66 DIFFERENTIAL EQUATIONS
then (9.1) can be rewritten as
dny
n = s n Y ( s) c0 s n 1 c1s n 2 L cn 2 s cn 1
(9.3)
dx
For the special cases of n = 1 and n = 2, Equation 9.3 simplies to
cfw_y ( x ) = sY ( s) c0
(9.4)
cfw_y ( x ) = s 2 Y ( s) c0 s c1
(9.5
48 DIFFERENTIAL EQUATIONS
Figure 7-1
Hookes law: The restoring force F of a spring is equal and opposite to
the forces applied to the spring and is proportional to the extension (contraction) l of the spring as a result of the applied force; that is, F =
46 DIFFERENTIAL EQUATIONS
y = c1e 2 x + c2 e x
3
1
sin 2 x +
cos 2 x
20
20
Solved Problem 6.3 Solve y + y = sec x.
This is a third-order equation with
yh = c1 + c2 cos x + c3 sin x
It follows from Equation 6.8 that
y p = v1 + v2 cos x + v3 sin x
(6.15)
H
CHAPTER 6: Linear Nonhomogeneous Equations
45
Solving this system, we nd that A2 = 2, A1 = 2, and A0 = 3. Hence
Equation 6.13 becomes
y p = 2 x 2 + 2 x 3
and the general solution is
y = yh + y p = c1e 2 x + c2 e x 2 x 2 + 2 x 3
Solved Problem 6.2 Solve y
44 DIFFERENTIAL EQUATIONS
those cases where both methods are applicable, the method of undetermined coefcients is usually the more efcient and, hence, preferable.
As a practical matter, the integration of vi( x ) may be impossible to
perform. In such an e
42 DIFFERENTIAL EQUATIONS
we take yp to be the sum (or difference) of the corresponding assumed solutions and algebraically combine arbitrary constants where possible.
Modications
If any term of the assumed solution, disregarding multiplicative constants,
CHAPTER 6: Linear Nonhomogeneous Equations
41
Dont Forget
yp = A sin bx + B cos bx in its entirety is assumed
for f(x) = k1 sin bx + k2 cos bx even when k1 or k2
is zero, because the derivatives of sines or cosines
involve both sines and cosines.
Generali
40 DIFFERENTIAL EQUATIONS
for obtaining yh when the differential equation has constant coefcients
are given in Chapter Five. In this chapter, we give methods for obtaining
a particular solution yp once yh is known.
The Method of Undetermined Coefcients
Si
38 DIFFERENTIAL EQUATIONS
y = c1e 2 x + d2 e ( 4 + i
2 )x
+ d3 e ( 4 i
2 )x
which can be rewritten, using Eulers relations
e ibx = cos bx + i sin bx and e ibx = cos bx i sin bx
as
y = c1e 2 x + d2 e 4 x e i
2x
+ d3 e 4 x e i
2x
y = c1e 2 x + d2 e 4 x (cos
CHAPTER 5: Linear Homogeneous Differential Equations
37
If the roots are distinct, but some are complex, then the solution is again given by 5.10.
As in the second-order equation, those terms
involving complex exponentials can be combined to yield terms i
36 DIFFERENTIAL EQUATIONS
y = c1e l1 x + c2 e l2 x
(5.6)
In the special case l2 = l1, the solution 5.6 can be rewritten as y =
k1 cosh l1x + k2 sinh l1x.
Case 2. l1 = a + ib, a complex number. Since a1 and a0 in 5.1 and
5.2 are assumed real, the roots of
CHAPTER 5: Linear Homogeneous Differential Equations
35
Example 5.2. The characteristic equation of y ( 4 ) 3 y + 2 y y = 0 is
l4 3l3 + 2l2 1 = 0. The characteristic equation of
d5x
d3x
dx
+ 5 7x = 0
5 3
dt
dt 3
dt
is
l 5 3l 3 + 5l 7 = 0
Caution!
Characte
CHAPTER 7: Second-Order Linear Differential Equations
49
proportionality. Note that the restoring force Fs always acts in a direction
that will tend to return the system to the equilibrium position: if the mass
is below the equilibrium position, then x is
50 DIFFERENTIAL EQUATIONS
Figure 7-2
Kirchhoffs loop law: The algebraic sum of the voltage drops in a simple closed electric circuit is zero.
It is known that the voltage drops across a resistor, a capacitor, and
an inductor are respectively RI, (1/C)q, a
CHAPTER 7: Second-Order Linear Differential Equations
The rst initial condition is I(0) = I0. The second initial condition is obtained from Equation 7.3 by solving for dI /dt and then setting t = 0. Thus,
dI
dt
t =0
=
1
R
1
E ( 0) I0
q0
L
L
LC
(7.8)
An e
CHAPTER 8: Laplace Transforms and Inverse Transforms
F( s) = cfw_ f ( x ) = cfw_x =
63
1
s2
and
cfw_e 4 x x = F( s 4) =
1
( s 4) 2
(c) Set f(x) = e4x. Using Property 8.3 with n = 1 and the results of
Problem 8.1, or alternatively, entry 7 of the Appendix
64 DIFFERENTIAL EQUATIONS
No function of this form appears in the Appendix. Using the results
of Problem 8.3 and Property 8.7, we obtain
s+3
5 1 1 2 1 1
1
=
( s 2)( s + 1) 3
s 2 3
s + 1
5
2
= e2 x e x
3
3
62 DIFFERENTIAL EQUATIONS
Theorem 8.4. If F(s) = cfw_ f(x), then
cfw_u( x c) f ( x c) = e cs F( s)
Conversely,
x<c
0
1cfw_e cs F( s) = u( x c) f ( x c) =
f ( x c) x c
Solved Problems
Solved Problem 8.1 Find cfw_eax.
Using Equation 8.1, we obtain
R
F(
CHAPTER 8: Laplace Transforms and Inverse Transforms
61
1
Theorem 8.3. cfw_u( x c) = e cs .
s
Translations
Given a function f(x) dened for x 0, the function
x<c
0
u( x c ) f ( x c ) =
f ( x c) x c
represents a shift, or translation, of the function f(x
60 DIFFERENTIAL EQUATIONS
You Need to Know
The inverse Laplace transform of a product is
computed using a convolution.
1cfw_F (s)G (s) = f (x ) g (x ) = g (x ) f (x )
(8.10)
If one of the two convolutions in Equation 8.10 is simpler to calculate,
then th
58 DIFFERENTIAL EQUATIONS
transformed into such a form by algebraic manipulation. Observe from
the Appendix that almost all Laplace transforms are quotients. The recommended procedure is to rst convert the denominator to a form that
appears in the Appendi
CHAPTER 8: Laplace Transforms and Inverse Transforms
59
Here Ai, Bj, and Ck (i = 1,2,., m; j, k = 1,2,., p) are constants which still
must be determined.
Set the original fraction a(s)/b(s) equal to the sum of the new fractions just constructed. Clear the
CHAPTER 8: Laplace Transforms and Inverse Transforms
Property 8.4. If cfw_ f(x) = F(s) and if lim
x0
x >0
57
f ( x)
exists, then
x
1
f ( x ) = F(t )dt
s
x
(8.6)
Property 8.5. If cfw_ f(x) = F(s), then
1
x
f (t )dt = F( s)
s
0
(8.7)
Property 8.6. If f
56 DIFFERENTIAL EQUATIONS
Denition of the Laplace Transform
Let f(x) be dened for 0 x < and let s denote an arbitrary real variable. The Laplace transform of f (x), designated by either cfw_ f(x) or F(s),
is
cfw_ f ( x ) = F( s) = e sx f ( x )dx
(8.1)
0
f
54 DIFFERENTIAL EQUATIONS
Solved Problems
Solved Problem 7.1 A 10-kg mass is attached to a spring, stretching it
0.7 m from its natural length. The mass is started in motion from the
equilibrium position with an initial velocity of 1 m/sec in the upward d
52 DIFFERENTIAL EQUATIONS
placed by the cylinder is pr2h, which provides a buoyant force of pr2hr
that must equal the weight of the cylinder mg. Thus,
pr 2 hr = mg
(7.9)
Motion will occur when the cylinder is displaced from its equilibrium position. We ar
CHAPTER 7: Second-Order Linear Differential Equations
undamped when f(t) 0 and a1 = 0. It is classied as free and damped when f(t) is identically
zero but a1 is not zero. For damped motion,
there are three separate cases to consider, according as the root
34 DIFFERENTIAL EQUATIONS
The Characteristic Equation
Second-Order Equations
Corresponding to the differential equation
y + a1 y + a0 y = 0
(5.1)
in which a1 and a0 are constants, is the algebraic equation
l 2 + a1 l + a0 = 0
(5.2)
which is obtained by su
30 DIFFERENTIAL EQUATIONS
If g(x) = 0, then Equation 4.1 is homogeneous; if not, 4.1 is nonhomogeneous. A linear differential equation has constant coefcients if all
the coefcients bj(x) in 4.1 are constants; if one or more of these coefcients is not cons