LECTURE 10: HOMOGENEOUS SECOND ORDER LINEAR ODES WITH CONSTANT
COEFFICIENTS
MINGFENG ZHAO
September 30, 2015
Theorem 1. Let p(x) and q(x) be continuous functions, y1 and y2 are two linearly independent solutions to a homogeneous equation y 00 + p(x)y 0 + q(x)y = 0. Then the general solution to y 00 + p(x)y 0 + q(x)y = 0 is:
y = C1 y1 (x) + C2 y2 (x).
Definition 1. Any two linearly independent solutions y1 and y2 to a homogeneous equation y 00 + p(x)y 0 + q(x)y = 0 is
called a fundamental set of solutions to this homogeneous equation y 00 + p(x)y 0 + q(x)y = 0.
Homogeneous second order linear ODEs with constant coefficients
Question 1. Let a, b and c be three constants, how can we solve the differential equation ay 00 + by 0 + cy = 0?
Let’s try y(x) = erx to be a solution, then y 0 = rerx and y 00 = r2 erx . So we get
ay 00 + by 0 + cy = ar2 erx + brerx + cerx = erx [ar2 + br + c].
So y(x) = erx is a solution to ay 00 + by 0 + cy = 0 if and only if r is a zero of ar2 + br + c = 0.
Definition 2. Let a, b and c be three constants, the characteristic equation of the differential equation ay 00 +by 0 +cy = 0
is:
ar2 + br + c = 0.
Theorem 2. Let a, b and c be three constants, the roots of ar2 + br + c = 0 are:
√
√
−b + b2 − 4ac
−b − b2 − 4ac
, and r2 =
.
r1 =
2a
2a
Then
1
2
MINGFENG ZHAO
1) If b2 − 4ac > 0, then r1 and r2 are two different real numbers, and
√
√
−b
b2 − 4ac
b2 − 4ac
−b
r1 =
+
, and r2 =
−
.
2a
2a
2a
2a
The general solution to ay 00 + by 0 + cy = 0 is:
y = C1 er1 x + C2 er2 x .
2) If b2 − 4ac = 0, then r1 = r2 = −
b
is a real number. The general solution to ay 00 + by 0 + cy = 0 is:
2a
−b
−b
y = C1 e 2a ·x + C2 xe 2a ·x .
3) If b2 − 4ac < 0, then r1 and r2 are two different complex numbers, and
√
√
4ac − b2
4ac − b2
b
b
, and r2 = − − i
r1 = − + i
2a
2a
2a
2a
The general solution to ay 00 + by 0 + cy = 0 is:
!
√
b
b
4ac − b2
− 2a
·x
y = C1 e
cos
· x + C2 e− 2a ·x sin
2a
√
!
4ac − b2
·x .
2a
Proof. If b2 − 4ac 6= 0, let r be a solution to the characteristic equation ar2 + br + c = 0 and y = erx , then y 0 = rerx
and y 00 = r2 erx . So we get
ay 00 + by 0 + cy
=
ar2 erx + brerx + cerx
=
erx ar2 + br + c
=
0,
Since ar2 + br + c = 0.
So y = erx is a solution to ay 00 + by 0 + cy = 0.
1) If b2 − 4ac > 0, the we have two different real zeros to ar2 + br + c = 0 are:
√
√
−b − b2 − 4ac
−b + b2 − 4ac
, and r2 =
.
r1 =
2a
2a
So we know that y = er1 x and y = er2 x are two linearly independent solutions to ay 00 + by 0 + cy = 0, which implies
that the general solution to ay 00 + by 0 + cy = 0 is:
√
y = C1 e
−b+
b2 −4ac
x
2a
√
+ C2 e
−b−
b2 −4ac
x
2a
.
b
. We have known that y = erx is a solution to
2a
ay 00 + by 0 + cy = 0, now let’s look for another linearly independent solution y2 . Let y = xerx , then
2) If b2 − 4ac = 0, then the only solution to ar2 + br + c = 0 is r = −
y 0 = erx + rxerx ,
and y 00 = rerx + rerx + r2 xerx = 2rerx + r2 xerx .
