Math 315 Homework 1 Solutions
§ 1.3 #26: A ball is dropped from rest at a height of 200 m. What is the velocity and position of
the ball 3 seconds later?
Solution: We start with
x00 (t) = −g
where g = 9.8 m/s2 is the acceleration due to gravity. Integrating, we get
x0 (t) = −gt + C1 .
Since the ball is dropped from rest, the initial velocity is zero, so x0 (0) = 0. So C1 = 0. So
x0 (t) = −gt.
Integrating again gives
1
x(t) = − gt2 + C2
2
The ball is dropped from a height of 200 m, so x(0) = 200. So C2 = 200. So
1
x(t) = − gt2 + 200.
2
Thus, the velocity of the ball 3 seconds after dropping it is given by
2
2
x0 (3) = (−9.8 m/s )(3s) = −29.4 m/s .
The position of the ball 3 seconds after dropping it is given by
1
2
x(3) = − (9.8 m/s )(3s)2 + 200 m = 155.9 m.
2
§ 2.1 #8
(a) Use implicit differentiation to show that t2 + y 2 = C 2 implicitly defines solutions of the differential equation t = yy 0 = 0.
Solution: Differentiating both sides of the equation t2 + y 2 = C 2 with respect to t gives
2t + 2yy 0 = 0.
Dividing both sides by 2, we get
t + yy 0 = 0.
Hence, t2 + y 2 = C 2 implicitly defines solutions of the differential equation t = yy 0 = 0.
(b) Solve t2 + y 2 = C 2 for y in terms of t to provide explicit solutions. Show that these functions
are also solutions of t + yy 0 = 0.
Solution: Note
t2 + y 2 = C 2
⇒
y=±
p
C 2 − t2 .
Note
y 0 (t) = ∓ √
t
C2
− t2
.
Thus,
t + yy 0 = t ±
p
C 2 − t2 · √
∓t
= 0.
C 2 − t2
(c) Discuss the interval of existence for each of the solutions in part (b).
√
Solution: The solutions are y = ± C 2 − t2 . Since we cannot take the square root of a
negative number, we need t2 ≤ C 2 , which implies that t ∈ [−C, C]. In order that y 0 = √C∓t
2 −t2
is defined, we also can’t have t = ±C. Thus, the interval of existence is (−C, C).
(d) Sketch solutions for C = 1, 2, 3, 4.
The sketch of solutions for C = 1, 2, 3, 4 should be circles of radius 1, 2, 3, and 4, respectively.
§ 2.2 #14: Find the exact solution of the initial value problem y 0 = −2t(1 + y 2 )/y, y(0) = 1.
Indicate the interval of existence.
Solution: Note
y0 = −
2t(1 + y 2 )
y
⇒
Z
y dy
= −2t
1 + y 2 dt
⇒
y
dy = −
1 + y2
Z
2tdt
Integrating, we get
1
ln(1 + y 2 ) = −t2 + C
2
⇒
ln(1 + y 2 ) = −2t2 + 2C
⇒
1 + y 2 = e−2t
⇒
y 2 = Ae−2t − 1
p
y(t) = ± Ae−2t2 − 1.
⇒
2
+2C
2
Now we use the initial condition y(0) = 1, which gives
√
1 = y(0) = ± A − 1 ⇒
2
= e2C e−2t = Ae−2t
A = 2.
2
Also, we must keep only the solution which contains the initial condition y(0) = 1, so we need the
+ sign in front of the square root. So the exact solution is
p
y(t) = 2e−2t2 − 1.
2
The square root exists provided that 2e−2t − 1 ≥ 0. Also, we don’t want y = 0 since otherwise the
original differential equation
y 0 (t) = −
2t(1 + y 2 )
y
is not defined. Thus, we need
2
2e−2t − 1 > 0
⇒
⇒
⇒
⇒
2
e−2t >
1
2
1
−2t2 > ln( ) = − ln(2)
2
1
2
t < ln(2)
2
r
1
ln(2).
|t| <
2
Thus, the interval of convergence is
r
−
1
ln(2),
2
r
1
ln(2) .
2