Math 211 Fall 2007: Solutions: HW #1
Instructor: S. Cautis
1. section 2.1, #8
(a) Implicit differentiation gives 2t + 2yy 0 = 0. So y satisfies t + yy 0 = 0.
√
(b) We√ have y 2 = C 2 − t2 . So we get two solutions y = C 2 − t2 and
y = − C 2 − t2 . In the first case we get y 0 = 12 (−2t)(C 2 − t2 )−1/2 and so
yy 0 = (C 2 − t2 )1/2 (−t)(C 2 − t2 )−1/2 = −t
so t + yy 0 = 0. Similarly in the second case.
(c) Both solutions are defined as long as C 2 − t2 ≥ 0 or equivalently
C 2 ≥ t2 . Take square roots this is the same as |C| ≥ |t|.
2. section 2.1, #10
−18
9
2
y 0 = (6t−11)
= −2 (6t−11)
Thus y 0 = −2y 2 . Moreover,
2 while −2y
2.
3
y(2) = 6·2−11 = 3. The solution is defined whenever 6t − 11 6= 0 or
equivalently t 6= 11/6.
The graph tends to −∞ as you approach 11/6 from the left and to ∞
as you approach from the right. As t → ±∞ the graph tends to zero.
The graph never crosses the t-axis, crosses the y-axis at −3/11 and passes
through (2, 3).
3. section 2.2, #6
The right side equals (y − 2)(ex + 1) so we can separate variables to get
y 0 /(y − 2) = ex + 1 which gives dy/(y − 2) = (ex + 1)dx. Integrating both
sides gives
ln(|y − 2|) = ex + x + C
or |y − 2| = eC · ee
x
y = Aee +x + 2.
x
+x
. Replacing eC by an appropriate constant this gives
4. section 2.2, #14
Separating variables gives
ydy
= −2tdt
1 + y2
Integrating both sides gives
1
ln(1 + y 2 ) = −t2 + C
2
(notice we don’t need absolute value signs since 1 + y 2 > 0 always). So
1 + y 2 = e2C · e−2t
2
which means
p
y = Ae−2t2 − 1
√
Condition y(0) = 1 implies 1 = A − 1 so A = 2. The interval of existence
2
2
is whenever 2e−2t −
1 ≥ 0 or e−2t ≥ 1/2 or −2t2 ≥ ln(1/2) = − ln(2) or
p
ln(2) ≥ 2t2 or |t| ≤ ln(2)/2.
5. section 2.2, #16 Separating variables gives
e−y dy = ex dx
and integrating we get −e−y = ex + C. Thus −y = ln(−ex − C) so
y = − ln(−ex − C). y(0) = 0 means 0 = − ln(−e0 − C) so −1 − C = 1
so C = −2. The interval of existence is when −ex + 2 > 0 which is when
ex < 2 or x < ln(2).
6. section 2.2, #40
Rx
The area under the curve is 0 y(t)dt. The area of the rectangle is xy. So
we get
Z x
1
y(t)dt = xy
4
0
Taking the derivative of both sides gives
1
(y + xy 0 )
4
Since y(0) = 0 get 4y = y + xy 0 or xy 0 = 3y. Separating variables gives
y(x) − y(0) =
3dx
dy
=
y
x
and integrating gives
ln(y) = 3 ln(x) + C
or y = Ae
3 ln x
3
= Ax .
7. section 2.3, #10
(a) We know terminal velocity is −mg/r = −20. So r = mg/20 =
(70)(9.8)/20 = 34.3. The velocity is given by v(t) = Ce−(34.3)t/70 − 20.
Since v(0) = 0 get C = 20. So the velocity at t = 2 is 20e−(34.3)2/70 − 20 =
20e−0.98 − 20.
The distance travelled is x(t) = −(70)(20)/34.3e−(34.3)t/70−(70)(9.8)t/34.3+
A. Since x(0) = 0 get A = 40.81. But
x(2) = −40.81e−(34.3)(2)/70 − (70)(9.8)(2)/34.3 + 40.81 = −14.5
so distance travelled after 2 secs is 14.5 meters downward.
(b) We want t such that 20e−34.3t/70 − 20 = −(0.8)(20) or e−0.49t = 0.2.
This gives t = ln(0.2)/(−0.49) = 3.285 seconds for the time when object
reaches 80 percent of its terminal velocity.