CALCULUS
421
Example 51.9. Assume that x is real variable with the property that x2 > 1. Then x2 > 12 .
Hence, we have that
Z
Z
1
1
dx =
dx
2
2
1−x
1 − x2
x
1
arccoth
+ C (by the 4th equation in the HFCIAT)
=
1
1
= arccoth(x) + C.
Example 51.10. Assume that x is a real variable in (0, 1). Then we have that
Z
Z
1
1
√
√
dx =
dx
2
2
x 1−x
x 1 − x2
x
−1
=
arcsech
+ C (by the 5th equation in the HFCIAT)
1
1
= −arcsech(x) + C.
Example 51.11. Assume that x is a real variable that is never 1. Then we have that
Z
Z
1
1
√
√
dx =
dx
x 1 + x2
x 12 + x2
−1
x
=
arccsch
+ C (by the 6th equation in the HFCIAT)
1
1
= −arccsch|x| + C.
Exercise 51.12. Prove the HFCIAT.
422
DANA D. CLAHANE
52. Introduction to Differential Equations
Definition 52.1 (Differential equation). An equation is called a differential equation iff it
contains at least one derivative of some function. Often, we abbreviate this term by “d.e.”
Example 52.2. By Definition 52.1, y 0 = 3y is a differential equation if y is a real-valued,
differentiable function of a real variable, since the equation contains y 0 , which means the
derivative of y.
Exercise 52.3. Assume that y is a real-valued, differentiable function of some real variable.
Is y 0 = 6y − 7 a d.e.?
Example 52.4. Suppose that y is a real-valued, differentiable function of some real variable.
Then y = 67x − 7 is not a d.e., by Definition 52.1, since this equation contains no derivatives
of any function.
Exercise 52.5. Suppose that x is a real variable. Is x3 − 67x2 − 5 = 0 a d.e.?
Definition 52.6 (Ordinary differential equation). Suppose that a d.e. contains derivatives
of a real-valued function y of a real variable. We call the d.e. ordinary and also call the
equation an o.d.e. if Dom(y) ⊂ R.
Example 52.7. The d.e. in Example 52.1 is an o.d.e., by definition, since we assumed in
that example that y is a real-valued function of a single real variable, thus implying that y’s
domain is a subset of R.
Exercise 52.8. Suppose that y : R → R and that y is differentiable. Is y 00 − 6y + y 0 = 90
an o.d.e.?
Definition 52.9 (Order of an ordinary differential equation). Suppose that an o.d.e. contains finitely many derivatives of some real-valued function y. We define the order of the
o.d.e. to be the highest order of any derivative of y that appears in the o.d.e..
Example 52.10. The order of y 0 = 3y in Example 52.1 is 1, by definition, since the highestorder derivative in the o.d.e. is y 0 , which is of derivative order 1.
Exercise 52.11. What is the order of
x00 − x0 + 2 = 0
if x is a real-valued, differentiable function of some real variable t?
Definition 52.12 (Linear ordinary differential equation). Suppose that a first-order o.d.e.
contains derivatives of some (unknown) function y : I → R, where I is a non-degenerate
interval in R. We call the o.d.e. linear iff ∃a, b, c : I → R such that
ay 0 + by + c = 0.
Example 52.13. Consider the first-order o.d.e. y 0 = 3y, where y is a real-valued, differentiable function with domain R as in Example 52.1. We can rewrite this equation as
1y 0 − 3y + 0 = 0.
Therefore, by definition, this first-orde o.d.e. is linear.
CALCULUS
423
Exercise 52.14. Is y 0 = 4(y − 98.6) a first-order linear o.d.e. if y is a real-valued, differentiable function with domain R?
Research Project 52.15. Prove that (y 0 )2 = y or y 0 = y 2 are non-linear, or even prove
that one of these is, where y is a real-valued, differentiable function with domain R.
Definition 52.16 (Solution of an ordinary differential equation). Suppose that a differential
equation contains derivatives of some function y : D → R, where D ⊂ R is an open interval.
