (t ) = e −γ t / 2 m ( A cos μ t + B sin μ t ), μ > 0
We examine this last case. Recall
A = R cos δ , B = R sin δ
Then
u (t ) = R e −γ t / 2 m cos(μ t − δ )
and hence
u (t ) ≤ R e −γ t / 2 m
(damped oscillation)
Damped Free Vibrations: Quasi Frequency
(3 of 8)
Thus we have damped oscillations:
u (t ) = R e −γ t / 2 m cos(μ t − δ ) ⇒
u (t ) ≤ R e −γ t / 2 m
Amplitude R depends on the initial conditions, since
u (t ) = e −γ t / 2 m ( A cos μ t + B sin μ t ), A = R cos δ , B = R sin δ
Although the motion is not periodic, the parameter μ
determines mass oscillation frequency.
Thus μ is called the quasi frequency.
Recall
4mk − γ 2
μ=
2m
Damped Free Vibrations: Quasi Period
(4 of 8)
Compare μ with ω0 , the frequency of undamped motion:
4km − γ 2
4km − γ 2
4km − γ 2
μ
γ2
=
=
=
= 1−
2
4km
ω0 2 m k / m
4km
4m k / m
For small γ
γ2
2
γ4
⎛
γ2 ⎞
γ2
⎟⎟ = 1 −
≅ 1−
+
= ⎜⎜1 −
2 2
4km 64k m
8km
⎝ 8km ⎠
Thus, small damping reduces oscillation frequency slightly.
Similarly, quasi period is defined as Td = 2π/μ. Then
Td
γ ⎞
2π / μ ω0 ⎛
⎟⎟
=
=
= ⎜⎜1 −
T 2π / ω0 μ ⎝ 4km ⎠
2
−1 / 2
−1
⎛
γ ⎞
γ2
⎟⎟ ≅ 1 +
≅ ⎜⎜1 −
8km
⎝ 8km ⎠
2
Thus, small damping increases quasi period.
Damped Free Vibrations:
Neglecting Damping for Small γ 2/4km
(5 of 8)
Consider again the comparisons between damped and
undamped frequency and period:
1/ 2
−1 / 2
2
2
Td ⎛
μ ⎛
γ ⎞
γ ⎞
⎟⎟
⎟⎟ ,
= ⎜⎜1 −
= ⎜⎜1 −
ω0 ⎝ 4km ⎠
T ⎝ 4km ⎠
Thus it turns out that a small γ is not as telling as a small
ratio γ 2/4km.
For small γ 2/4km, we can neglect effect of damping when
calculating quasi frequency and quasi period of motion. But
if we want a detailed description of motion of mass, then we
cannot neglect damping force, no matter how small.
Damped Free Vibrations:
Frequency, Period (6 of 8)
Ratios of damped and undamped frequency, period:
1/ 2
Td ⎛
μ ⎛
γ2 ⎞
γ2 ⎞
⎟⎟ ,
⎟⎟
= ⎜⎜1 −
= ⎜⎜1 −
ω0 ⎝ 4km ⎠
T ⎝ 4km ⎠
−1 / 2
Thus
lim μ = 0 and lim Td = ∞
γ → 2 km
γ → 2 km
The importance of the relationship between γ2 and 4km is
supported by our previous equations:
γ 2 − 4mk > 0 : u (t ) = Ae r t + Be r t , r1 < 0, r2 < 0
1
2
γ 2 − 4mk = 0 : u (t ) = ( A + Bt )e −γ t / 2 m , γ / 2m > 0
γ 2 − 4mk < 0 : u (t ) = e −γ t / 2 m ( A cos μ t + B sin μ t ), μ > 0
Damped Free Vibrations:
Critical Damping Value (7 of 8)
Thus the nature of the solution changes as γ passes through
the value 2 km .
This value of γ is known as the critical damping value, and
for larger values of γ the motion is said to be overdamped.
Thus for the solutions given by these cases,
γ 2 − 4mk > 0 : u (t ) = Ae r t + Be r t , r1 < 0, r2 < 0
1
2
(1)
γ 2 − 4mk = 0 : u (t ) = ( A + Bt )e −γ t / 2 m , γ / 2m > 0
(2)
γ 2 − 4mk < 0 : u (t ) = e −γ t / 2 m ( A cos μ t + B sin μ t ), μ > 0
(3)
we see that the mass creeps back to its equilibrium position
for solutions (1) and (2), but does not oscillate about it, as
for small γ in solution (3).
