Section 2.1
Introduction:
Second-Order Linear Equations
Second Order Differential Equations: F(x, y, y′, y″) = 0
2nd Order Linear Differential Equations: A(x) y″ + B(x) y′ + C(x) y = f (x)
often these will be in the form y″ + P(x) y′ + Q(x) y = f (x)
nth Order Linear Differential Equations
Homogeneous Linear Differential Equations: A(x) y″ + B(x) y′ + C(x) y = 0
Solutions to homogeneous linear Differential Equations form vector spaces!
Theorem
Let y1 and y2 be solutions to y″ + P(x) y′ + Q(x) y = 0. Then functions of the
form f (x) = c1 y1 + c2 y2 (with c1 and c2 being any constants) are also solutions.
General Solutions
Initial Conditions
Theorem
Suppose the P(x) and Q(x) are continuous on an interval J containing xo. Then
the initial value problem y″ + P(x) y′ + Q(x) y = 0, y′(xo) = a , y(xo) = b has
a unique solution on J.
Ex. 1 Solve y″ + y = 0,
y 4  
2
2
,
y  4  
5 2
2
Suppose we've found two linearly independent solutions, y1 and y2, to the
initial value problem y″ + P(x) y′ + Q(x) y = 0, y′(xo) = a, y(xo) = b.
How do we find the general solution to this differential equation?
Recall:
Two functions f (x) and g(x) are said to be linearly independent if one
function cannot be written as a multiple of the other
(or equivalently, f (x) and g(x) are linearly independent if the equation
c1 f (x) + c2 g(x) = 0 has only a solution of c1 = c2 = 0).
Ex. 2 Show that y1 = sin(x) and y2 = 3sin(x) are both solutions to y″ + y = 0,
y 4  
2
2
,
y  4  
5 2
2
but we cannot use these to find our particular solution.
Definition:
The Wronskian of f (x) and g(x) is defined as:
W ( f ,g) 
f
g
f
g
Ex. 3 Calculate the Wronskian of f (x) = sin(x) and g(x) = x2.
Theorem:
Suppose y1 and y2 are solutions of y″ + P(x) y′ + Q(x) y = 0 on an interval J in which
P(x) and Q(x) are continuous.
(a) If y1 and y2 are linearly dependent then W(f, g) = 0 for all x on the interval J.
(b) If y1 and y2 are linearly independent then W(f, g) ≠ 0 for any x on the interval J.
Ex. 4
(a) What does this theorem imply regarding the solutions y1 = sin(x) and
y2 = cos(x) to the differential equation y″ + y = 0?
Ex. 4
(b) What does this theorem imply regarding the solutions y1 = sin(x) and
y2 = 3sin(x) to the differential equation y″ + y = 0?
Theorem:
Suppose y1 and y2 are linearly independent solutions of y″ + P(x) y′ + Q(x) y = 0
where P(x) and Q(x) are continuous on an interval J. If g(x) is any solution to
this differential equation then g(x) = c1 y1 + c2 y2 for some choices of constants
c1 and c2.
Big Important Question:
How do we determine two linearly independent solutions, y1 and y2?
We shall examine this question for the differential equation ay″ + by′ + cy = 0
(where a, b, and c are constants).
Claim: y = erx is a solution (for some constant r) to ay″ + by′ + cy = 0. We
just have to figure out what value of r will work.
Ex. 5 Solve y″ – 5y′ + 6y = 0, y(0) = 1, y′(0) = –1.
Theorem:
If the roots of the characteristic equation ar2 + br + c = 0 are r1 and r2 with
r1 ≠ r2 then y1 = _________ and y2 = _________ are two linearly independent
solutions to the differential equation ay″ + by′ + cy = 0.
Thus, y = _________________ is the general solution to ay″ + by′ + cy = 0.
What if there is a repeated root of the characteristic equation?
Ex. 6 Solve y″ – 8y′ + 16y = 0, y(0) = 5, y′(0) = 21.
Theorem:
If r1 is a repeated root of the characteristic equation ar2 + br + c = 0 then
y1 = _________ and y2 = _________ are two linearly independent solutions
to the differential equation ay″ + by′ + cy = 0. Thus,
y = _________________ is the general solution to ay″ + by′ + cy = 0.
Ex. 6 Solve y″ – 8y′ + 16y = 0, y(0) = 5, y′(0) = 21.
Why does this theorem work? Let's prove that it produces the correct general
solution for this past problem. (Introduction to the linear operator D.)
Why does this theorem work? Let's prove that it produces the correct general
solution for this past problem. (Introduction to the linear operator D.)
Why does this theorem work? Let's prove that it produces the correct general
solution for this past problem. (Introduction to the linear operator D.)
Why does y″ + y = 0 have solutions of y1 = sin(x) and y2 = cos(x)????
Our previous theorems couldn't have possibly generated these solutions.
Ex. 7 Use our past two theorems to solve y″ + y = 0.
Recall that
ez = 1 
1
1!
z
1
2!
z 
2
1
3!
z 
3
1
4!
z 
4
1
5!
z 
5

