Quiz 3 - Math 105, Section 204
Name:
Date: Mar 7, 2012
SID:
Time: 20 mins
This quiz has two problems. Look overleaf for the second.
1. Solve the initial value problem
y 0 = tet y −1 ,
y(0) = −5.
(10 points)
Solution: The differential equation is separable, so we rewrite it as
dy
tet
=
dt
y
y dy = tet dt.
(3 points)
Integrating both sides, we get
y2
+C
2
on the left. For the right, we can integrate by parts. Let
u = t, dv = et dt. Then du = dt, v = et , so integration by parts
gives us
Z
t
te − et dt = tet − et + C.
That gives
y2
= tet − et + C.
2
(3 points)
Since y(0) = −5, that tells us y = −5 when t = 0. Hence, we get
25
= −1 + C,
2
that is
27
C= .
2
(3 points)
Finally, we solve for y. We have
√
y = ± 2tet − 2et + 27,
and since y < 0 when t = 0, we must pick the negative sign. So we
get
√
y = − 2tet − 2et + 27.
(1 points)
1
2
2. Determine whether each of the integrals below is finite. If so, find
its value.
Z 3
Z ∞
8
8
dx
and
dx.
3
(5 − 2x)3
0 (5 − 2x)
3
(10 points)
Solution: First, we evaluate the indefinite integral. Using the
substitution u = 5 − 2x, we get du = −2 dx, so
Z
Z
1
2
8 du
1
1
2
= −4
du = −4 · 2 · − + C = 2 + C =
+ C.
3
3
u −2
u
u
2
u
(5 − 2x)2
(2 points)
For the first integral, we see that x = 5/2 is an asymptote, so the
integral is improper and we write it as a sum of limits:
Z 3
Z a
Z 3
8
8
8
dx = lim −
dx + lim +
dx.
3
3
3
a→5/2
b→5/2
0 (5 − 2x)
0 (5 − 2x)
b (5 − 2x)
(3 points)
We get
Z a
2
2
8
dx = lim −
−
= ∞.
lim −
3
2
a→5/2 (5 − 2a)
a→5/2
(5 − 2(0))2
0 (5 − 2x)
Hence the integral is divergent.
(1 points)
For the second integral, the infinite endpoint makes the integral
improper, so we write it as a limit:
Z ∞
Z a
8
8
dx = lim
dx.
3
a→∞ 3 (5 − 2x)3
(5 − 2x)
3
(3 points)
We get
Z a
8
2
2
lim
dx = lim
−
= −2.
3
2
a→∞ 3 (5 − 2x)
a→∞ (5 − 2a)
(5 − 2(3))2
Hence the integral is finite.
(1 points)