Soln (1) is overdamped and soln (2) is critically damped.
Damped Free Vibrations:
Characterization of Vibration
(8 of 8)
Mass creeps back to equilibrium position for solns (1) & (2),
but does not oscillate about it, as for small γ in solution (3).
γ 2 − 4mk > 0 : u (t ) = Ae r t + Be r t , r1 < 0, r2 < 0
(Green)
(1)
γ 2 − 4mk = 0 : u (t ) = ( A + Bt )e −γ t / 2 m , γ / 2m > 0
(Red, Black)
( 2)
γ 2 − 4mk < 0 : u (t ) = e −γ t / 2 m ( A cos μ t + B sin μ t )
(Blue)
(3)
1
2
Soln (1) is overdamped and soln (2) is critically damped.
Example 3: Initial Value Problem
(1 of 4)
Suppose that the motion of a spring-mass system is governed
by the initial value problem
u ′′ + 0.125u ′ + u = 0, u (0) = 2, u ′(0) = 0
Find the following:
(a) quasi frequency and quasi period;
(b) time at which mass passes through equilibrium position;
(c) time τ such that |u(t)| < 0.1 for all t > τ.
For Part (a), using methods of this chapter we obtain:
u (t ) = e
− t / 16
⎛
⎛ 255
⎞
255
2
255 ⎞
32 − t /16
⎜ 2 cos
⎟
⎜
t+
t⎟ =
e
t − δ ⎟⎟
sin
cos⎜
⎜
16
16 ⎠
255
255
⎝
⎝ 16
⎠
where
tan δ =
1
⇒ δ ≅ 0.06254 (recall A = R cos δ , B = R sin δ )
255
Example 3: Quasi Frequency & Period
(2 of 4)
The solution to the initial value problem is:
u (t ) = e
− t / 16
⎛
⎛ 255
⎞
255 ⎞
32 − t /16
255
2
⎜ 2 cos
⎟
⎜
t+
t⎟ =
e
t − δ ⎟⎟
cos⎜
sin
⎜
16
16 ⎠
255
255
⎝
⎝ 16
⎠
The graph of this solution, along with solution to the
corresponding undamped problem, is given below.
The quasi frequency is
μ = 255 / 16 ≅ 0.998
and quasi period
Td = 2π / μ ≅ 6.295
For undamped case:
ω0 = 1, T = 2π ≅ 6.283
Example 3: Quasi Frequency & Period
(3 of 4)
The damping coefficient is γ = 0.125 = 1/8, and this is 1/16 of
the critical value 2 km = 2
Thus damping is small relative to mass and spring stiffness.
Nevertheless the oscillation amplitude diminishes quickly.
Using a solver, we find that |u(t)| < 0.1 for t > τ ≈ 47.515 sec
Example 3: Quasi Frequency & Period
(4 of 4)
To find the time at which the mass first passes through the
equilibrium position, we must solve
⎞
⎛ 255
32 − t /16
⎜
u (t ) =
e
t − δ ⎟⎟ = 0
cos⎜
255
⎠
⎝ 16
Or more simply, solve
255
π
t −δ =
16
2
16 ⎛ π
⎞
⇒t =
+
δ
⎜
⎟ ≅ 1.637 sec
255 ⎝ 2
⎠
Electric Circuits
The flow of current in certain basic electrical circuits is
modeled by second order linear ODEs with constant
coefficients:
1
I (t ) = E ′(t )
C
I (0) = I 0 , I ′(0) = I 0′
L I ′′(t ) + R I ′(t ) +
It is interesting that the flow of current in this circuit is
mathematically equivalent to motion of spring-mass system.
For more details, see text.