1
n!
z 
n
Recall that
ez = 1 
1
1
z
1!
z 
2
2!
1
3!
z 
3
1
4!
z 
4
1
z 
5
5!
  1 2 n
z
 2 n !
n!
n
cos(z) = 1 
1
2!
z 
2
1
4!
z 
4


1

z 
n
Recall that
1
ez = 1 
1
z
1!
1
z 
2
2!
z 
3
3!
1
z 
4
4!
1
z 

5
5!
n!
  1 2 n
z
 2 n !
n
cos(z) = 1 
sin(z) =
1
1!
1
z 
2
2!
z
1
z 
4
4!
1
3!
z 
3
1
5!
z 
5


  1
1

n
 2 n  1!
z
2 n 1

z 
n
Recall that
1
ez = 1 
1
z
1!
1
z 
2
2!
z 
3
3!
1
z 
4
4!
1
z 

5
5!
n!
  1 2 n
z
 2 n !
n
cos(z) = 1 
sin(z) =
1
1!
1
z 
2
2!
z
1
z 
4
4!
1
3!
z 
3
1
5!
z 
5


  1
1

n
 2 n  1!
z
2 n 1

z 
n
Ex. 8 Solve y″ – 6y′ + 13y = 0
Theorem:
If the roots of the characteristic equation ar2 + br + c = 0 are A ± Bi then
y1 = eAx sin(Bx) and y2 = eAx cos(Bx) are two linearly independent solutions
to the differential equation ay″ + by′ + y = 0. Thus,
y = c1 eAx sin(Bx) + c2 eAx cos(Bx) is the general solution to ay″ + by′ + y = 0.
Section 2.2
General Solutions of Linear Equations
Higher order linear differential equations:
y(n) + pn–1(x) y(n–1) + pn–2(x) y(n–2) + · · · · · + p2(x) y″ + p1(x) y′ + p0(x) y = f
(x)
If y1, y2, . . . , yn are linearly independent solutions to the homogeneous diff. eq.
y(n) + pn–1(x) y(n–1) + pn–2(x) y(n–2) + · · · · · + p1(x) y′ + p0(x) y = 0,
then the general solution to y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = 0 is
y = c1y1 + c2y2 + · · · + cnyn.
To prove this we needed the following definitions and theorems:
Theorem:
If y1, y2, . . . , yn are solutions to y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = 0
then so is y = c1y1 + c2y2 + · · · + cnyn
Theorem:
Suppose y1, y2, .., yn are solutions to y(n) + pn–1(x) y(n–1) + ··· + p1(x) y′ + p0(x) y = 0
on an interval J where the pi(x) are continuous.
(a) If y1, y2, . . . , yn are linearly dependent then W(y1, y2, . . . , yn) = 0 at each
point on the interval J.
(b) If y1, y2, . . . , yn are linearly independent then W(y1, y2, . . . , yn) ≠ 0 at each
point on the interval J.
Theorem:
If y1, y2, . . . , yn are solutions to y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y =
0, then so is y = c1y1 + c2y2 + · · · + cnyn for any choice of constants c1, c2, . . . ,
cn.
Theorem: If y1, y2, . . . , yn are linearly independent solutions to
y(n–1) + pn–2(x) y(n–2) + · · · · · + p1(x) y′ + p0(x) y = 0, then general solution is
y=c y +c y +···+c y .
Ex. 1 Suppose that the following functions are all particular solutions to
y(3) + 3y″ + 4y′ + 12y = 0. We would like to have a set of three linearly
independent solutions. If this can be done then which three can we use?
If it cannot be done then show that it cannot.
y1 = sin(2x)
y2 = cos(2x)
y3 = 5 sin(2x) + 3 cos(2x)
y4 = sin(x) cos(x)
y5 = cos2(x) – sin2(x)
y6 = e–3x
Ex. 2 Use the definition of linear dependence to show that the following
