Math 1220
Quiz 6 - Take-home
October 9, 2015
Answer the first 4 questions below; question 5 is for extra-credit only. The
value of each question is indicated at the beginning of it. Due on October
22
Name:
UID:
1. (5 pts) Consider the integral
Z
2
∞
1
√
dx
x+2−1
1. Do you expect it to converge or not? Hint: how does the integrand behave as
x → ∞?.
2. Compute the integral directly and verify your answer in (a).
3. Use the comparison theorem to verify your previous answers.
2. (5 pts) Consider the integral
∞
Z
x4
1
5x + 2
dx
+ 8x2 + 4
1. Do you expect it to converge or not? Hint: how does the integrand behave as
x → ∞?.
2. Use the comparison theorem to justify your previous guess.
3. (5 pts) Consider the integral
Z
1
∞
x2 + 1
dx
x3 + 3x + 2
1. Do you expect it to converge or not?
2. Use the comparison theorem to justify your previous guess. Illuminating hint: since
x ≥ 1, then 3x ≤ 3x3 and 2 ≤ 2x3
Page 2
4. (5 pts) In class we defined the Laplace transform of a function and we illustrated its
power by solving a differential equation of second order. In this exercise I’m asking you
to compute a few properties of the Laplace transform (I did all them except for the last
one in class) and in the next one I’ll ask you to use them to solve a more complicated
differential equation.
Remember that the Laplace transform of a function f (x) is another function defined as
Z ∞
(Lf )(t) =
f (x)e−xt dx
(so this is a function of t)
0
We will assume throughout that limx→∞ f (x)e−tx = limx→∞ f 0 (x)e−tx = 0 for any t.
(i) Use integration by parts to show that (Lf 0 )(t) = −f (0) + t(Lf )(t)
(ii) Use integration by parts to show that (Lf 00 )(t) = −f 0 (0) − tf (0) + t2 Lf
Page 3
(iii) Show that L(eαx )(t) =
1
for
t−α
αx
Remark: we also say that e
1
transform of t−α
.
(iv) Show that L(eαx sin x)(t) =
t > α.
= L−1
1
(t−α)2 +1
1
t−α
(x), namely, that eαx is the inverse Laplace
for t > α.
Hint: if you use integration by parts twice, you will get the integral that you started with,
times a constant.
Page 4
5. (Extra Credit) Use the Laplace transform to solve the following differential equation
y 00 (x) + 4y 0 (x) + 5y(x) = 10ex ,
y(0) = 1,
y 0 (0) = 2
You should follow these steps (as I did in class)
(i) Apply the Laplace transform on both sides of the equation and use the previous
exercise to conclude that
t2 (Ly) − ty(0) − y 0 (0) + 4t(Ly) − 4y(0) + 5(Ly) =
10
t−1
(ii) Use that y(0) = 1 and y 0 (0) = 2 and solve for Ly to conclude that
Ly =
t2 + 5t + 4
(t − 1)(t2 + 4t + 5)
(iii) Use the method of partial fractions and then complete the square to show that
Ly =
1
1
+
2
(t + 2) + 1 t − 1
Page 5
(iv) Finally take the inverse Laplace transform to conclude that
y(x) = e2x sin x + ex
Solution: You don’t need to do anything here other than read. From part (iii) we
have that
1
1
Ly(t) =
+
2
(t + 2) + 1 t − 1
and we have seen in the previous exercise that
1
1
x
−1
−1
=e ,
L
= e−2x sin x
L
t−1
(t + 2)2 + 1
so the solution to our differential equation is
y(x) = ex + e−2x sin x
In conclusion: given any linear differential equation for y(x) involving derivatives
of arbitrarily high order:
(a) Applying the Laplace transform to it we obtain a rational function for L(y(x)).
(b) Using partial fraction decomposition, we can rewrite this rational function as
a combination of simpler rational functions.
(c) One finally looks up these simpler rational functions in a table of Laplace
transforms and inverts them.
• In this exercise we didn’t need the table because I made you compute the
two functions you needed in the previous exercise.
• In practice, once you have obtained the decomposition of Ly in partial
1
fractions (in our case (t+2)1 2 +1 and t−1
) you would look them up in the table
on the next page.
• Each numbered item in that table contains a function followed by its
Laplace transform. In exercise 4, you showed items (2) and (19).
• Note that in class I used the variable x for the original function and t for its
transform. The table on the next page uses t for the original function, and
s for the transform. This is just notation, and it shouldn’t be a problem.
1
in number (2) with a = 1 and you can find
• Note that you can find t−1
1
in number (19), with a = −2 and b = 1.
(t+2)2 +1
Page 6