Families of curves

1. 1
n-parameter families of curves
For purposes of this discussion, acurvewill mean any equation involvingx,y, and no other
variables. Some examples of curves are
x
2
+ (y¡3)
2
= 9 circle with radius 3, centered at (0;3)
x
2
=yparabola
x=y
2
another parabola
x
3
e
y
+ sin(xy) = 2 ln(x
2
+y
2
+ 1)
I don't know what this looks
like, but it's still a curve
Some curves can obviously be viewed as functions. For instance,
²y=x
2
obviously makesyinto a function ofx, sody=dxmakes sense anywhere along that
curve. We say thatyis expressedexplicitlyas a function ofx; that means we have a formula
of the form
y=
Ã
some expression
not involvingy
!
:
²x=y
2
makes x into a function ofy, sodx=dymakes sense anywhere along that curve. We say
thatxis expressedexplicitlyas a function ofy; that means we have a formula of the form
x=
Ã
some expression
not involvingx
!
:
The circlex
2
+ (y¡3)
2
= 9 satis¯es neither of those conditions | it is neither explicit foryinx,
nor explicit forxiny. We say that the relation betweenxandyis representedimplicitly.
However, even the circle gives us an explicit relationlocally, at most points. For instance,
the upper half-circle
n
(x; y) :x
2
+ (y¡3)
2
= 9; y >3
o
can be rewritten asy=
3 +
p
9¡x
2
, makingyinto a function ofx, sody=dxmakes sense. (Remember that the
symbol
p
means he nonnegative square root of," so it always takes a value greater
than or equal to zero. To make it two-valued you have to put§in front of it.)
In a similar fashion, the lower half circle givesyas a function ofx, wheredy=dxalso makes sense;
and the right or left half circle givesxas a function ofy, wheredx=dymakes sense. Thus both
derivatives are de¯ned everywhere on the circle except at four points. That's enough for most
purposes of this course, which is primarily concerned with thelocal behaviorof functions | i.e.,
what happens near a point. (Global behavior is considered in more advanced courses.)
In a similar fashion, on \most" curves (with exceptions such as a vertical line or a horizontal
line), the derivatives are de¯ned at \most" points (with exceptions at corners, local maxima or
minima, etc.). Rather than thinking of one ofx; yas a unction" of the other, it may be more
helpful to think ofxandyas tworelatedquantities: When one changes, then the other changes in
a corresponding fashion. The derivativesdy=dxanddx=dygive the relative rates of those changes.
Many curves can also be expressedparametrically| i.e., withxandyboth given as functions
of a third variable. For instance, the circlex
2
+(y¡3)
2
= 9 (given by one equation in two variables)
can also be expressed as
x(µ) = 3 cosµ; y (µ) = 3 + 3 sinµ (0·µ·2¼)

2. 2
(two equations in 3 variables). But we won't be using that kind of parametrization in this discussion
(except once near the very end when we consider Clairaut's equation). The discussion in this
document will be largely concerned with another, very di®erent use of the word \parameter":
Aone-parameter family of curvesis the collection of curves we get by taking an equation
involvingx,y, and one other variable | for instance,c(though any other letter will do just as
well). Plugging in di®erent numbers forcgives us di®erent curves; in many cases those curves are
related in some way that is visually simple. Here are some examples:
²The equationx
2
+ (y¡c)
2
= 9 represents a one-parameter family of curves. That is, it
represents in¯nitely many di®erent curves. Indeed, it represents the collection of all circles
that have radius 3 and that are centered at points along the linex= 0. A few of the curves
it represents are
c= 3 x
2
+ (y¡3)
2
= 9 circle centered at (0,3)
c=¡1 x
2
+ (y+ 1)
2
= 9 circle centered at (0;¡1)
c=¼+
p
17x
2
+ (y¡¼¡
p
17)
2
= 9 circle centered at (0; ¼+
p
17)
For comparison, note that this family can also be written asx
2
+ (y¡c)
2
¡9 = 0.
²More generally, iff(x; y) is some function of two variables, thenf(x; y¡c) = 0 is a one-
parameter family of curves. To get the graph of one of these curves from the graph of another,
just translate (move) it vertically.
Similarly,f(x+b; y) = 0 represents a one-parameter family; its curves can be obtained from
each other by horizontal translation.
²The equationy=ax
2
represents in¯nitely many functions ofx. For each value ofa, we gety
equal to one function ofx. The curves represented are the parabolas with axis of symmetry
along they-axis and with vertex at (0;0). They di®er from one another in that they are
stretched vertically by a factor ofa(or stretched horizontally by
p
a).
Similarly, atwo-parameter family of curvesis the collection of curves we get by taking an
equation involvingx,y, and two other variables,provided that that family of curves cannot also be
represented using just one parameter. That last clause is a bit subtle, and will be illustrated by the
third example below. Some examples:
²The equation (x¡a)
2
+(y¡b)
2
= 9 is a 2-parameter family of curves. It represents the collection
of all circles of radius 3. Plugging in some number foraand some number forbgives us one
particular circle of radius 3, the one centered at (a; b). Note that the numbersaandbcan be
chosenindependentlyof each other. We can rewrite the equation as (x¡a)
2
+(y¡b)
2
¡9 = 0.
More generally, the equationf(x¡a; y¡b) = 0 represents the set of all translates (horizontal
and vertical combined) of the curvef(x; y) = 0.
²The equationy=ax+brepresents all straight lines. This is a 2-parameter family.
²At ¯rst glance, the equationy=ax+bxmight look like a 2-parameter family of curves to
beginners. But it is actually a 1-parameter family of curves; it simply has not been written
in the most economical form.