We call g : D → R a solution of the differential equation iff g is differentiable and the
equation is true when y is replaced by g.
Example 52.17. Let f : R → R be given by f (t) = e3t . Substituting f for y in the above
example, we see that y 0 (t) = f 0 (t) = 3e3t = 3f (t) = 3y. Therefore, by Definition 52.16, f is
a solution to the d.e. in Example 52.1.
Example 52.18. Let’s show that sin isn’t a solution of the d.e. given in Example 52.1:
Proof. Suppose to the contrary that sin is a solution of y 0 = 3y, assuming that y is a realvalued, differentiable function of a real variable. Then sin0 = 3 sin, by Definition 52.16.
Therefore, sin0 (t) = 3 sin t ∀t ∈ R. Hence we have that cos t = 3 sin t ∀t ∈ R. In particular,
1 = cos 0 = 3 sin 0 = 3(0) = 0. Hence, 1 = 0, and we have arrived at a contradiction.
Therefore, sin isn’t a solution of y 0 = 3y.
Theorem 52.19 (First-Order Linear ODE Solution Theorem). Let I be an interval in R
that is not a single point. Suppose that y : I → (0, ∞) is differentiable, and let M ∈ R be
such that y(t) 6= M ∀t ∈ I. Let k ∈ (−∞, 0) ∪ (0, ∞). Then the following statements are
equivalent:
(i) For all t ∈ I, we have that y 0 (t) = k[y(t) − M ];
(ii) There are C, k ∈ (−∞, 0) ∪ (0, ∞) such that for all t ∈ I, we have
y(t) = M + Cekt .
We will now prove the First-Order Linear ODE Solution Theorem:
Proof. Let I be an interval in R that is not a single point. Suppose that y : I → (0, ∞) is
differentiable, and let M ∈ R be such that y(t) 6= M ∀t ∈ I. Let k ∈ (−∞, 0) ∪ (0, ∞).
To show that (i) and (ii) are equivalent means to show that
(a) If (i) holds, then (ii) holds.
(b) If (ii) holds, then (i) holds.
Assume first that (i) holds. Then we have that ∀t ∈ I,
dy
= k[y(t) − M ].
dt
Denoting y(t) by y, we can rewrite the above equation in the form
dy
= k[y − M ],
dt
424
DANA D. CLAHANE
which implies that
dy = k(y − M )dt.
The above equation can be rewritten in the form
dy
= kdt.
y−M
Thus we have that
Z
dy
= kdt.
y−M
d
(y − M )dy = (1 − 0)dy = 1dy = dy. Thus the above
If we let u = y − M , then du = dy
equation of integrals can be rewritten in the form
Z
Z
du
= kdt,
u
Z
which implies, since k 6= 0 that for some constants C1 and C2 , we have
ln |u| + C1 = kt + C2 .
That is, for some C1 , C2 ∈ R, we have that
ln |y − M | + C1 = kt + C2 ,
which can be rewritten as
ln |y − M | = kt + C2 − C1 .
Thus it must be the case that
eln |y−M | = ekt+C2 −C1 .
It follows that
|y − M | = ekt eC2 −C1 .
Since C2 and C1 can be any real numbers, eC2 −C1 can be any positive number, so let D
denote this real number. Then we have that ∃D > 0 such that
|y − M | = Dekt
Therefore, we must have that
y − M = ±Dekt .
Denote this non-zero constant by C. Then we have that ∃C ∈ (−∞, 0) ∪ (0, ∞) such that
y − M = Cekt .
The above equation can be rewritten as
y(t) = M + Cekt ,
which holds for some constant real-number C ∈ (−∞, 0) ∪ (0, ∞) and ∀t ∈ I. That is, we
have shown that if (i) holds, then (ii) holds.
Next, we show that if (ii) holds, then (i) holds: Suppose that
y(t) = M + Cekt ,
CALCULUS
425
holds for some constant C ∈ (−∞, 0) ∪ (0, ∞) and ∀t ∈ I. Then we have that ∀ such t,
y 0 (t) =
=
=
=
=
d
[M + Cekt ]
dt
d
d
(M ) + C ekt
dt
dt
0 + Cekt k
kCekt
k(M − M + Cekt )
= k(M + Cekt − M )
= k[y(t) − M ].
