CHAPTER
7
Systems of First Order Linear
Equations
7.1
2. As in Example 1, let x1 = u and x2 = u0 . Then x01 = x2 and x02 = u00 = 3 sin t
3u0 2u = 2x1 3x2 + 3 sin t.
4. In this case let x1 = u, x2 = u0 , x3 = u00 , and x4 = u000 . The last equation gives

CHAPTER
5
Series Solutions of Second Order
Linear Equations
5.1
2. Use the ratio test:
lim
n
(n + 1)xn+1 /3n+1
|nxn /3n |
= lim
n
n+11
|x|
|x| =
.
n 3
3
Therefore the series converges absolutely for |x| < 3. For x = 3 and x = 3 the nth
term does not appr

CHAPTER
3
Second Order Linear Equations
3.1
3. Let y = ert , so that y 0 = r ert and y 00 = r2 ert . Direct substitution into the
differential equation yields (12r2 r 1)ert = 0 . Since ert 6= 0, the characteristic
equation is 12r2 r 1 = 0 . The roots of t

CHAPTER
6
The Laplace Transform
6.1
1. The graph of f (t) is shown. Since the function is continuous on each interval,
but has a jump discontinuity at t = 1, f (t) is piecewise continuous.
2. The function is neither continuous nor piecewise continuous. Ob

CHAPTER
4
Higher Order Linear Equations
4.1
2. We will first rewrite the equation as y 000 + (sin t/t)y 00 + (4/t)y = cos t/t. Since
the coefficient functions p1 (t) = sin t/t, p2 (t) = 4/t and g(t) = cos t/t are continuous
for all t 6= 0, the solution is