3. 3
Indeed, rewrite it asy= (a+b)x. You'll see that it is a straight line, passing through (0;0),
with slope equal toa+b. The same straight line is represented byy=mxif we simply take
m=a+b. Here are some of the curves that belong to this family of curves:
a b curvem
3 2 y= 5x5
1 7 y= 8x8
12¡7y= 5x5
12¡4y= 8x8
Choosinga= 3 andb= 2 yields thesame curveas choosinga= 12 andb=¡7; both those
curves are represented byy=mxwithm= 5. In other words,
the set of curves that we can get fromy=ax+bxby plugging in some number for
aand some number forb
is the same as
the set of curves that we can get fromy=mxby plugging in some number form.
More generally, for any positive integern, ann-parameter family of curvesis the collection
of curves we get by taking an equation involvingx,y, andnother variables, provided that that
family of curves cannot be represented with fewer parameters.
Once you've understood the de¯nitions, it should be fairly obvious thaty=ax
2
+bx+cis a
3-parameter family whiley=ax+bxis only a 1-parameter family. But some examples are not so
obvious. For instance,
(¤) y=asin
2
x+bcos
2
x+c
looks like a 3-parameter family of curves, because the functions sin
2
xand cos
2
xand 1 (the constant
function) are di®erent functions. But we can rewrite this equation as
y= (a¡b) sin
2
x+ (b+c):
The curves we can get by plugging particular numbers in fora; b; care the same as the curves we
can get from
y=psin
2
x+q
by plugging in particular numbers forpandq. Thus, (¤) is actually a 2-parameter family of curves.
Exercises on families of curves.Represent each of the following as ann-parameter family of curves
| i.e., represent it by an equation inxandywithnadditional variables, wherenis as low as
possible. Simplify as much as you can; then circle your ¯nal answer.
(A)All horizontal lines.
(B)All circles in the plane. (Hint: Translations and size.)
(C)Withyequal to a function ofx: All functions that are polynomials of degree 3 or less.