Ch 3.9: Forced Vibrations
We continue the discussion of the last section, and now
consider the presence of a periodic external force:
m u ′′(t ) + γ u ′(t ) + k u (t ) = F0 cos ω t
Forced Vibrations with Damping
Consider the equation below for damped motion and external
forcing funcion F0cosωt.
mu ′′(t ) + γ u ′(t ) + ku (t ) = F0 cos ω t
The general solution of this equation has the form
u (t ) = c1u1 (t ) + c2u 2 (t ) + A cos(ω t ) + B sin (ω t ) = uC (t ) + U (t )
where the general solution of the homogeneous equation is
uC (t ) = c1u1 (t ) + c2u 2 (t )
and the particular solution of the nonhomogeneous equation is
U (t ) = A cos(ω t ) + B sin (ω t )
Homogeneous Solution
The homogeneous solutions u1 and u2 depend on the roots r1
and r2 of the characteristic equation:
2
−
γ
±
γ
− 4mk
2
mr + γ r + kr = 0 ⇒ r =
2m
Since m, γ, and k are are all positive constants, it follows that
r1 and r2 are either real and negative, or complex conjugates
with negative real part. In the first case,
(
)
lim uC (t ) = lim c1e r1t + c2 e r2t = 0,
t →∞
t →∞
while in the second case
(
)
lim uC (t ) = lim c1e λ t cos μt + c2 e λ t sin μt = 0.
t →∞
t →∞
Thus in either case,
lim uC (t ) = 0
t →∞
Transient and Steady-State Solutions
Thus for the following equation and its general solution,
mu ′′(t ) + γ u ′(t ) + ku (t ) = F0 cos ω t
u (t ) = c1u1 (t ) + c2u2 (t ) + A cos(ω t ) + B sin (ω t ),
1442443 144424443
uC ( t )
U (t )
we have
lim uC (t ) = lim(c1u1 (t ) + c2u 2 (t ) ) = 0
t →∞
t →∞
Thus uC(t) is called the transient solution. Note however that
U (t ) = A cos(ω t ) + B sin (ω t )
is a steady oscillation with same frequency as forcing function.
For this reason, U(t) is called the steady-state solution, or
forced response.
Transient Solution and Initial Conditions
For the following equation and its general solution,
mu ′′(t ) + γ u ′(t ) + ku (t ) = F0 cos ω t
u (t ) = c1u1 (t ) + c2u2 (t ) + A cos(ω t ) + B sin (ω t )
1442443 144424443
uC ( t )
U (t )
the transient solution uC(t) enables us to satisfy whatever initial
conditions might be imposed.
With increasing time, the energy put into system by initial
displacement and velocity is dissipated through damping force.
The motion then becomes the response U(t) of the system to
the external force F0cosωt.
Without damping, the effect of the initial conditions would
persist for all time.
Rewriting Forced Response
Using trigonometric identities, it can be shown that
U (t ) = A cos(ω t ) + B sin (ω t )
can be rewritten as
U (t ) = R cos(ω t − δ )
It can also be shown that
R=
F0
m (ω − ω ) + γ ω
cos δ =
where
2
2
0
2 2
2
,
2
m(ω02 − ω 2 )
m (ω − ω ) + γ ω
2
2
0
ω02 = k / m
2 2
2
2
, sin δ =
γω
m 2 (ω02 − ω 2 ) 2 + γ 2ω 2
Amplitude Analysis of Forced Response
The amplitude R of the steady state solution
R=
F0
m (ω − ω ) + γ ω
2
2
0
2 2
2
2
,
depends on the driving frequency ω. For low-frequency
excitation we have
F0
F0
F0
lim R = lim
=
=
2
2
2
2 2
2 2
ω →0
ω →0
mω 0
k
m (ω0 − ω ) + γ ω
where we recall (ω0)2 = k /m. Note that F0 /k is the static
displacement of the spring produced by force F0.
For high frequency excitation,
lim R = lim
ω →∞
ω →∞
F0
m (ω − ω ) + γ ω
2
2
0
2 2
2
2
=0
Maximum Amplitude of Forced Response
Thus
lim R = F0 k , lim R = 0
ω →0
ω →∞
At an intermediate value of ω, the amplitude R may have a
maximum value. To find this frequency ω, differentiate R and
set the result equal to zero. Solving for ωmax, we obtain
ω
2
max
⎛
γ2 ⎞
⎟⎟
=ω −
= ω ⎜⎜1 −
2
2m
⎝ 2mk ⎠
2
0
γ2
2
0
where (ω0)2 = k /m. Note ωmax < ω0, and ωmax is close to ω0
for small γ. The maximum value of R is
Rmax =
F0
γω0 1 − (γ 2 4mk )
Maximum Amplitude for Imaginary ωmax
We have
ω
and
Rmax =
2
max
⎛
γ2 ⎞
⎟⎟
= ω ⎜⎜1 −
⎝ 2mk ⎠
2
0
F0
γω0 1 − (γ 2
F0 ⎛
γ2 ⎞
⎜⎜1 +
⎟⎟
≅
4mk ) γω0 ⎝ 8mk ⎠
where the last expression is an approximation for small γ. If
γ 2 /(mk) > 2, then ωmax is imaginary. In this case, Rmax= F0 /k,
which occurs at ω = 0, and R is a monotone decreasing
function of ω. Recall from Section 3.8 that critical damping
occurs when γ 2 /(mk) = 4.