In other words, (i) holds.
Example 52.20. Suppose y is a positive, real-valued, differentiable function such that
y 0 (t) = 3y(t) and y(t) 6= 0 ∀t ∈ R. The theorem above gives that there is a C ∈
(−∞, 0) ∪ (0, ∞) such that for all t ∈ R,
y(t) = 0 + Ce3t = Ce3t .
(In this case, M is 0.) Let’s verify this without using the theorem so that we can get an
appreciation for the power of the theorem: So assume again that y 0 = 3y. Hence, we have
that
dy
= 3y.
dt
Therefore, by definition of the differential dy, we have that
dy = 3ydt.
Since y(t) 6= 0 ∀t ∈ R, we have that
dy
= 3dt.
y
Since both sides of this equation are equal and represent functions that depend on t, the
indefinite integrals of both sides must be the same, and we can use y as the dummy variable
of integration on the left side, while using t as the dummy variable of integration on the
right side, obtaining
Z
Z
dy
= 3dt.
y
Therefore, there is a constant c ∈ R such that
ln |y| = 3t + c ∀t ∈ R.
Hence,
|y| = e3t+c ,
for some arbitrary constant c ∈ R and ∀t ∈ R. It follows that
y = ±e3t+c ∀t ∈ R.
426
DANA D. CLAHANE
Letting C 0 = ec , and noting that it cannot be 0, we have that
y = ±C 0 e3t ,
where C 0 can be any positive constant. However, if C 0 can be any positive constant, then ±C 0
can be any real constant. That is, we have that for some non-zero real constant C = ±C 0 ,
we have that
y(t) = Ce3t .
On the other hand, if y is given as in the above equation, where C 6= 0, note that y 0 (t) =
3Ce3t = 3y(t).
The above example with M = 0 also exemplifies the following special case of the above
theorem.
Corollary 52.21. Let D ⊂ R. Suppose that y : D → (0, ∞) is differentiable. Then the
following statements are equivalent:
(i.) There is a k ∈ R with k 6= 0 such that for all t ∈ D, y 0 (t) = ky(t);
(ii.) There are C, k ∈ (−∞, 0) ∪ (0, ∞) such that ∀t ∈ D, we have
y = Cekt .
Exercise 52.22. Solve the initial value problem y 0 = −5y, y(0) = 2.
Definition 52.23 (Exponential growth/decay). Suppose that y : I → (0, ∞), where I ⊂ R
is an open interval, and assume that y is differentiable. If in this case, there is a C ∈ R and
a k ∈ R with k 6= 0, such that y(t) = Cekt ∀t ∈ I, then we call C the initial value of y, and
we call k the proportionality constant of y. In this scenario, we say that y has exponential
growth iff k > 0 and that y has exponential decay iff k < 0.
Example 52.24. In the example above, C = Ce0 and is therefore the initial value of y. 3
is the proportionality constant here, and since 3 > 0, we must have that the quantity y has
exponential growth. If in contrast, we change 3 in the above example to −3, then in this
case y has exponential decay, and we have that for some constant C ∈ R,
y = Ce−3t .
Theorem 52.25 (Newton’s Law of Cooling). Suppose that an object whose points are all at
the same temperature T (t) at any given time t. Suppose that the object is surrounded by a
liquid or gas medium that is at a constant temperature M . Then the rate of change of the
temperature T (t) of all points in the object at time t, with respect to t, is directly proportional
to T (t) − M .
Example 52.26. Suppose that a person is killed and that their body is found in a room
which is known to be held at a constant temperature of 70◦ F. When the body is found, its
temperature is 90◦ F. One hour later, the temperature of the body is found to be at 80◦ F.
How long at that point had the person been dead?