4. 4
(D)All parabolas that have vertical lines for their axes of symmetry. (Hint: Translations and
vertical stretching, or think in terms of polynomials.)
(E)All ellipses whose axes of symmetry are parallel to thex- andy- coordinate axes. (Hint: Start
with a circle; apply translations, horizontal stretching, and vertical stretching.)
Making Di®erential Equations
Adi®erential equationis an equation involving derivatives. Theorderof a di®erential
equation is the highest number of derivations that have been applied to any term in the equation.
For instance,
d
3
y
dx
3
+ 3x
3
d
2
y
dx
2
+ 2xy
dy
dx
+ 16e
x
y= 0
is a third-order di®erential equation, because it involves
d
3
y
dx
3
but does not involve any higher deriva-
tives.
(This should not be confused with thedegreeof a di®erential equation. That's harder to de¯ne,
but here's an example: The equationy
2
dy
dx
= sin
3
xcould be said to be of third degree, because the
termy
2
(dy=dx)
1
has exponents summing to 3.)
Tosolvea di®erential equation means, roughly, to ¯nd the set of all the functions which satisfy
that di®erential equation. It turns out that, for the most part,
the general solution of annth
order di®erential equation is
ann-parameter family of curves.
(We'll come back to that or the most part" business later, in the last section of this document.)
Most of this course is concerned with solving di®erential equations | i.e.,
Given a di®erential equation, ¯nd the solution.
But, just to help us master the concept of \solution," we'll start with thisreverse problem, which
is much easier:
Given the solution, ¯nd the corresponding di®erential equation.
The procedure is quite simple to describe (though in some problems the the actual computations
get messy):
1.Algebraically solve for one of the parameters. For instance, if one of the parameters isa,
rewrite the equation so that it is in the form
a=
Ã
some expression
not involvinga
!
:

5. 5
(If there is more than one parameter, it doesn't matter which one you start with; pick
whichever one seems most convenient.) Actually, you could instead solve for something of
the form Ã
some expression involving
aand no other letters
!
=
Ã
some expression
not involvinga
!
:
The left side of that equation might be 1=a, or 2a
2
, or 3a
3
+sin(a), or something like that. Any
equation of this sort will do just as well, and one of these variants may be more convenient
than another | i.e., it might allow us a simpler choice on the right side of the equation, and
that will be advantageous in subsequent steps.
2.Di®erentiate both sides of the equation with respect tox. (Or with respect to some other
variable, if that is more convenient.) Keep in mind thatyis a function ofx, so its derivative
isy
0
. Use the chain rule wherever needed. The derivative of any constant is 0. Thus, we get
0 =
Ã
some new expression
not involvinga
!
:
We have eliminatedafrom the equation. But we have also raised the order of the di®erential
equation by one.
3.Now pick another parameter | sayb| and repeat the process. Continue until all the
parameters are gone. Each repetition reduces the number of parameters by one, and raises
the order of the di®erential equation by one; thus ann-parameter solution yields annth-order
di®erential equation.
4.Simplify the result as much as possible.
Following are some examples.
An example with parabolas.Find the di®erential equation whose solution isy=ax
2
.
Answer. Rewrite asa=x
¡2
y. Di®erentiate both sides with respect tox, viewingyas a function
ofx. On the left side we'll just get 0. On the right side, we have to use the rule for the derivative
of a product of two functions:
d
dx
³
x
¡2
¢y
´
=
Ã
d
dx
x
¡2
!
y+x
¡2
Ã
d
dx
y
!
=¡2x
¡3
y+ x
¡2
y
0
:
Thus we get the di®erential equation 0 =¡2x
¡3
y+x
¡2
y
0
. Simplify, to
xy
0
= 2y or
dy
dx
= 2
y
x
:
Checking the answer. Ify=ax
2
, theny
0
= 2ax. Do those satisfy the answer we arrived at? I.e.,
do we havexy
0?
= 2y? Well, that equation simpli¯es tox¢2ax
?
= 2¢ax
2
, which is indeed true.
An example with some bizarre curve.3x
2
y+xcos(xy) +e
x
lny=c. I have no idea what this
curve looks like, but its di®erential equation is not hard to ¯nd.

6. 6
Answer. It's already solved forc; all we have to do is di®erentiate both sides. That's going to
confuse some students, who need to review \implicit di®erentiation" from calculus. I'll break up
into little steps for you:
3x
2
y+xcos(xy) +e
x
lny=c
Di®erentiate both sides;
d
dx
³
3x
2
y
´
+
d
dx
(xcos(xy)) +
d
dx
(e
x
lny) = 0:
Use the product rule, (uv)
0
=u
0
v+uv
0
. Thus
"
d
dx
³
3x
2
´
#
y+ 3x
2
"
d
dx
y
#
+
"
dx
dx
#
cos(xy) +x
"
d
dx
cos(xy)
#
+
"
d
dx
e
x
#
lny+e
x
"
d
dx
lny
#
= 0:
That simpli¯es a little, to
6xy+ 3x
2
y
0
+ cos(xy) +x
"
d
dx
cos(xy)
#
+e
x
lny+e
x
"
d
dx
lny
#
= 0:
We're going to need the chain rule. It helps to think ofyas a function ofx; write it asy(x) if you
like.
dlny
dx
=
dlny
dy
¢
dy
dx
=
1
y
¢y
0
:
Similarly,xyis a function ofx; it could be written asx¢y(x) if you like. Using the chain rule again,
and also the product rule:
dcos(xy)
dx
=
dcos(xy)
d(xy)
¢
d(xy)
dx
= [¡sin(xy)]¢
"
dx
dx
¢y+x¢
dy
dx
#
=¡sin(xy)(y+xy
0
):
Substituting those results into our di®erential equation yields
6xy+ 3x
2
y
0
+ cos(xy)¡xsin(xy)(y+xy
0
) +e
x
lny+e
x
y
¡1
y
0
= 0:
If you want to emphasize the di®erential equation aspect of this, you could group all they
0
terms
together:6xy+ cos(xy)¡xysin(xy) +e
x
lny+ [3x
2
y
0
¡x
2
sin(xy) +e
x
y
¡1
]y
0
= 0. Or, moving
the di®erentials to opposite sides of the equation,
[xysin(xy)¡6xy¡cos(xy)¡e
x
lny]dx= [3x
2
y
0
¡x
2
sin(xy) +e
x
y
¡1
]dy:
An example: Ellipses or hyperbolas with axes on the coordinate axes. That can be
written as
(1) ax
2
+by
2
= 1:
What is the di®erential equation?
Answer. Let's begin by solving fora; we get
(1
0
) a=x
¡2
(1¡by
2
):

7. 7
Now comes our ¯rst di®erentiation. We want to di®erentiate both sides with respect tox. Herea
andbare constants, and we will viewyas a function ofx. Here are some facts from calculus that
we will need. (You may need to review your calculus.)
d
dx
(a) = 0
d
dx
³
x
¡2
´
= ¡2x
¡3
d
dx
(uv) = u
dv
dx
+ v
du
dx
d
dx
³
y
2
´
= 2y
dy
dx
= 2 yy
0
d
dx
³
x
¡2
y
2
´
=x
¡2
d
dx
³
y
2
´
+y
2
d
dx
³
x
¡2
´
= 2x
¡2
yy
0
¡ 2x
¡3
y
2
Thus, di®erentiating equation (1
0
) yields
(2) 0 = ¡2x
¡3
(1¡by
2
) +x
¡2
(¡2byy
0
):
To simplify a little, multiply through by¡x
3
=2; we obtain
(2
0
) 0 = 1 ¡by
2
+bxyy
0
:
Solving forbyields
(2
00
) b=
1
y
2
¡xyy
0
:
If we di®erentiate both sides of that, the left side will get replaced by 0, but the right side will
get messy. Here's a trick to save some work: Before di®erentiating, modify both sides so that the
di®erentiation will be easier. In this example, that can be accomplished by taking the reciprocal on
both sides. Thus we obtain
(2
000
)
1
b
=y
2
¡xyy
0
:
Nowdi®erentiate both sides. The left side still gets replaced by 0, but we save a lot of work on
di®erentiating the right side. The result is
(3) 0 = 2 yy
0
¡yy
0
¡xy
0
y
0
¡xyy
00
:
Simplify toyy
0
=xy
0
y
0
+xyy
00
. This is a second-order di®erential equation, since it involvesy
00
.
But checking this one is messy.
An example: more hyperbolas.Find the di®erential equation for (x¡a)(y¡b) = 4.
Answer. We can writeyas an explicit function ofx:
y=b+
4
x¡a
:

8. 8
Di®erentiating once gets rid ofb; we havey
0
=¡4(x¡a)
¡2
. (Note thaty
0
is less than 0; the function
b+
4
x¡a
is decreasing on each of the two intervals where it is de¯ned.) Now solve that equation
algebraically fora; we get
a=x§(¡
1
4
y
0
)
¡1=2
:
Di®erentiating both sides of that yields
0 = 1§
µ
¡1
2
¶ µ
¡1
4
y
0
¶¡3=2µ
¡1
4
y
00
¶
:
Simplifying,
µ
¡1
4
y
0
¶3=2
=§
1
8
y
00
Square both sides and simplify;(y
00
)
2
+ (y
0
)
3
= 0.
Checking:y
0
=¡4(x¡a)
¡2
andy
00
= 8(x¡a)
¡3
, hence 0
?
= (y
00
)
2
+ (y
0
)
3
= [8(x¡a)
¡3
]
2
+
[¡4(x¡a)
¡2
]
3
= (64¡64)(x¡a)
¡6
p
= 0.
Exercises on making di®erential equations.Find the di®erential equations for these families of
curves. Simplify; then circle your ¯nal answer. (You are urged to check your answers, too.)
(F)Straight lines with slope 2/3. These have equationy=
2
3
x+b. You should get a ¯rst-order
di®erential equation.Optional hint:If you look ahead to the next section, you'll see that this
one-parameter family is additive with a nonzero particular term, so the resulting di®erential
equation should be linear nonhomgeneous.
(G)All straight lines. That'sy=mx+b. Eliminate bothmandb, to obtain a second-order
di®erential equation.Optional hint:This is additive with two homogeneous terms and no
particular term, so the di®erential equation should be linear homogeneous.
(H)The circles tangent to thex-axis at (0;0) have equationx
2
+ (y¡b)
2
=b
2
. Find the corre-
sponding ¯rst-order di®erential equation.Optional hint: Not additive; answer is not linear.
(I)y=ax
2
+bx
¡1=3
.Optional hint: Linear homogeneous.
(J)y=asinx+bcosx.Hint: Instead of solving foraorband then di®erentiating, use this alter-
nate method, which works well for this problem but not for many other problems: Di®erentiate
the given equation to obtain formulas fory
0
andy
00
. Thus we have three equations:
y=asinx+bcosx; y
0
= [what]; y
00
= [what];
where the right sides involvea; b; x, but noty. Then algebraically eliminateaandbfrom
those three equations, to ¯nd one second-order di®erential equation involving some or all of
x; y; y
0
; y
00
but not explicitly mentioningaorb.
(K)y=asin
p
x+bcos
p
x. Hint: Similar to previous problem, but there's more computation.
Don't forget to use the chain rule.

9. 9
Envelope Solutions
(It might be best to postpone this topic for a while | e.g., until one has covered separation of variables
and other elementary methods for solving ¯rst order di®erential equations.)
I stated earlier that,for the most part, the solution of annth order di®erential equation is
ann-parameter family of curves. That's exactly right when we work withlinearequations | our
topic for most of the semester | but we get exceptions with some nonlinear di®erential equations.
The next few pages consider some of the simplest exceptions. (I will postpone until later in the
semester a discussion of what \linear" means in this context.)
Theorem on envelopes.LetG(x; y:c) be a function of three variables. Thus the
equationG(x; y; c) = 0 represents a one-parameter family of curves, if we viewcas a
parameter. Let us denote byEcthe curve given by parameter valuec.
Suppose that that one parameter family of curves satis¯es some ¯rst-order di®erential
equation. Then there may also be an additional curve satisfying that di®erential equa-
tion, which is tangent to eachEcat one point. This curve envelopes theEc's, so it
is called theirenvelope. (In some books it is also called asingular solutionto the
di®erential equation.)
If an envelope solution exists, it can be found by this procedure:
(¤) G(x; y; c) = 0 and
@G
@c
(x; y; c) = 0
is a system of two equations in the three variablesx; y; c. By algebraic means, we may
eliminatec, thus obtaining one equation in just the variablesxandy. That equation is
the envelope curve.
The proof uses the vector version of the chain rule. We won't go into any of the details of that, but
we will give several examples.
Example 1.The nonlinear di®erential equation
dy
dx
=x
q
y¡1 can be solved by the method of
separation of variables, discussed later in this course. The general one-parameter solution turns out
to bey= 1+(
1
4
x
2
+c)
2
, a family of fourth-degree polynomials. We can rewrite that asG(x; y; c) = 0,
if we use the functionG(x; y; c) = 1 + (
1
4
x
2
+c)
2
¡y.
Now compute
@G
@c
=
1
2
x
2
+ 2c. We want to eliminatecfrom the system of two equations (¤)
mentioned in the theorem | that is, the system of these two equations:
1 +
µ
1
4
x
2
+c
¶2
¡y= 0 and
1
2
x
2
+ 2c= 0:
Probably the easiest way to do that is to solve the second equation forc; we getc=¡
1
4
x
2
. Now
plug that in forcin the ¯rst equation. We end up with the horizontal liney= 1 for the envelope
curve in this example.

10. 10
Example 2.The nonlinear di®erential equationy=xy
0
+(y
0
)
2
can be shown to have one parameter
family of solutionsy=cx+c
2
, a family of straight lines. (Thosec's are the same. If you prefer
to see theconly once, write the family instead asy= (c+
1
2
x)
2
¡
1
4
x
2
. But that would make the
computations below more complicated.)
We can represent the solution family asG(x; y; c) = 0 if we use the functionG(x; y; c) =
cx+c
2
¡y.
Now compute
@G
@c
=x+ 2c. We want to eliminatecfrom the system of two equations (¤)
mentioned in the theorem | that is, the system of these two equations:
cx+c
2
¡y= 0 and x+ 2c= 0:
The second of those equations yieldsc=¡
1
2
x. Plug that into the ¯rst equation; thus we obtain
(¡
1
2
x)x+ (¡
1
2
x)
2
¡y= 0. That simpli¯es toy=¡x
2
=4, a parabola.
Actually, we do have to chooseGwith some skill. For instance, we could have represented that
one-parameter family of straight line solutions asy= (c+
1
2
x)
2
¡
1
4
x
2
, or even as
q
y+
1
4
x
2
=c+
1
2
x.
Thus, we could have used the functionG(x; y; c) =
q
y+
1
4
x
2
¡c¡
1
2
x. But that yields@G=@c=¡1,
and so the second equation in the system (¤) is¡1 = 0. Thus, there is no point (x; y) in the
plane that satisfy satis¯es both equations of (¤), since no point satis¯es the second equation. This
approach does not yield the envelope solution | i.e., it yields the wrong answer. [I haven't yet
¯gured out how to avoid this di±culty, except to recommend that you sketch a graph when possible
to see if there might be an envelope solution. Watch for a later edition of this document.]

11. 11
Example 3.The preceding example is a particular instance (obtained withh(c) =c
2
) of an entire
broad class of examples:
Lethbeanycontinuously di®erentiable function of one variable (other than a constant function).
Then the equationy=xy
0
+h(y
0
) is known asClairaut's equation. It's not hard to verify that
the straight linesy=cx+h(c) form a one-parameter solution to Clairaut's equation. After a bit
of computation, we arrive at the envelope solution, which can be expressed parametrically:
x(t) =¡h
0
(t); y (t) =¡th
0
(t) +h(t):
Example 4.The di®erential equationxdy+ydx= 0 has general solutionx
2
+y
2
=c. That's
the set of all circles centered at 0 | a family of concentric circles. Sketching the graph (not shown
here) makes it obvious that there are no envelope solutions. The theorem on envelopes agrees with
that conclusion | it tells us to look for a simultaneous solution of the two equations
x
2
+y
2
=c and ¡1 = 0:
Obviously those have no simultaneous solution | i.e., there are no points (x; y) in the plane that
satisfy both these equations. In this case, that's correct, as we can see by looking at the graph.
Exercises on evelopes.Find the envelope for each of the following families of curves. (You don't
need to ¯nd the di®erential equation.)
(L)y=e
¡x
sin(x+c).Hint:The envelope consists of two curves.
(M)y= sin(
1
2
x
2
+c).Hint:The envelope consists of two straight lines.
(N)(x¡c)(y¡c) =¡1.Hint:The envelope consists of two straight lines.
(O)y¡x= 2(c¡x)(1¡c).
This problem has a history for me personally: When I was a child,
one of the toys I played with was a square frame lined with evenly
spaced pegs, over which one could loop some colored elastic bands.
One of my favorite patterns consisted of the arrangement shown
in the accompanying ¯gure. The straight line for parameterc
passes through the points (c¡1;1¡c) and (c; c); that gives us
the one-parameter family indicated above. I could see a curve be-
ing formed, but I didn't know much geometry back then. The only
curves I knew well were circles, so I guessed the envelope curve
was a quarter of a circle. (I didn't know it was called an envelope.)
Actually, it turns out to be a parabola, not a circle. You just have
to ¯nd the parabola's equation.