Math 2C
§ 4.3 Homogeneous Linear Equations with Constant Coefficients
Introduction: In this section we consider equations of the form
ay′′ + by′ + cy = 0
(1)
where a, b and c are constants. Let’s assume that all solutions of this equation are of the form y = emx .
Then
and
y′ =
y′′ =
Plug these into equation (1) and we get
Since emx is never zero, the only way the above equation can be satisfied is if
am2 + bm + c = 0
This equation is called the characteristic equation (or auxiliary equation) of equation (1).
Let m1 and m2 be the roots of this quadratic equation. We consider three cases:
•
m1 and m2 real and distinct ( b2 − 4ac > 0 )
•
m1 and m2 real and equal ( b2 − 4ac = 0 )
•
m1 and m2 conjugate complex numbers ( b2 − 4ac < 0 )
Case I: Distinct Real Roots
If the characteristic equation has two unequal real roots m1 and m2 , we get two solutions
(
)
y1 = __________ and y2 = __________ . These functions are linearly independent on −∞,∞ and
hence form a fundamental set. The general solution of (1) on this interval is y = c1e
m1x
+ c2 em2 x .
Case II: Repeated Real Roots
b
(because b2 − 4ac = 0 ). Using the
2a
formula we derived in section 4.2, we get a second solution:
e2m1x
y2 = em1x ∫ 2m x dx = em1x ∫ dx = xem1x
e 1
mx
mx
The general solution is then y = c1e 1 + c2 xe 2 .
When m1 = m2 we get one solution y1 = e
Note: ay′′ + by′ + cy = 0 → y′′ +
m1x
, where m1 = −
b
c
b
b
b
y′ + y = 0 so P ( x ) = . Since m1 = −
then 2m1 = − and
a
a
a
2a
a
b
− ∫ P ( x ) dx = − ∫ dx = ∫ 2m1 dx = 2m1x
a
Zill/Wright – 8e
1
Case III: Conjugate Complex Roots
If m1 and m2 are complex, then we can write m1 = α + iβ and m2 = α − iβ . We get two solutions
y1 = __________ and y2 = __________ . So y = C1e(
α +iβ ) x
+ C2 e(
α −iβ ) x
.
Since we started with only real numbers in our differential equation we would like our solution to only
involve real numbers. To do this we’ll need Euler’s Formula:
eiθ = cosθ + isin θ
A nice variant of Euler’s Formula that we’ll need is: e− iθ = cos ( −θ ) + isin ( −θ ) = cosθ − isin θ
So
eiβ x =
e− iβ x =
and
Adding these two equations gives:
Subtracting these two equations gives:
Since y = C1e(
α +iβ ) x
+ C2 e(
α −iβ ) x
is a solution of (1) for any choice of C1 and C2 , we can choose
C1 = C2 = 1 for the first equation and C1 = 1 , C2 = −1 for the second equation:
y1 = e(
α +iβ ) x
+ e(
α −iβ ) x
(
)
(
)
But
y1 = eα x eiβ x + e− iβ x =
and
y2 = eα x eiβ x − e− iβ x =
and
y2 = e(
α +iβ ) x
− e(
α −iβ ) x
The last two results show that eα x cos β x and eα x sin β x are real solutions of (1). These solutions form
a fundamental set on ( −∞,∞ ) .
The general solution is then
Zill/Wright – 8e
y = c1eα x cos β x + c2 eα x sin β x = eα x ( c1 cos β x + c2 sin β x ) .
2
Example: Solve the following differential equations.
a) 3 y′′ + 2 y′ − 8y = 0
b) y′′ + 14 y′ + 49 y = 0
c) y′′ − 4 y′ + 9 y = 0
Zill/Wright – 8e
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Example: Solve the IVP.
⎛π⎞
⎛π⎞
y′′ + 16 y = 0 ; y ⎜ ⎟ = −10 , y′ ⎜ ⎟ = 3
⎝ 2⎠
⎝ 2⎠
Higher-Order Equations:
Example: Solve the following differential equations.
a)
d4y
d2y
−
7
− 18y = 0
dx 4
dx 2
Zill/Wright – 8e
4
5
4
b) y ( ) − y ( ) + 4 y′′′ + 28 y′′ + 35 y′ + 13y = 0
Zill/Wright – 8e
5