Quiz 8
Math 2250 - Differential Equations & Linear Algebra
Name:
November 20, 2015
Quiz Score:
/10
Answer each question completely in the area below. Show all work and explain your reasoning.
If the work is at all ambiguous, it is considered incorrect. No phones, calculators, or notes are
allowed. Anyone found violating these rules will be asked to leave immediately. Point values are in
the square to the left of the question. If there are any other issues, please ask the instructor.
7
1. Using the definition, compute the Laplace transform of the following function, f (x):
fHxL
1.0
0.8
0.6
0.4
0.2
1
2
3
4
x
Solution: We know that the definition of the Laplace transform is:
Z ∞
F (s) = L {f (x)} :=
e −sx f (x) dx.
0
Here, we can see that f (x) can be effectively broken down into three parts:

0≤x <1

x
f (x) = 2 − x 1 ≤ x < 2 .


0
x ≥2
Thus, this integral reduces to
Z ∞
Z
−sx
e
f (x) =
0
1
xe
Z
−sx
dx +
0
2
(2 − x)e −sx dx,
1
since everywhere else, the integral is 0. We have to compue this by parts, but it’s not too bad:
1
Z
xe
0
−sx
x=1
xe −sx
dx = −
s
1 −sx
1 e
dx = 2 1 − e −s (s + 1) .
s
s
Z
2
+
x=0
Similarly, for the second term, we get
Z 2
Z
(2 − x)e −sx dx = 2
1
1
Z
1
0
2
e −sx −
1
xe −sx =
e −2s s
[e (s − 1) + 1] .
s2
where we had already done the same integration by parts (making this a little easier).Thus, the total
Laplace transform is just the sum of these two parts.
F (s) =
e −2s s
1 −s
1
−
e
(s
+
1)
+ 2 [e (s − 1) + 1] .
s2
s
1/2
Quiz 8
3
Math 2250 - Differential Equations & Linear Algebra
November 20, 2015
2. Find an explicit expression for X(s), the Laplace transform of x(t), but do not invert the transform
to find x(t):
x 00 + 3x 0 + x = 0,
x(0) = 1,
x 0 (0) = −2.
Solution: Here, we use the linearity of the Laplace transform along with the properties we have
below. To start, we Laplace transform both sides of the equation:
L x 00 + 3x 0 + x = L{0}
L x 00 + 3L x 0 + L {x} = 0.
Note that we’ve used both the linearity of the transform and the fact that L {0} = 0. Now, using
the hints at the bottom individually:
L x 00 = s 2 X(s) − sx(0) − x 0 (0) = s 2 X(s) − s + 2,
L x 0 = sX(s) − x(0) = sX(s) − 1,
L {x} = X(s).
Thus, we’re left with
2
s X(s) − s + 2 + 3 [sX(s) − 1] + X(s) = 0.
Rearranging this a bit yields
2
s + 3s + 1 X(s) = s + 1,
=⇒ X(s) =
s2
s +1
.
+ 3s + 1
We could, in theory, invert this to find our original solution x(t), but this was not required for the
problem.
Useful facts:
L f 0 (t) = sF (s) − f (0),
L f 00 (t) = s 2 F (s) − sf (0) − f 0 (0).
2/2