LECTURE 10: HOMOGENEOUS SECOND ORDER LINEAR ODES WITH CONSTANT COEFFICIENTS
3
So we get
ay 00 + by 0 + cy
= a[2rerx + r2 xerx ] + b[erx + rxerx ] + cxerx
= erx 2ar + ar2 x + b + brx + cx
= erx (2ar + b) + x(ar2 + br + c)
= erx (2ar + b)
=
0
Since ar2 + br + c = 0
Since r = −
b
.
2a
So y = xerx is also a solution to ay 00 + by 0 + cy = 0. Hence the general solution to ay 00 + by 0 + cy = 0 is:
−b
−b
y = C1 e 2a ·x + C2 xe 2a ·x .
3) If b2 − 4ac < 0, the we have two different complex zeros to ar2 + br + c = 0 are:
√
√
√
−b + b2 − 4ac
b
4ac − b2
b
4ac − b2
r1 =
=− +i
, and r2 = − − i
.
2a
2a
2a
2a
2a
So we know that y = er1 x and y = er2 x are two linearly independent solutions to ay 00 + by 0 + cy = 0. Since r1 and r2
are complex numbers, then y = er1 x and y = er2 x are two complex functions. Notice that
e
r1 x
= e
√
b
+i
− 2a
√
=
b
e− 2a x ei
"
4ac−b2
2a
4ac−b2
2a
x
x
√
!
!#
√
4ac − b2
4ac − b2
cos
x + i sin
x
By Euler’s identity
= e
2a
2a
!
!
√
√
b
b
4ac − b2
4ac − b2
− 2a
x
x
− 2a
= e
cos
sin
x + ie
x
2a
2a
b
x
− 2a
e
r2 x
= e
√
b
− 2a
−i
4ac−b2
2a
√
b
= e− 2a x e−i
"
4ac−b2
2a
x
x
√
!
!#
√
4ac − b2
4ac − b2
= e
cos
x − i sin
x
By Euler’s identity
2a
2a
!
!
√
√
2
2
b
b
4ac
−
b
4ac
−
b
= e− 2a x cos
x − ie− 2a x sin
x .
2a
2a
b
− 2a
x
So we get
e r1 x + e r2 x
2
√
=
e
b
− 2a
x
cos
4ac − b2
x
2a
!
4
MINGFENG ZHAO
e r1 x − e r2 x
2i
√
=
e
b
x
− 2a
sin
!
4ac − b2
x .
2a
√
!
!
√
2
b
4ac − b2
4ac
−
b
cos
x and y = e− 2a x sin
x are two linearly independent
Hence we know that y = e
2a
2a
solutions to ay 00 + by 0 + cy = 0. Hence we know that the general solution to ay 00 + by 0 + cy = 0 is:
!
!
√
√
b
b
4ac − b2
4ac − b2
·x
·x
− 2a
− 2a
cos
sin
y = C1 e
· x + C2 e
·x .
2a
2a
b
x
− 2a
Remark 1. Let i be a number which is called the imaginary unit, that is, i2 = −1. A complex number is the form of:
a + ib,
for some real numbers a and b.
The arithmetic laws of complex numbers: Let a, b, c and d be four real numbers, then
• Addition:
(a + bi) + (c + di) = (a + c) + i(b + d).
• Subtraction:
(a + bi) − (c + di) = (a − c) + i(b − d).
• Multiplication:
(a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + i(ad + bc).
• Division:
(a + bi)(c − di)
ac − adi + bci − bdi2
(ac + bd) − i(ad + bc)
a + bi
=
=
=
.
c + di
(c + di)(c − di)
c2 + d2
c2 + d2
• Conjugation:
a + bi = a − bi.
• Euler’s indentity:
ea+ib = ea · eib = ea [cos(b) + i sin(b)] = ea cos(b) + iea sin(b).
Example 1. Solve y 00 − y 0 − y = 0.
Let y = erx be a solution to y 00 − y 0 − y = 0, then
y 0 = rerx ,
and y 00 = r2 erx .
LECTURE 10: HOMOGENEOUS SECOND ORDER LINEAR ODES WITH CONSTANT COEFFICIENTS
5
So we get
0
=
y 00 − y 0 − y
=
r2 erx − rerx − erx
=
erx [r2 − r − 1].
So we get r2 − r − 1 = 0, which implies that
√
1+ 5
r1 =
,
2
√
1− 5
and r2 =
.
