Math 308, Sections 301, 302,
Summer 2008
Review Before Test II.
06/23/2008
Chapter 4. Linear Second Order Equations
Section 4.2 Linear Differential Operators
Definition A linear second order equation is an equation that
can be written in the form
a2 (x)
d 2y
dy
+ a0 (x)y = b(x).
+ a1 (x)
dx 2
dx
We will assume that a0 (x), a1 (x), a2 (x), b(x) are continuous
functions of x on an interval I . When a0 , a1 , a2 , b are constants,
we say the equation has constant coefficients, otherwise it has
variable coefficients.
We are interested in those linear equations for which a2 (x) is never
zero on I . In that case we can rewrite (??) in the standard form
dy
d 2y
+ p(x)
+ q(x)y = g (x),
2
dx
dx
where p(x) = a1 (x)/a2 (x), q(x) = a0 (x)/a2 (x) and
g (x) = b(x)/a2 (x) are continuous on I .
(1)
Associated with equation (1) is the equation
y ′′ + p(x)y ′ + q(x)y = 0,
(2)
which is obtained from (1) by replacing g (x) with zero. We say
that equation (1) is a nonhomogeneous equation and that (2) is
the corresponding homogeneous equation.
Given any function y with a continuous second derivative on the
interval I , then y ′′ (x) + p(x)y ′ (x) + q(x)y (x) generates a new
function
L[y ] = y ′′ (x) + p(x)y ′ (x) + q(x)y (x).
What we have done is to associate with each function y the
function L[y ]. This function L is defined on a set of functions. Its
domain is the collection of functions with continuous second
derivatives. We will call this mappings operators. Because L
involves differentiation, we refer to L as a differential operator.
The image of a function t under the operator L is the function
L[y ]. If we want to evaluate this image function at some point x,
we write L[y ](x).
Lemma Let L[x] = y ′′ (x) + p(x)y ′ (x) + q(x)y (x). If y , y1 , and
y2 are any twice-differentiable functions on the interval I and if c is
any constant, then
(a)L[y1 + y2 ] = L[y1 ] + L[y2 ],
(b)L[cy ] = cL[y ].
Any operator that satisfied satisfies both properties from Lemma
for any constant c and any functions y , y1 , and y2 in its domain is
called a linear operator and we can say that ”L preserves linear
combination”. If properties fails to hold, the operator is nonlinear.
Theorem (linear combination of solutions). Let y1 and y2 be
solutions to the homogeneous equation (2). Then any linear
combination C1 y1 + C2 y2 of y1 and y2 , where C1 and C2 are
constants, is also the solution to (2).
Theorem (existence and uniqueness of solution). Suppose
p(x), q(x), and g (x) are continuous on some interval (a, b) that
contains the point x0 . Then, for any choice of initial values y0 , y1
there exists a unique solution y (x) on the whole interval (a, b) to
the initial value problem
y ′′ + p(x)y ′ + q(x)y = g (x),
y (x0 ) = y0 , y ′ (x0 ) = y1 .
Section 4.3. Fundamental solutions of homogeneous equations
Theorem Let y1 and y2 denote two solutions on I to
y ′′ + p(x)y ′ + q(x)y = 0
where p(x) and q(x) are continuous on I . Suppose at some point
x0 ∈ I these solutions satisfy
y1 (x0 )y2′ (x0 ) − y1′ (x0 )y2 (x0 ) 6= 0.
Then every solution to (2) on I can be expressed in the form
y (x) = C1 y1 (x) + C2 y2 (x)
where C1 and C2 are constants.
Definition For any two differentiable functions y1 and y2 , the
determinant
y1 (x) y2 (x) = y1 (x)y ′ (x) − y ′ (x)y2 (x)
W [y1 , y2 ](x) = ′
1
2
y1 (x) y2′ (x) is called the Wronskian of y1 and y2 .
(3)
Procedure for solving homogeneous equations
To determine all solutions to y ′′ + p(x)y ′ + q(x)y = 0:
(a) Find two solutions y1 and y2 that constitute a fundamental
solution set.
(b) Form the linear combination
y (x) = C1 y1 (x) + C2 y2 (x),
to obtain the general solution.
Definition Two functions y1 and y2 are said to be linearly
dependent on I if there exist constants C1 and C2 , not both zero,
such that
C1 y1 (x) + C2 y2 (x) = 0
for all x ∈ I . If two functions are not linearly dependent, they are
said to be linearly independent.
Theorem Let y1 and y2 be solutions to the equation
y ′′ + p(x)y ′ + q(x)y = 0 on I , and let x0 ∈ I . Then y1 and y2 are
linearly dependent on I if and only if the constant vectors
y1 (x0 )
y2 (x0 )
and
y1′ (x0 )
y2′ (x0 )
are linearly dependent.
Corollary If y1 and y2 are solutions to y ′′ + p(x)y ′ + q(x)y = 0
on I , then the following statements are equivalent:
(i) {y1 , y2 } is a fundamental solution set on I .
(ii) y1 and y2 are linearly independent on I .
(iii) W [y1 , y2 ] is never zero on I .
Another representation of the Wronskian for two solutions y1 (x)
and y2 (x) to the equation y ′′ + py ′ + qy = 0 on (a, b) is Abel’s
identity:
Z
x
W [y1 , y2 ](x) = C exp −
p(t)dt ,
x0
where x0 ∈ (a, b) and C is a constant that depends on y1 and y2 .
Section 4.4 Reduction of Order
A general solution to a linear second order homogeneous equation
is given by a linear combination of two linearly independent
solutions.
Let f be nontrivial solution to equation
y ′′ + p(x)y ′ + q(x)y = 0.
Let’s try to find solution of the form
y (x) = v (x)f (x),
where v (x) is an unknown function.
y ′ = v ′ f + vf ′ ,
y ′′ = v ′′ f + 2v ′ f ′ + vf ′′ .
Substituting these expression into equation gives
v ′′ f + 2v ′ f ′ + vf ′′ + p(v ′ f + vf ′ ) + qvf = 0
or
(f ′′ + pf ′ + qf )v + fv ′′ + (2f ′ + pf )v ′ = 0.
Since f is the solution, fv ′′ + (2f ′ + pf )v ′ = 0.
Let’s w (x) = v ′ (x), then we have
fw ′ + (2f ′ + pf )w = 0,
Separating the variables and integrating gives
dw
f′
= (−2 − p)dx,
w
f
Z ′
Z
Z
f
dw
= −2
dx − pdx,
w
f
w = v′ = ±
e−
R
pdx
,
f2
which holds on any interval where f (x) 6= 0.
Z − R p(x)dx
e
v =±
.
[f (x)]2
Section 4.5 Homogeneous Linear Equations with Constant
Coefficients
For the equation
ay ′′ + by ′ + cy = 0
where a, b, c are constants, we try to find a solution of the form
y = erx . If we substitute y = erx into (??), we obtain
ar 2 + br + c = 0
This quadratic equation is called the auxiliary equation or
characteristic equation.
y = erx is a solution to ay ′′ + by ′ + cy = 0 if an only if r satisfies
ar 2 + br + c = 0.
The auxiliary equation has two roots
√
√
−b − b 2 − 4ac
−b + b 2 − 4ac
, r2 =
r1 =
2a
2a
When b2 − 4ac > 0, then r1 , r2 ∈ R and r1 6= r2 . So, y1 (x) = er1 x
and y2 (x) = er2 x are two LI solutions to ay ′′ + by ′ + cy = 0 and
y (x) = c1 er1 x + c2 er2 x
is the general solution to ay ′′ + by ′ + cy = 0.
If b2 − 4ac = 0, then the auxiliary equation has a repeated root
b
r ∈ R, r = − 2a
. In this case, y1 (x) = erx and y2 (x) = xerx are two
LI solutions to ay ′′ + by ′ + cy = 0 and
y (x) = c1 erx + c2 xerx = (c1 + c2 x)erx
is the general solution to ay ′′ + by ′ + cy = 0.
If b2 − 4ac < 0, then the equation (2) has two complex conjugate
roots
√
b
ac − b 2
r1 = − + i
= α + i β,
2a
2a
√
b
ac − b 2
r2 = − − i
= α − i β = r¯1 ,
2a
2a
√
2
b
here i 2 = −1, α = − 2a
, β = ac−b
, α, β ∈ R.
2a
Then two linearly independent solutions to ay ′′ + by ′ + cy = 0 are
eαx cos βx and eαx sin βx and a general solution is
y (x) = c1 eαx cos βx + c2 eαx sin βx,
where c1 and c2 are arbitrary constants.
Cauchy-Euler Equations
A linear second order differential equation that can be expressed in
the form
d 2y
dy
ax 2 2 + bx
+ cy = 0
dx
dx
where a, b, and c are constants is called a homogeneous
Cauchy-Euler equation.
To solve a homogeneous Cauchy-Euler equation, we make the
substitution x(t) = et . Substituting x(t) = et into an equation
gives
2
dy
dy
d y
−
+ cy = 0,
+b
a
2
dt
dt
dt
dy
d 2y
+ cy = 0,
+ (b − a)
2
dt
dt
which is linear second order differential equation with constant
coefficients.
a
Section 4.7 Superposition and nonhomogeneous equations
Let L be a linear differential operator
L[y ](x) = y ′′ + p(x)y ′ + q(x)y .
L maps function y into the function y ′′ + p(x)y ′ + q(x)y .
Suppose that y1 (x) an y2 (x) are two functions, and
L[y1 ](x) = y1′′ (x) + p(x)y1′ (x) + q(x)y1 (x) = g1 (x),
L[y2 ](x) = y2′′ (x) + p(x)y2′ (x) + q(x)y2 (x) = g2 (x).
L[c1 y1 + c2 y2 ](x) = c1 L[y1 ](x) + c2 L[y2 ](x) = c1 g1 (x) + c2 g2 (x).
Theorem 1 (superposition principle) Let y1 be a solution to a
differential equation
L[y1 ](x) = y1′′ (x) + p(x)y1′ (x) + q(x)y1 (x) = g1 (x)
and let y2 be a solution to a differential equation
L[y2 ](x) = y2′′ (x) + p(x)y2′ (x) + q(x)y2 (x) = g2 (x),
where L is a linear differential operator. Then for any constant c1
and c2 , the function c1 y1 + c2 y2 is the solution to the differential
equation
L[y ](x) = c1 g1 (x) + c2 g2 (x).
Theorem 2 (representation of solutions for nonhomogeneous
equations). Let yp (x) be particular solution to the
nonhomogeneous equation
y ′′ + p(x)y ′ + q(x)y = g (x)
on the interval (a, b) and let y1 (x) and y2 (x) be linearly
independent solutions on (a, b) of the corresponding homogeneous
equation
y ′′ + p(x)y ′ + q(x)y = 0.
Procedure for solving nonhomogeneous equations.
To solve y ′′ + p(x)y ′ + q(x)y = g (x):
(a) Determine the general solution c1 y1 (x) + c2 y2 (x) of the
corresponding homogeneous equation.
(b) Find the particular solution yp (x) of the given
nonhomogeneous equation.
(c) Form the sum of the particular solution and a general solution
to the homogeneous equation
y (x) = yp (x) + c1 y1 (x) + c2 y2 (x),
to obtain the general solution to the given equation.
Section 4.8 Method of Undetermined Coefficients
In this section, we give a simple procedure for finding a particular
solution to the equation
ay ′′ + by ′ + cy = g (x),
(4)
when the nonhomogeneous term g (x) is of a special form
g (x) = eαx (Pm1 (x) cos βx + Qm2 (x) sin βx),
where
Pm1 (x) = p0 x m1 + p1 x m1 −1 + p2 x m1 −2 + . . . + pm1 −1 x + pm1
is a polynomial of degree m1 and
Qm2 (x) = q0 x m2 + q1 x m2 −1 + q2 x m2 −2 + . . . + qm2 −1 x + qm2
is a polynomial of degree m2 , α, β ∈ R.
To apply the method of undetermined coefficients, we first have to
solve the auxiliary equation for the corresponding homogeneous
equation
ar 2 + br + c = 0
Particular solutions to ay ′′ + by ′ + cy = g (x)
Type g (x)
I
II
III
IV
V
VI
VII
xk
x k−1
p0 + p1
+ . . . + pk
deαx
eαx (p0 x k + ... + pk )
d cos βx + f sin βx
Pm1 (x) cos βx+
+Qm2 (x) sin βx
eαx (d cos βx + f sin βx)
eαx [Pm1 (x) cos βx+
+Qm2 (x) sin βx]
yp (x)
x s (Ax k + Bx k−1 + ... + F )
x s Aeαx
x s eαx (Ax k + Bx k−1 + . . . + F )
x s (A cos βx + B sin βx)
x s {(A0 x m + ... + Am ) cos βx+
+(B0 x m + ... + Bm ) sin βx}
x s eαx (A cos βx + B sin βx)
x s eαx [(A0 x m + ... + Am ) cos βx+
+(B0 x m + ... + Bm ) sin βx]
In this table s = 0, when α + i β is not a root to the auxiliary
equation, s = 1, when α + i β is one of two roots to the auxiliary
equation, and s = 2, when β = 0 and α is a repeated root to the
auxiliary equation; m = max{m1 , m2 },
Pm1 (x) = p0 x m1 + p1 x m1 −1 + p2 x m1 −2 + . . . + pm1 −1 x + pm1 ,
Qm2 (x) = q0 x m2 + q1 x m2 −1 + q2 x m2 −2 + . . . + qm2 −1 x + qm2
Section 4.9 Variation of Parameters
Consider the nonhomogeneous linear second order differential
equation
y ′′ + p(x)y ′ + q(x)y = g (x).
(5)
Let {y1 (x), y2 (x)} be a fundamental solution set to the
corresponding homogeneous equation
y ′′ + p(x)y ′ + q(x)y = 0.
The general solution to this homogeneous equation is
yh (x) = c1 y1 (x) + c2 y2 (x), where c1 and c2 are constants. To find
a particular solution to (1) we assume that c1 = c1 (x) and
c2 = c2 (x) are functions of x and we seek a particular solution
yp (x) in form
yp (x) = c1 (x)y1 (x) + c2 (x)y2 (x).
Let’s substitute yp (x), yp′ (x), and yp′′ (x) into (1).
yp′ (x) = c1′ (x)y1 (x) + c2′ (x)y2 (x) + c1 (x)y1′ (x) + c2 (x)y2′ (x).
Assume that c1′ (x)y1 (x) + c2′ (x)y2 (x) = 0 , then
yp′ (x) = c1 (x)y1′ (x) + c2 (x)y2′ (x).
yp′′ (x) = c1′ (x)y1′ (x) + c2′ (x)y2′ (x) + c1 (x)y1′′ (x) + c2 (x)y2′′ (x).
yp′′ + p(x)yp′ + q(x)yp = c1′ (x)y1′ (x) + c2′ (x)y2′ (x) = g (x).
We can find c1 (x) and c2 (x) solving the system
′
c1 (x)y1 (x) + c2′ (x)y2 (x) = 0
c1′ (x)y1′ (x) + c2′ (x)y2′ (x) = g (x)
for c1′ (x) and c2′ (x).
Z
Z
−g (x)y2 (x)
g (x)y1 (x)
c1 (x) =
dx , c2 (x) =
dx
W [y1 , y2 ](x)
W [y1 , y2 ](x)
Section 4.11 A closer look at free mechanical vibrations
A damped mass-spring oscillator consists of a mass m attached to
a spring fixed at one end.
Model for the motion of the mass is expressed by the initial value
problem
my ′′ + by ′ + ky = Fexternal ,
y (0) = y0 , y ′ (0) = v0 ,
where m is a mass, b is the damping coefficient, k is the stiffness.
Let’s Fexternal = 0
Undamped free case: b = 0
The equation reduces to
my ′′ + ky = 0| : m
y ′′ + ω 2 y = 0
where ω =
q
k
m.
The solution of this equation is
y (t) = C1 cos ωt + C2 sin ωt
y (t) = A sin(ωt + φ),
where A =
q
C12 + C22 , tan φ =
The motion is periodic with
period 2π/ω
natural frequency ω/2π
angular frequency ω
amplitude A.
C1
C2 .
Underdamped or oscillatory motion (b 2 < 4mk)
The solution to the equation
my ′′ + by ′ + ky = 0
is
y (t) = e αt (C1 cos βt + C2 sin βt) = Ae αt sin(βt + φ),
where
q
b
C1
1 p
4mk − b 2 , A = C12 + C22 , tan φ =
α=−
,β=
.
2m
2m
C2
The solution y (t) varies between −Ae αt and Ae αt with
4πm
2π
=√
and quasifrequency 1/P.
quasiperiod P =
β
4mk − b 2
y (t) → 0 as t → ∞.
An exponential factor Ae αt is called a damping factor.
The system is called underdamped because there is not enough
damping present (b is too small) to prevent the system from
oscillating.
Overdamped motion (b 2 > 4mk)
The solution to the equation
my ′′ + by ′ + ky = 0
is
y (t) = c1 er1 t + c2 er2 t ,
b
b
1 p
1 p
where r1 = −
4mk − b 2 , r2 = −
4mk − b 2
+
−
2m 2m
2m 2m
r2 < 0, and since b 2 > b 2 − 4mk, r1 < 0. y (t) → 0 as t → ∞.
Critically damped motion (b 2 = 4mk) The solution to the
equation
my ′′ + by ′ + ky = 0
is
b
y (t) = (c1 + c2 t)e− 2m t .
y (t) → 0 as t → ∞.