functions are linearly dependent.
f (x) = x2 + 3x, g(x) = 7x2, h(x) = 31x2 + 6x.
The Wronskian of the n functions f1(x), f2(x), . . . , fn(x):
W
 f1 ,
f2 , f3 ,
, fn  
f1
f2
f3
fn
f 1
f 2
f 3
f n
f 1
f 2 
f 3
f n 
f1
(n)
(n)
f2
(n)
f3
(n)
fn
Theorem:
Suppose y1, y2, ..., yn are solutions of y(n) + pn–1(x) y(n–1) + ··· + p1(x) y′ + p0(x) y = 0
on an interval J in which pn–1(x), pn–2(x), · · · p1(x), p0(x) are continuous.
(a) If y1, y2, . . . , yn are linearly dependent then W(y1, y2, . . . , yn) = 0 for all x on
the interval J.
(b) If y1, y2, . . . , yn are linearly independent then W(y1, y2, . . . , yn) ≠ 0 for all x on
the interval J.
Ex. 3
(a) y1 = sin(2x), y2 = cos(2x), y3 = e–3x are solutions to y(3) + 3y″ + 4y′ + 12y = 0.
Use the Wronskian to determine whether or not they are linearly independent.
Ex. 3
(b) y1 = sin(2x), y2 = cos(2x), y3 = e–3x are solutions to y(3) + 3y″ + 4y′ + 12y = 0.
Use the Wronskian to determine whether or not they are linearly independent.
Ex. 3
(c) y1 = x, y2 = x ln(x), y3 = x2 are solutions to x3y(3) – x2y″ + 2xy′ – 2y = 0. Use
the Wronskian to determine whether or not they are linearly independent.
Suppose we can find three linearly independent solutions to
y(3) + 3 y″ + 4y′ + 12y = 0.
Are we guaranteed a solution to the initial valued problem with given initial
conditions of y(0) = 1, y′(0) = 2, y″(0) = 3? Why??
Theorem:
Suppose pn–1(x), pn–2(x), · · · p1(x), p0(x) are continuous on an interval J
which contains xo. If we can find n linearly independent solutions to the diff. eq.
y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = 0
then we are guaranteed a solution to the initial value problem of
y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = 0
y(x0) = b0
y′(x0) = b1
y″(x0) = b2
:
:
y(n)(x0) = bn
Theorem:
If y1, y2, . . . , yn are linearly independent solutions to
y(n) + pn–1(x) y(n–1) + · · · + p1(x) y′ + p0(x) y = 0,
then general solution is y = c1y1 + c2y2 + · · · + cnyn.
Nonhomogeneous linear Differential Equations:
y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = f (x)
We solve a nonhomogeneous differential equation by first examining the
corresponding homogeneous differential equation.
Review:
Theorem:
Suppose that yc = c1y1 + c2y2 + · · · + cnyn is the general solution to the
homogeneous differential equation y(n) + pn–1(x) y(n–1) + ··· + p1(x) y′ + p0(x) y =
0. Further suppose that yp is any particular solution to the nonhomogeneous
differential equation y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = f (x). Then
the general solution to the nonhomogeneous differential equation
y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = f (x) is:
y = yp + yc
= yp + c1y1 + c2y2 + · · · + cnyn
Ex. 4
(a) Verify that y = (–1/4) x is a particular solution to the diff. eq. y″ – 4y = x.