2
So the general solution to y 00 − y 0 − y = 0 is:
y(x) = C1 e
√
1+ 5
·x
2
+ C2 e
√
1− 5
·x
2
.
Example 2. Let a be a real constant, solve y 00 − 4y 0 − 5y = 0, y(0) = 6 and y 0 (0) = 6a. Then find the value of a such
that the solution approaches 0 as x → ∞.
Let y = erx be a solution to y 00 − 4y 0 − 5y = 0, then
y0
=
rerx
y 00
=
r2 erx
0
=
y 00 − 4y 0 − 5y
=
r2 erx − 4rerx − 5erx
=
erx (r2 − 4r − 5).
So r2 − 4r − 5 = 0, that is,
r1 = −1,
and r2 = 5.
So the general solution to y 00 − 4y 0 − 5y = 0 is:
y(x) = C1 e−x + C2 e5x .
Then
y 0 (x) = −C1 e−x + 5C2 e5x .
Since y(0) = 6 and y 0 (0) = 6a, then
C1 + C2 = 6,
and
− C1 + 5C2 = 6a.
6
MINGFENG ZHAO
Solve the above system, we get
C1 = 5 − a,
and C2 = 1 + a.
So we know that the solution to y 00 − 4y 0 − 5y = 0, y(0) = 6 and y 0 (0) = 6a is:
y(x) = (5 − a)e−x + (1 + a)e5x .
Moreover, if we assume y(x) → 0 as x → ∞, then we must have 1 + a = 0, that is, a = −1 .
Example 3. Solve y 00 − 8y 0 + 16y = 0, y(0) = 1 and y 0 (0) = 6.
Let y = erx be a solution to y 00 − 8y 0 + 16y = 0, then y 0 = rerx and y 00 = r2 erx . So we have
0
=
y 00 − 8y 0 + 16y
=
r2 erx − 8rerx + 16erx
=
erx (r2 − 8r + 16).
So r2 − 8r + 16 = 0, that is,
r1 = r2 = 4.
So the general solution to y 00 − 8y 0 + 16y = 0 is:
y(x) = C1 e4x + C2 xe4x .
Since y(0) = 1, then C1 = 1 and
y(x) = e4x + C2 xe4x .
So
y 0 (x) = 4e4x + C2 e4x + 4C2 xe4x .
Since y 0 (0) = 6, then 6 = 4 + C2 , that is, C2 = 2. So the solution to y 00 − 8y 0 + 16y = 0, y(0) = 1 and y 0 (0) = 6 is:
y(x) = e4x + 2xe4x .
Example 4. Find the solution of y 00 − 6y 0 + 13y = 0, y(0) = 0 and y 0 (0) = 10.
Let y = erx be a solution to y 00 − 6y 0 + 13y = 0, then
y0
= rerx
y 00
= r2 erx
0
= y 00 − 6y 0 + 13y
LECTURE 10: HOMOGENEOUS SECOND ORDER LINEAR ODES WITH CONSTANT COEFFICIENTS
7
= r2 erx − 6rerx + 13erx
= erx (r2 − 6r + 13).
So r2 − 6r + 13 = 0, that is,
r1 =
6+
√
√
36 − 4 · 13
6 + −13
=
= 3 + 2i,
2
2
and r2 = 3 − 2i.
Notice that
e r1 x
=
e(3+2i)x = e3x+2ix = e3x · ei(2x)
=
e3x [cos(2x) + i sin(2x)]
=
e3x cos(2x) + ie3x sin(2x).
By Euler’s identity
So the general solution to y 00 − 6y 0 + 13y = 0 is:
y(x) = C1 e3x cos(2x) + C2 e3x sin(2x).
Since y(0) = 0, then C1 = 0, that is, y(x) = C2 e3x sin(2x). So we get
y 0 (x) = 3C2 e3x sin(2x) + 2C2 e3x cos(2x).
Since y 0 (0) = 10, then 10 = y 0 (0) = 2C2 that is, C2 = 5. Therefore, the solution of y 00 − 6y 0 + 13y = 0, y(0) = 0 and
y 0 (0) = 10 is y(x) = 5e3x sin(2x) .
Problems you can do:
Lebl’s Book [2]: All exercises on Page 55 and Page 56.
Braun’s Book [1]: All Exercises on Page 140, Page 144, Page 145, Page 149 and Page 150. Read all materials
in Section 2.2.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca