LECTURE 10: CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES
MINGFENG ZHAO
September 24, 2014
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).
In this case, {y1 (x), y2 (x)} is called a fundamental set of solutions to y 00 + p(x)y 0 + q(x)y = 0.
To find the general solution to a constant coefficient second order liner differential equation ay 00 + by 0 + cy = 0:
1) Write the characteristic equation of ay 00 + by 0 + cy = 0:
ar2 + br + c = 0.
2) Find the solutions to the characteristic equation ar2 + br + c = 0.
– Compute b2 − 4ac.
– Solve ar2 + br + c = 0:
∗ If ∆ = b2 − 4ac > 0, then r1 , r2 are two different real numbers, and
√
√
−b
b2 − 4ac
b2 − 4ac
−b
+
, and r2 =
−
.
r1 =
2a
2a
2a
2a
∗ If b2 − 4ac = 0, then we have the same real root, and
r1 = r2 = −
b
.
2a
∗ If b2 − 4ac < 0, then r1 and r2 are two different complex numbers, and
√
√
b
b
4ac − b2
4ac − b2
r1 = − + i
, and r2 = − − i
2a
2a
2a
2a
3) Write the general solution to ay 00 + by 0 + cy = 0:
– If b2 − 4ac > 0, then
y = C1 er1 x + C2 er2 x .
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MINGFENG ZHAO
– If b2 − 4ac = 0, then
−b
−b
y = C1 e 2a ·x + C2 xe 2a ·x .
– If b2 − 4ac < 0, then
√
y = C1 e
b
·x
− 2a
cos
4ac − b2
x
2a
√
!
b
·x
− 2a
+ C2 e
sin
!
4ac − b2
x .
2a
Complex roots
Recall the quadratic formula: the solution to ar2 + br + c = 0 is:
√
−b ± b2 − 4ac
.
r1,2 =
2a
Question 1. How about b2 − 4ac < 0?
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.
Theorem 2 (Euler’s identity). Let a and b be two real numbers, then
ea+ib = ea cos(b) + iea sin(b).
LECTURE 10: CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES
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Consider the equation ar2 + br + c = 0, if b2 − 4ac < 0, then the roots are:
√
√
b
4ac − b2
4ac − b2
b
r1 = − + i
, and r2 = − − i
.
2a
2a
2a
2a
So
√
e
r1 x
e
r2 x
= e
b
− 2a
·x
cos
√
=
b
− 2a
·x
e
cos
4ac − b2
·x
2a
!
4ac − b2
·x
2a
!
√
+ ie
b
− 2a
·x
sin
√
− ie
b
− 2a
·x
sin
4ac − b2
·x
2a
!
!
4ac − b2
·x .
2a
Therefore, we have
√
e
b
− 2a
·x
cos
√
e
b
− 2a
·x
sin
4ac − b2
·x
2a
!
4ac − b2
·x
2a
!
=
er1 x + er2 x
2
=
er1 x − er2 x
.
2i
Example 1. Solve y 00 − y 0 − y = 0.
The characteristic equation of y 00 − y 0 − y = 0 is:
r2 − r − 1 = 0.
Solve r2 − r − 1 = 0, we get
√
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 → ∞.
The characteristic equation of y 00 − 4y 0 − 5y = 0 is:
r2 − 4r − 5 = 0.
Solve r2 − 4r − 5 = 0, we have
r = −1,
and r = 5.
So the general solution to y 00 − 4y 0 − 5y = 0 is:
y(x) = C1 e−x + C2 e5x .
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MINGFENG ZHAO
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.
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. Find the general solution to 16y 00 + 24y 0 + 9y = 0.
The characteristic equation of 16y 00 + 24y 0 + 9y = 0 is:
16r2 + 24r + 9 = 0.
Solve 16r2 + 24r + 9 = 0,then
3
r=− .
4
So the general solution to 16y 00 + 24y 0 + 9y = 0 is:
3
3
y(x) = C1 e− 4 x + C2 xe− 4 x .
Example 4. Solve y 00 − 8y 0 + 16y = 0, y(0) = 1 and y 0 (0) = 6.
The characteristic equation of y 00 − 8y 0 + 16y = 0 is:
r2 − 8r + 16 = 0.
Solve r2 − 8r + 16 = 0, we have
r1 = r2 = 4.
So the general solution to y 00 − 8y 0 + 16y = 0 is:
y(x) = C1 e4x + C2 xe4x .
LECTURE 10: CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES
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 5. Find the general solution to y 00 + k 2 y = 0, where k > 0.
The characteristic equation of y 00 + k 2 y = 0 is:
r2 + k 2 = 0.
Solve r2 + k 2 = 0, then
r1 = ki,
and r2 = −ki.
Then the general solution to y 00 + k 2 y = 0 is:
y = C1 cos(kx) + C2 sin(kx) .
Example 6. Solve y 00 + 2y 0 + 5y = 0, y(0) = 3, y 0 (0) = 1.
The characteristic equation of y 00 + 2y 0 + 5y = 0 is:
r2 + 2r + 5 = 0.
Solve r2 + 2r + 5 = 0, we get
r1 =
−2 +
√
−2 + −16
4−4·5
=
= −1 + 2i,
2
2
√
and r2 = −1 − 2i.
So the general solution to y 00 + 2y 0 + 5y = 0 is:
y(x) = C1 e−x cos(2x) + C2 e−x sin(2x).
Since y(0) = 3, then C1 = 3 and
y(x) = 3e−1 cos(2x) + C2 e−x sin(2x).
So
y 0 (x) = −3e−x cos(2x) − 6e−x sin(2x) + C2 e−x sin(2x) + 2C2 e−x cos(2x).
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MINGFENG ZHAO
Since y 0 (0) = 1, −3 + 2C2 = 1, then C2 = 2. So the solution to y 00 + 2y 0 + 5y = 0, y(0) = 3, y 0 (0) = 1 is:
y(x) = 3e−x cos(2x) + 2e−x sin(2x) .
Example 7. Find the solution of y 00 − 6y 0 + 13y = 0, y(0) = 0 and y 0 (0) = 10.
The characteristic equation of y 00 − 6y 0 + 13y = 0 is:
r2 − 6r + 13 = 0.
Solve r2 − 6r + 13 = 0, we get
r1 =
6+
√
√
36 − 4 · 12
6 + −16
=
= 3 + 2i,
2
2
and r2 = 3 − 2i.
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) .
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