Ex. 4
(b) Verify that y = (–1/4) x + e2x is another particular solution to the diff. eq.
y″ – 4y = x.
Ex. 4
(c) What is the general solution to the differential equation y″ – 4y = x ?
Ex. 4
(d) Find a solution to the initial value problem y″ – 4y = x, y(0) = 11, y′(0) = –2.
Ex. 5
(a) y = 2x is one solution to the differential equation y″ – 10y′ + 21y = 42x – 20.
Use this to find the general solution.
Ex. 5
(b) y = 2x + 5e3x is another solution to the differential equation
y″ – 10y′ + 21y = 42x – 20. Use this to find the general solution.
Section 2.3
Homogeneous Equations
with Constant Coefficients
Find the general solution to y(3) + y″ – 17y′ + 15y = 0
Recall:
If r1 and r2 are distinct roots of the characteristic equation ar2 + br + c = 0,
then y1  e r x and y 2  e r x are linearly independent solutions to the
differential equation ay″ + by′ + cy = 0, and thus y  c1 e r x  c 2 e r x is the
general solution to ay″ + by′ + cy = 0.
1
2
1
2
Recall:
If r1 and r2 are distinct roots of the characteristic equation ar2 + br + c = 0,
then y1  e r x and y 2  e r x are linearly independent solutions to the
differential equation ay″ + by′ + cy = 0, and thus y  c1 e r x  c 2 e r x is the
general solution to ay″ + by′ + cy = 0.
1
2
1
2
Now:
Theorem: If r1, r2, . . . , rk are distinct roots of the characteristic equation
anrn + an–1rn–1 + · · · + a1r + a0 = 0, then y1  e r x , y 2  e r x , y 3  e r x , , y k  e r x
are linearly independent solutions to the differential equation
any(n) + an–1 y(n–1) + · · · + a1y′ + a0y = 0.
1
2
3
k
Ex. 1 Find the general solution to y(3) + y″ – 17y′ + 15y = 0.
Ex. 2 Find the particular solution to y(3) + y″ – 17y′ + 15y = 0
y(0) = 10
y′(0) = 8
y″(0) = 50
Ex. 3 Find the general solution to y(5) – 3y(4) + 2y(3) + 2y″ – 3y′ + y = 0.
Theorem:
If r1 is a root of multiplicity k in the characteristic equation
anrn + an–1rn–1 + · · · + a1r + a0 = 0, then
rx
rx
2 rx
k 1 r x
are linearly independent
y1  e , y 2  xe , y 3  x e , , y k  x e
solutions to the differential equation any(n) + an–1 y(n–1) + · · · + a1y′ + a0y = 0.
1
1
1
1
Ex. 4 Find the general solution to
y(9) – 25y(8) + 264y(7) – 1520y(6) + 5120y(5) – 9984y(4) + 10240y(3) – 4096y″ = 0.
Section 2.5
Nonhomogeneous Equations &
Undetermined Coefficients
Recall:
Theorem:
Suppose that yc = c1y1 + c2y2 + · · · + cnyn is the general solution to the
homogeneous differential equation y(n) + pn–1(x) y(n–1) + ··· + p1(x) y′ + p0(x) y =
0. Further suppose that yp is any particular solution to the nonhomogeneous
differential equation y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = f (x). Then
the general solution to the nonhomogeneous differential equation
y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = f (x) is:
y = yp + yc
= yp + c1y1 + c2y2 + · · · + cnyn
Recall:
Ex. 5
(a) y = 2x is one solution to the differential equation y″ – 10y′ + 21y = 42x – 20.
Use this to find the general solution.
How do we determine yp though?
The Method of Undetermined Coefficients
(Using an educated guess to determine yp)
Ex. 1 y″ + 2y′ – 3y = 3e2x
(a) Make a guess on what yp should look like for this differential equation?
(What "form" should yp have?)
Ex. 1 y″ + 2y′ – 3y = 3e2x
(b) Determine, specifically, what yp is.
Ex. 1 y″ + 2y′ – 3y = 3e2x
(c) Give the general solution to this nonhomogeneous differential equation.
Ex. 2 y″ + 2y′ – 3y = 30 cos(x)
(a) Make a guess on what yp should look like for this differential equation?
(What "form" should yp have?)
Ex. 2 y″ + 2y′ – 3y = 30 cos(x)
(b) Determine, specifically, what yp is.
Ex. 2 y″ + 2y′ – 3y = 30 cos(x)
(c) Give the general solution to this nonhomogeneous differential equation.
Ex. 3 Find the general solution to 2y″ + 4y′ + 7y = x2
Ex. 4 Find the general solution to y″ – 4y = 2e2x
Ex. 4 Find the general solution to y″ – 4y = 2e2x
Ex. 5 Find the form of yp for the nonhomogeneous differential equation:
y″ – 4y = 2xe2x
Ex. 6 Find the form of yp for the nonhomogeneous differential equation:
y(3) + 9y′ = x sin(7x) + x2e2x
Ex. 7 Find the form of yp for the nonhomogeneous differential equation:
(D – 2)(D – 5)3 y = xe2x + x2e5x + cos(x)
Lemma 1: For any function u(x) we have: (D – r)[u(x)erx] = (D [u(x)]) (erx)
Proof:
(D – r)[u(x)erx] =
D[u(x)erx]
– r[u(x)erx]
= D[u(x)] erx + u(x)D[erx] – ru(x)erx
= D[u(x)] erx + ru(x)erx
= D[u(x)] erx
– ru(x)erx
Lemma 2: For any function u(x) we have: (D – r)k [u(x)erx] = (Dk [u(x)]) (erx)
Question:
Find the form of yc for the differential equation (D – r)3 y = (–3 + 2x)erx
Solution:
We compute yc : yc = c1erx + c2xerx + c3x2erx
Our initial guess for yp : yp = Aerx + Bxerx
Our adjusted guess for yp : yp = Ax3erx + Bx4erx
Why did this second guess work?
This is the correct form for yp, but why? We have seen why our first guess at
yp won't work. But why does throwing in higher powers of x into these terms
give us the correct guess for yp?
Before we answer this we must first note the following:
Note that if we determined the values of A and B we would have our general
solution to this nonhomogeneous equation be
y = c1erx + c2xerx + c3x2erx + Ax3erx + Bx4erx
Now let's look at the question again, this time without relying on our theorem
from chapter 2 which indicates that throwing in higher powers of x into these
terms give us the correct guess for yp .
Question:
Find the form of yp for the differential equation (D – r)3 y = (–3 + 2x)erx
Solution:
We transform this difficult nonhomogeneous diff. eq. into an easy homogeneous
differential equation:
Question:
Find the form of yp for the differential equation (D – r)3 y = (–3 + 2x)erx
Solution:
If y is a solution to (D – r)3 y = (–3 + 2x)erx
then y is a solution to (D – r)2 [(D – r)3 y]
then y is a solution to
(D – r)5 y
then y is a solution to
(D – r)5 y
then y is a solution to
(D – r)5 y
then y is a solution to
(D – r)5 y
Thus y takes the following form:
=
=
=
=
=
(D – r)2 [(–3 + 2x)erx]
(D – r)2 [(–3 + 2x)erx]
D2 [(–3 + 2x)] erx
0
erx
0
y = c1erx + c2xerx + c3x2erx + c4x3erx + c5x4erx
y = c1erx + c2xerx + c3x2erx + Ax3erx + Bx4erx
yc
yp
Since we know that this first part comes from yc, the second part must come from
yp. Therefore the form of yp is yp = Ax3erx + Bx4erx .
Ex. 8 Find the general solution to (D – 2)5 y = e2x + 3xe2x
The Method of Variation of Parameters
Try to use the method of undetermined coefficients to solve: y″ + y = tan(x)
Suppose we are given the differential equation y″ + P(x) y′ + Q(x) y = f (x)
and we have determined that yc = c1y1 + c2y2. It turns out that yp can be
formed from y1 and y2. In fact there exists some functions u1(x) and u2(x) so
that yp = u1y1 + u2y2.
We are given the differential equation y″ + P(x) y′ + Q(x) y = f (x), and we
have determined that yc = c1y1 + c2y2. Suppose yp = u1y1 + u2y2 is a
particular solution to the nonhomogeneous equation (where u1 and u2 are
some functions unknown to us). We must determine what u1 and u2 are.
yp = u1y1 + u2y2
yp′ = (u1′ y1 + u1y1′ ) + (u2′ y2 + u2y2′ )
= (u1y1′ + u2y2′ ) + (u1′ y1 + u2′ y2 )
We shall now impose a restriction of u1′ y1 + u2′ y2 = 0
So, yp′ = u1 y1′ + u2 y2′
yp″ = (u1 y1′ + u2 y2′ )′
yp″ = (u1′ y1′ + u1 y1″ ) + (u2′ y2′ + u2 y2″ )
yp″ = u1′ y1′ + u2′ y2′ + u1 y1″ + u2 y2″
yp″ = u1′ y1′ + u2′ y2′ + u1[–P(x) y1′ – Q(x) y1] + u2 [–P(x) y2′ – Q(x) y2]
yp″ = u1′ y1′ + u2′ y2′ – P(x) [u1 y1′ + u2 y2′] – Q(x) [u1y1 + u2y2]
yp″ = u1′ y1′ + u2′ y2′ – P(x) [yp′ ] – Q(x) [yp]
yp″ + P(x) yp′ + Q(x) yp = u1′ y1′ + u2′ y2′
f (x) = u1′ y1′ + u2′ y2′
Now we have two conditions placed on u1 and u2:
u1′ y1 + u2′ y2 = 0
u1′ y1′ + u2′ y2′ = f (x)
Now we have two conditions placed on u1 and u2:
u1′ y1 + u2′ y2 = 0
u1′ y1′ + u2′ y2′ = f (x)
Cramer's rule states that we can solve this for u1′ and u2′ as long as
y1
y2
y1
y 2
 0.
Now we have two conditions placed on u1 and u2:
u1′ y1 + u2′ y2 = 0
u1′ y1′ + u2′ y2′ = f (x)
Cramer's rule states that we can solve this for u1′ and u2′ as long as
Cramer's rule states
u 1 
u 2 
0
y2
f (x)
y 2
y1
y2
y1
y 2
y1
0
y1
f (x)
y1
y2
y1
y 2
y1
y2
y1
y 2
 0.
Now we have two conditions placed on u1 and u2:
u1′ y1 + u2′ y2 = 0
u1′ y1′ + u2′ y2′ = f (x)
Cramer's rule states that we can solve this for u1′ and u2′ as long as
Cramer's rule states
u 1 
u 2 
0
y2
f (x)
y 2
y1
y2
y1
y 2
y1
0
y1
f (x)
y1
y2
y1
y 2

