§4.5 The Superposition Principle and Undetermined Coefficients Revisited
Superposition Principle: Let y1 be a solution to the differential equation
ay 00 + by 0 + cy = f1 (t)
and let y2 be a solution to the differential equation
ay 00 + by 0 + cy = f2 (t).
Then for any constants c1 and c2 , the function c1 y1 + c2 y2 is a solution to the differential
equation
ay 00 + by 0 + cy = c1 f1 (t) + c2 f2 (t).
Method of Undetermined Coefficients (Revisited):
To find a particular solution to the differential equation
ay 00 + by 0 + cy = Pm (t)ert ,
where Pm (t) is a polynomial of degree m, use the form
yp (t) = ts (Am tm + · · · + A1 t + A0 )ert ,
with
1. s = 0 if r is not a root of the associated auxiliary equation;
2. s = 1 if r is a simple root of the associated auxiliary equation;
3. s = 2 if r is a double root of the associated auxiliary equation.
To find a particular solution to the differential equation
ay 00 + by 0 + cy = Pm (t)eαt cos βt + Qn (t)eαt sin βt,
where Pm (t) is a polynomial of degree m and Qn (t) is a polynomial of degree n, use the form
yp (t) = ts (Ak tk + · · · + A1 t + A0 )eαt cos βt + ts (Bk tk + · · · + B1 t + B0 )eαt sin βt,
where k is the larger of m and n, with
1. s = 0 if α + iβ is not a root of the associated auxiliary equation;
2. s = 1 if α + iβ is a root of the associated auxiliary equation.
§4.6 Variation of Parameters
Method of Variation of Parameters:
To determine a particular solution to the differential equation
ay 00 + by 0 + cy = g :
1. Find two linear independent solution {y1 (t), y2 (t)} to the corresponding homogeneous
equation ay 00 + by 0 + cy = 0 and take
yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t).
1
2. Determine v1 (t) and v2 (t) by solving the system
y1 v10 + y2 v20 = 0,
g
y10 v10 + y20 v20 = ,
a
for v10 and v20 and integrating.
3. Substitute v1 (t) and v2 (t) into the expression for yp (t) to obtain a particular solution.
§4.10 Forced Mechanical Vibrations
Suppose that a mass-spring system satisfies the differential equation
m
d2 y
dy
+ b + ky = F0 cos γt,
2
dt
dt
(1)
where b is the damping constant, k is the spring constant (or stiffness) and F0 cos γt is the
external force. Then a particular solution to (1) is given by
yp (t) =
or
F0
[(k − mγ 2 ) cos γt + bγ sin γt]
2
2
2
2
(k − mγ ) + b γ
F0
yp (t) = p
sin(γt + θ)
(k − mγ 2 )2 + b2 γ 2
where tan θ = A1 /A2 = (k − mγ 2 )/(bγ) and the quadrant in which θ lies is determined by the
sign of A1 and A2 .
If a mass is hung vertically and comes to rest at its equilibrium position. At time t = 0, an
external force F (t) = F0 cos γt is applied to the system. Then the steady-state solution for the
system is given by yp (t).
§7.2 Definition of the Laplace Transform:
Definition: Let f (t) be a function on [0, ∞). The Laplace transform of f is the function
defined by the integral
Z ∞
F (s) =:
e−st f (t)dt.
(2)
0
The domain of F (s) is all the values of s for which the integral in (2) exists. The Laplace
transform of f is denoted by both F and L {f }.
Linearity of the Transform: Let f, f1 and f2 be functions whose Laplace transforms exist
for s > α and let c be a constant. Then for s > α,
L {f1 + f2 } = L {f1 } + L {f2 },
L {cf } = cL {f }.
Definition: A function f (t) is said to be of exponential order α if there exists positive constants
T and M such that
|f (t)| ≤ M eαt , for all t ≥ T.
(3)
Condition for existence of the Transform: If f (t) is piecewise continuous on [0, ∞) and
of exponential order α, then L {f }(s) exists for s > α.
Brief Table of Laplace Transforms
2
f (t)
F (s) = L {f }(s)
1
, s>0
s
1
, s>a
s−a
n!
, s>0
n+1
s
b
, s>0
2
s + b2
s
, s>0
2
s + b2
n!
, s>a
(s − a)n+1
b
, s>a
(s − a)2 + b2
s−a
, s>a
(s − a)2 + b2
1
eat
tn ,
n = 1, 2, · · ·
sin bt
cos bt
eat tn ,
n = 1, 2, · · ·
eat sin bt
eat cos bt
§7.3 Properties of the Laplace Transform
Translation in s: If the Laplace transform L {f }(s) = F (s) exists for s > α, then
L {eat f (t)}(s) = F (s − a)
(4)
for s > α + a.
Laplace Transform of the Derivative: Let f (t) be continuous on [0, ∞) and f 0 (t) be
piecewise continuous on [0, ∞), with both of exponential order α. Then, for s > α,
L {f 0 }(s) = sL {f }(s) − f (0).
(5)
Laplace Transform of Higher-Order Derivative: Let f (t), f 0 (t), · · · , f (n−1) (t) be continuous on [0, ∞) and f (n) (t) be piecewise continuous on [0, ∞), with all these functions of
exponential order α. Then, for s > α,
L {f (n) }(s) = sn L {f }(s) − sn−1 f (0) − sn−2 f 0 (0) − · · · − f (n−1) (0).
(6)
Derivative of the Laplace Transform: Let F (s) = L {f }(s) and assume f (t) is piecewise
continuous on [0, ∞) and of exponential order α. Then, for s > α,
L {tn f (t)}(s) = (−1)n
dn F
(s).
dsn
Properties of Laplace Transforms
L {f + g} = L {f } + L {g}
L {cf } = cL {f } for any constant c
L {eat f (t)}(s) = L {f }(s − a)
L {f 0 }(s) = sL {f }(s) − f (0)
L {f 00 }(s) = s2 L {f }(s) − sf (0) − f 0 (0)
L {f (n) }(s) = sn L {f }(s) − sn−1 f (0) − sn−2 f 0 (0) − · · · − f (n−1) (0)
dn
L {tn f (t)}(s) = (−1)n n (L {f }(s))
ds
3
(7)
§7.4 Inverse Laplace Transform
Definition: Given a function F (s), if there is a function f (t) that is continuous on [0, ∞) and
satisfies
L {f } = F,
(8)
then we say that f (t) is the inverse Laplace transform of F (s) and employ the notation
f = L −1 {F }.
Linearity of the Inverse Transform: Assume that L −1 {F }, L −1 {F1 } and L −1 {F2 } exist
and are continuous on [0, ∞) and let c be a constant. Then
L −1 {F1 + F2 } = L −1 {F1 } + L −1 {F2 },
L −1 {cF } = cL −1 {F }.
4