Resonance
From the expression
Rmax =
F0
γω0 1 − (γ 2
F0 ⎛
γ2 ⎞
⎜⎜1 +
⎟⎟
≅
4mk ) γω0 ⎝ 8mk ⎠
we see that Rmax≅ F0 /(γ ω0) for small γ.
Thus for lightly damped systems, the amplitude R of the forced
response is large for ω near ω0, since ωmax ≅ ω0 for small γ.
This is true even for relatively small external forces, and the
smaller the γ the greater the effect.
This phenomena is known as resonance. Resonance can be
either good or bad, depending on circumstances; for example,
when building bridges or designing seismographs.
Graphical Analysis of Quantities
To get a better understanding of the quantities we have been
examining, we graph the ratios R/(F0/k) vs. ω/ω0 for several
values of Γ = γ 2 /(mk), as shown below.
Note that the peaks tend to get higher as damping decreases.
As damping decreases to zero, the values of R/(F0/k) become
asymptotic to ω = ω0. Also, if γ 2 /(mk) > 2, then Rmax= F0 /k,
which occurs at ω = 0.
Analysis of Phase Angle
Recall that the phase angle δ given in the forced response
U (t ) = R cos(ω t − δ )
is characterized by the equations
cos δ =
m(ω02 − ω 2 )
m 2 (ω02 − ω 2 ) 2 + γ 2ω 2
, sin δ =
γω
m 2 (ω02 − ω 2 ) 2 + γ 2ω 2
If ω ≅ 0, then cosδ ≅ 1, sinδ ≅ 0, and hence δ ≅ 0. Thus the
response is nearly in phase with the excitation.
If ω = ω0, then cosδ = 0, sinδ = 1, and hence δ ≅ π /2. Thus
response lags behind excitation by nearly π /2 radians.
If ω large, then cosδ ≅ -1, sinδ = 0, and hence δ ≅ π . Thus
response lags behind excitation by nearly π radians, and
hence they are nearly out of phase with each other.
Example 1:
Forced Vibrations with Damping
(1 of 4)
Consider the initial value problem
u ′′(t ) + 0.125 u ′(t ) + u (t ) = 3 cos 2 t , u (0) = 2, u ′(0) = 0
Then ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625.
The unforced motion of this system was discussed in Ch 3.8,
with the graph of the solution given below, along with the
graph of the ratios R/(F0/k) vs. ω/ω0 for different values of Γ.
Example 1:
Forced Vibrations with Damping
(2 of 4)
Recall that ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625.
The solution for the low frequency case ω = 0.3 is graphed
below, along with the forcing function.
After the transient response is substantially damped out, the
steady-state response is essentially in phase with excitation,
and response amplitude is larger than static displacement.
Specifically, R ≅ 3.2939 > F0/k = 3, and δ ≅ 0.041185.
Example 1:
Forced Vibrations with Damping
(3 of 4)
Recall that ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625.
The solution for the resonant case ω = 1 is graphed below,
along with the forcing function.
The steady-state response amplitude is eight times the static
displacement, and the response lags excitation by π /2 radians,
as predicted. Specifically, R = 24 > F0/k = 3, and δ = π /2.
Example 1:
Forced Vibrations with Damping
(4 of 4)
Recall that ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625.
The solution for the relatively high frequency case ω = 2 is
graphed below, along with the forcing function.
The steady-state response is out of phase with excitation, and
response amplitude is about one third the static displacement.
Specifically, R ≅ 0.99655 ≅ F0/k = 3, and δ ≅ 3.0585 ≅ π.
Undamped Equation:
General Solution for the Case ω0 ≠ ω
Suppose there is no damping term. Then our equation is
mu ′′(t ) + ku (t ) = F0 cos ω t
Assuming ω0 ≠ ω, then the method of undetermined
coefficients can be use to show that the general solution is
F0
u (t ) = c1 cos ω0t + c2 sin ω0t +
cos ω t
2
2
m(ω0 − ω )
Undamped Equation:
Mass Initially at Rest
(1 of 3)
If the mass is initially at rest, then the corresponding initial
value problem is
mu ′′(t ) + ku (t ) = F0 cos ω t , u (0) = 0, u ′(0) = 0
Recall that the general solution to the differential equation is
F0
u (t ) = c1 cos ω0t + c2 sin ω0t +
cos ω t
2
2
m(ω0 − ω )
Using the initial conditions to solve for c1 and c2, we obtain
c1 = −
F0
, c2 = 0
2
2
m(ω0 − ω )
Hence
u (t ) =
F0
(cos ω t − cos ω0t )
2
2
m(ω0 − ω )
Undamped Equation:
Solution to Initial Value Problem
(2 of 3)
Thus our solution is
u (t ) =
F0
(cos ω t − cos ω0t )
2
2
m(ω0 − ω )
To simplify the solution even further, let A = (ω0 + ω)/2 and
B = (ω0 - ω)/2. Then A + B = ω0t and A - B = ωt. Using the
trigonometric identity
cos( A ± B) = cos A cos B m sin A sin B,
it follows that
cos ω t = cos A cos B + sin A sin B
cos ω0t = cos A cos B − sin A sin B
and hence
cos ω t − cos ω0t = 2 sin A sin B
Undamped Equation: Beats
(3 of 3)
Using the results of the previous slide, it follows that
⎡
(ω0 − ω )t ⎤ sin (ω0 + ω )t
2 F0
u (t ) = ⎢
sin
⎥
2
2
(
)
2
2
m
ω
−
ω
0
⎣
⎦
When |ω0 - ω| ≅ 0, ω0 + ω is much larger than ω0 - ω, and
sin[(ω0 + ω)t/2] oscillates more rapidly than sin[(ω0 - ω)t/2].
Thus motion is a rapid oscillation with frequency (ω0 + ω)/2,
but with slowly varying sinusoidal amplitude given by
(ω0 − ω )t
2 F0
sin
2
m ω02 − ω 2
This phenomena is called a beat.
Beats occur with two tuning forks of
nearly equal frequency.
Example 2: Undamped Equation,
Mass Initially at Rest
(1 of 2)
Consider the initial value problem
u ′′(t ) + u (t ) = 0.5 cos 0.8 t , u (0) = 0, u ′(0) = 0
Then ω0 = 1, ω = 0.8, and F0 = 0.5, and hence the solution is
u (t ) = 2.77778(sin 0.1t )(sin 0.9 t )
The displacement of the spring–mass system oscillates with a
frequency of 0.9, slightly less than natural frequency ω0 = 1.
The amplitude variation has a slow
frequency of 0.1 and period of 20π.
A half-period of 10π corresponds to
a single cycle of increasing and then
decreasing amplitude.
Example 2: Increased Frequency
(2 of 2)
Recall our initial value problem
u ′′(t ) + u (t ) = 0.5 cos 0.8 t , u (0) = 0, u ′(0) = 0
If driving frequency ω is increased to ω = 0.9, then the slow
frequency is halved to 0.05 with half-period doubled to 20π.
The multiplier 2.77778 is increased to 5.2632, and the fast
frequency only marginally increased, to 0.095.
Undamped Equation:
General Solution for the Case ω0 = ω
(1 of 2)
Recall our equation for the undamped case:
mu ′′(t ) + ku (t ) = F0 cos ω t
If forcing frequency equals natural frequency of system, i.e.,
ω = ω0 , then nonhomogeneous term F0cosωt is a solution of
homogeneous equation. It can then be shown that
F0
u (t ) = c1 cos ω0t + c2 sin ω0t +
t sin ω0t
2 mω 0
Thus solution u becomes unbounded as t → ∞.
Note: Model invalid when u gets
large, since we assume small
oscillations u.
Undamped Equation: Resonance
(2 of 2)
If forcing frequency equals natural frequency of system, i.e.,
ω = ω0 , then our solution is
F0
u (t ) = c1 cos ω0t + c2 sin ω0t +
t sin ω0t
2 mω 0
Motion u remains bounded if damping present. However,
response u to input F0cosωt may be large if damping is
small and |ω0 - ω| ≅ 0, in which case we have resonance.