 f ( x) y2
W ( y1 , y 2 )

f ( x ) y1
W ( y1 , y 2 )
y1
y2
y1
y 2
 0.
Now we have two conditions placed on u1 and u2:
u1′ y1 + u2′ y2 = 0
u1′ y1′ + u2′ y2′ = f (x)
Cramer's rule states that we can solve this for u1′ and u2′ as long as
Cramer's rule states
u 1 
u 2 
0
y2
f (x)
y 2
y1
y2
y1
y 2
y1
0
y1
f (x)
y1
y2
y1
y 2
Now, integrate to determine u1 and u2.

y1
y2
y1
y 2
 0.
 f ( x) y2
W ( y1 , y 2 )

f ( x ) y1
W ( y1 , y 2 )
u1 
 f ( x) y2
W (y , y
1
2
)
dx
u2 
f ( x ) y1
W (y , y
1
2
dx
)
Ex. 9 Use variation of parameters to solve: y″ + y = tan(x)
Ex. 9 Use variation of parameters to solve: y″ + y = tan(x)
Ex. 10 Use variation of parameters to solve: y″ – 4y = 2e2x
Ex. 10 Use variation of parameters to solve: y″ – 4y = 2e2x
Section 2.4
Mechanical Vibrations
Forces Involved:
FS = –kx (k > 0) Restorative Force (Hooke's Law)
FR = –cx′ (c > 0) Force due to dashpot
Total Force = FR + FS
mx″ = –cx′ – kx
mx″ + cx′ + kx = 0
Later we will include external force and get the differential equation
mx″ + cx′ + kx = F(t)
Terminology:
If c = 0 the motion is undamped
If c > 0 the motion is damped
If F(t) = 0 the motion is free
If F(t) ≠ 0 the motion is forced
Free Undamped Motion
mx″ + kx = 0
:
:
x(t) = A cos(ωot) + B sin(ωot)
where A and B are some constants and  o 
k
m
:
:
x(t) = C cos(ωot – α)
where C and α are some constants and  o 
k
m
Free Undamped Motion
mx″ + kx = 0
:
:
x(t) = A cos(ωot) + B sin(ωot)
where A and B are some constants and  o 
k
m
:
:
x(t) = C cos(ωot – α)
where C and α are some constants and  o 
k
m
Free Undamped Motion
mx″ + kx = 0
:
:
x(t) = A cos(ωot) + B sin(ωot)
where A and B are some constants and  o 
k
m
:
:
x(t) = C cos(ωot – α)
where C and α are some constants and  o 
For this last equation we have the following interpretations:
C
Amplitude
ωo
(Circular) Frequency
α
Phase Angle
o
2
2
o
Frequency
Period
k
m
Ex. 1 A body with mass m = (1/2)kg is attached to the end of a spring that is
stretched 2m by a force of 100 N. It is set in motion with initial position xo = 0.5m
and initial velocity vo = –10m/s. Find the position function of the body as well as
the amplitude, frequency, period of oscillation, and phase angle of its motion.
Ex. 1 A body with mass m = (1/2)kg is attached to the end of a spring that is
stretched 2m by a force of 100 N. It is set in motion with initial position xo = 0.5m
and initial velocity vo = –10m/s. Find the position function of the body as well as
the amplitude, frequency, period of oscillation, and phase angle of its motion.
Damped Free Motion
mx″ + cx′ + kx = 0
Damped Free Motion
mx″ + cx′ + kx = 0
The characteristic equation has roots:
c 
c  4 km
2
2m
Damped Free Motion
mx″ + cx′ + kx = 0
The characteristic equation has roots:
c 
c  4 km
2
2m
Whether these roots are real, complex, or repeated depends on the sign of c2 – 4km.
So, the sign of c2 – 4km really amounts to whether
c = 4 km , c < 4 km , or c > 4 km
Damped Free Motion
mx″ + cx′ + kx = 0
The characteristic equation has roots:
c 
c  4 km
2
2m
Whether these roots are real, complex, or repeated depends on the sign of c2 – 4km.
So, the sign of c2 – 4km really amounts to whether
c = 4 km , c < 4 km , or c > 4 km
Critical damping:
c
(results in repeated roots)
4 km
y  c1 e 1  c 2 xe 1
rx
0.15
0.1
0.05
2
-0.05
-0.1
4
6
8
10
rx
Damped Free Motion
mx″ + cx′ + kx = 0
c 
The characteristic equation has roots:
c  4 km
2
2m
Whether these roots are real, complex, or repeated depends on the sign of c2 – 4km.
So, the sign of c2 – 4km really amounts to whether
c = 4 km , c < 4 km , or c > 4 km
Overdamping:
c
(results in distinct real roots)
4 km
y  c1 e 1  c 2 e
rx
0.2
0.15
0.1
0.05
2
-0.05
-0.1
4
6
8
10
r2 x
Damped Free Motion
mx″ + cx′ + kx = 0
The characteristic equation has roots:
c 
c  4 km
2
2m
Whether these roots are real, complex, or repeated depends on the sign of c2 – 4km.
So, the sign of c2 – 4km really amounts to whether
c = 4 km , c < 4 km , or c > 4 km
Underdamping:
c
(results in complex roots)
4 km
1
y = eAx(c1 cos(Bx) + c2 sin(Bx))
0.75
0.5
0.25
2
-0.25
-0.5
-0.75
-1
4
6
8
10
Section 2.6
Forced Oscillations & Resonance
mx″ + cx′ + kx = F(t)
where F(t) = Fo cos(ωt) or F(t) = Fo sin(ωt)
Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
xc = c1 cos(ωot) + c2 sin(ωot)
(recall  o 
k
m
)
If ω ≠ ωo then we shall guess that xp = A cos(ωt).
Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
(recall  o 
xc = c1 cos(ωot) + c2 sin(ωot)
k
m
)
If ω ≠ ωo then we shall guess that xp = A cos(ωt).
Using our guess of xp = A cos(ωt) in the diff. eq. mx″ + kx = Fo cos(ωt)
we find that A 
Fo
m  o  
2
2

Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
(recall  o 
xc = c1 cos(ωot) + c2 sin(ωot)
k
m
)
If ω ≠ ωo then we shall guess that xp = A cos(ωt).
Using our guess of xp = A cos(ωt) in the diff. eq. mx″ + kx = Fo cos(ωt)
we find that A 
So,
xp 
Fo
m  o  
2
Fo
m   
2
o
2

2

cos   t 
Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
(recall  o 
xc = c1 cos(ωot) + c2 sin(ωot)
k
m
)
If ω ≠ ωo then we shall guess that xp = A cos(ωt).
Using our guess of xp = A cos(ωt) in the diff. eq. mx″ + kx = Fo cos(ωt)
we find that A 
So,
xp 
Fo
m  o  
2
Fo
m   
2
o
2

2

cos   t 
Thus, the general solution is
x  C cos   o t    
Fo
m   
2
o
2

cos   t 
Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
xc = c1 cos(ωot) + c2 sin(ωot)
(recall  o 
k
m
)
If ω ≠ ωo then we shall guess that xp = A cos(ωt).
Question - What happens when ω = ωo?
Answer - Resonance
Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
xc = c1 cos(ωot) + c2 sin(ωot)
(recall  o 
k
m
)
Suppose ω = ωo .
Our guess for xp would be: xp = At cos(ωot) + Bt sin(ωot).
Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
xc = c1 cos(ωot) + c2 sin(ωot)
(recall  o 
k
m
)
Suppose ω = ωo .
Our guess for xp would be: xp = At cos(ωot) + Bt sin(ωot).
:
A=0
B 
F0
2 m 0
Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
xc = c1 cos(ωot) + c2 sin(ωot)
(recall  o 
k
m
)
Suppose ω = ωo .
Our guess for xp would be: xp = At cos(ωot) + Bt sin(ωot).
:
A=0
B 
So,
xp 
F0
2 m 0
F0
2 m 0
t sin   o t 
Undamped Forced Oscillations
mx″ + kx = F(t)
Let's examine mx″ + kx = Fo cos(ωt)
xc = c1 cos(ωot) + c2 sin(ωot)
(recall  o 
k
m
)
Suppose ω = ωo .
Our guess for xp would be: xp = At cos(ωot) + Bt sin(ωot).
:
A=0
B 
So,
xp 
F0
2 m 0
F0
2 m 0
t sin   o t 
Thus, the general solution is x  C co s   o t    
F0
2 m 0
t sin   o t 