CALCULUS
427
By Newton’s Law of Cooling, we must have that the difference between the temperature
T (t) of the body t hours after death and the temperature of the room, 70◦ F, is directly
proportional to T 0 (t). That is, there is a constant k ∈ R such that ∀t ≥ 0, we have that
T 0 (t) = k(T (t) − 70).
By Theorem 52.19, there is a non-zero, real constant C such that
T (t) = 70 + Cekt ∀t ∈ [0, ∞).
We know that the temperature of the body, at the time that the person was killed, was
98.6◦ F. It therefore follows from the equation above with t = 0 that
98.6 = 70 + C.
That is, we have that C = 98.6 − 70 = 28.6. Hence, there is a k ∈ R such that ∀t ≥ 0, we
have that
T (t) = 70 + 28.6ekt .
(80)
Now we also know that if t is the number of hours that elapsed between the time of death
and the first temperature reading, then
90 = 70 + 28.6ekt .
Thus we have that
20 = 28.6ekt ,
which implies that
20
= ekt .
28.6
We now can take ln of both sides:
ln
20
= ln ekt ,
28.6
which can be rewritten as
ln 20 − ln 28.6 = kt.
Dividing both sides by k yields
ln 20 − ln 28.6
.
k
Now we use the second temperature reading 1 hour later, at time t + 1 hours after death:
(81)
(82)
t=
T (t + 1) = 70 + 28.6ek(t+1) ,
which implies, since T (t + 1) = 80, that
80 = 70 + 28.6ek(t+1) .
Subtracting 70 from both sides, we obtain
10 = 28.6ek(t+1) ,
which can be rewritten in the form
10
= ek(t+1) .
28.6
428
DANA D. CLAHANE
Taking ln of both sides yields
ln
10
= ln ek(t+1) ;
28.6
that is,
(83)
ln 10 − ln 28.6 = k(t + 1).
Dividing both sides by k, and then subtracting 1 from both sides of the resulting equation
yields
ln 10 − ln 28.6
(84)
t=
− 1.
k
Since t appears on the left sides of Equations (81) and (84), we can set the right sides of
those two equations equal to each other:
ln 10 − ln 28.6
ln 20 − ln 28.6
=
− 1.
k
k
Multiplying both sides of the above equation by k, we get that
ln 20 − ln 28.6 = ln 10 − ln 28.6 − k.
Therefore,
ln 20 = ln 10 − k.
We then obtain that k + ln 20 = ln 10, so that
k = ln 10 − ln 20 = ln(10/20) = ln(1/2) = ln 2−1 = − ln 2.
Substituting this now known value of k where k appears in Equation (84) above, and adding
1 to both sides of that equation, we have that
ln 10 − ln 28.6
t+1=
∼ 1.52 hours,
− ln 2
which implies that 0.52 hours is an approximate elapsed time from death to the first 90◦ F
temperature measurement of the body. Thus, when the second temperature 80◦ F was taken,
the body had been dead for approximately 1 + .52 = 1.52 hours.
Exercise 52.27. Solve the initial value problem y 0 = 6(y − 4), y(0) = 1.
Exercise 52.28. Find a first-order differential equation that is satisfied by y : R → R given
by
y(t) = 32 + 10e7t .
Exercise 52.29. Johnny hates to cook and is really lazy about even microwaving things.
His wife calls him an hour before getting home and asks him to heat up her leftover pizza
so that its temperature reaches 140◦ F at precisely 15 minutes before she gets home. Assume
that once the microwave is turned off precisely 15 minutes before his wife gets home, in order
to let the pizza cool, that the inside of the microwave oven is held at room temperature 72◦ F.
When Johnny’s wife Penelope gets home, she immediately touches the pizza and feels it is
too cool. She asks Johnny if he heated the pizza up, and he says, “Yes!” The problem is,
Penelope knows calculus and can verify if Johnny is lying. When she first touches the pizza,
its temperature is 74◦ F. Fifteen minutes later, she rechecks the temperature of the pizza and
CALCULUS
429
finds that it is 73◦ F. Did Johnny lie by saying that he ensured that the pizza’s temperature
when the microwave was turned off, was 140◦ F?