24
2 .4
CHAPTER 2. FIRST-ORDER EQUATIONS
Functions of Two Variables
The study of a type of first-order equation called exact, to be defined in Section 2.5,
requires that certain basic properties of functions of two variables be established first.
Definition 2.8. A function f of two variables assigns to any pair (x, y) in its domain
a real number denoted byf ( x, y), where the domain of f is either specified or taken
to be all points (x, y) in the plane where f ( x, y) is defined.
Several such functions have already been encountered in the preceding sections.
For example,
f(x, y) = x 2 + xy - y3, g(x, y) = - + 1 and h(x, y) = Jx - y
y
X
are functions of two variables. In the.above, the domain off is the entire plane IR2,
that of g is the set of all points (x, y) with y i= 0, and that of his the set of all points
(x, y) with x - y 2'. 0, i.e., x 2'. y.
It is important to note that, in the definition of a function of two variables, y is
not a function of .1:, and that x and y may be assigned values independently. In other
words, x and y are independent variables. For example,
J(x, y) = y - x 2 + 1 ⇒ J(l, 2) = 2, J(2, 1) = -2, and f(0, 0) = 1.
It is only a relation, such as f(x, y) = 0 or f(x, y) = c, which defines y as a function
of x. For example, f(x, y) = y - x 2 + 1 = 0 ⇒ y = x 2 - 1.
2.4.1
Partial Derivatives
The (first-order) derivative of a functionf of a single variable x is the function f' or
df
defined by
dx
f
J(x + h) - J(x).
J'(x) = d (x) = lim
1i-o
h
dx
If z = f(x), then dz = J'(x) is the rate of change of z with respect to x.
dx
In contrast to functions of one variable, there are different ways in which functions
of two variables can be differentiated.
2.4. FUNCTIONS OF TWO VARIABLES
25
Definition 2.9. Let f be a function of two variables x and y. The partial derivative
off with respect to x is the function j� or �� of two variables defined by
f(x+h,y)-J(x,y)
_of x, y -l·
(
) - Im
!
X ( x, y) .
h->O
h
uX
!:}
The partial derivative off with respect to y is the function J� or �� of two variables
defined by
f
. J(x,y + h)-J(x,y)
- a (x,y ) _
- 1Im
fy (x,y ) _
.
h->O
h
uy
!:}
oz
= fx (x,y) is the rate of change of z with respect to x with
ox
y held constant, and �: = j�(x, y) is the rate of change of z with respect toy with
If z = J(x,y), then
x held constant.
Example 2.18. Let f(x,y) = x4y2 + 2x- 3y+ e x - ln(y) - sin(y) + cos(x) + 2. Then
fx (x,y) = 4x3y2 + 2 + e x - sin(x) and fy (x,y) = 2x4y - 3 - � - cos(y).
y
In the computation of fx , y is regarded as a constant, and all expressions which are
'independent of x, such as y, ln(y) and sin(y) have derivative with respect to x equal to
0. In the computation of fy , xis regarded as a constant, and all expressions which are
independent of y, such as x, e x and cos(x) have derivative with respect toy equal to 0.
A function J of two variables has four second-order partial derivatives, namely,
the first-order partial derivatives of the two functions fx and fy • The four possibilities
are as follows:
fxx
a Jx = O a f = 3 f 1s. the partia. 1 denvat1ve
. h respect to x,
.
. of fx wit
2
ox ox
ox2
f
a2 f .
.
� fx = �� = � 1s tl1e partia
. 1 denvative
.
. of 1·x with respect toy,
uy
uy ux uyux
f
o2 f .
.
.
. i denvative
. of fy with respect to x,
1s the partia
fy
ox· = ox oy = oxoy
f
a2 f .
.
��
. i denvative
.
. of fy with respect toy.
1s the partia
=
=
1
·
y
�
2
uy
uy uy
uy
ox
a
aa
a
aa
a
aa
!:}
26
CHAPTER 2. FIRST-ORDER EQUATIONS
Example 2.19. Let f(x,y) = x3 y4 + e 2x ln(y) + x2 - y3 - 3. Then
3x2 y4 + 2e2x ln(y) + 2x,
e2x
4x3 y3 + - - 3y2 ,
y
4
6xy + 4e2x ln(y) + 2,
fx (x,y)
fy (x,y)
fxx (x,y)
2e 2x
12x2 y3 + --,
y
2x
?e
12x2 y3 + -�-,
y
2x
e
12x3 y 2 - - 6y.
y2
fxy (x,y)
Note that, in this example, fxy = fyx · Although this is not true in general, the
conditions under which it is true are given by the following:
Theorem 2.1. If fxy and fyx are continuous in an open region R of the (x,y)-plane
(i.e., R excludes its boundary), then f�y (x,y) = fyx (x,y) for all (x,y) in R.
Continuity of a function of two variables is defined in a manner analogous to the
definition for a function of a single variable:
Definition 2.10. A function f of two variables is continuous at a point (a, b) if
lim
(x,y)-+(a,b)
f(x,y) = f(a, b).
f is continuous in a region R if f is continuous at every point (a, b) in R.
2.4.2
The Chain Rule
The chain rule for functions of a single variable states that, if z = f ( x) and x = g ( t),
then the composition h(t) = f(g(t)) has derivative
h'(t) =
.
.
.
!
h(t) = J'(g(t))g'(i).
�
��
In other words, 1f z 1s a funct10n of x and xis a function oft, then - = --.
dx dt
dt
Several versions of the chain rule exist for functions of two (or more) variables.
The one of interest in the present study is the following:
2.4. FUNCTIONS OF TWO VARIABLES
27
Theorem 2.2. (The Chain Rule) If f is a function of two variables and x and
y are functions of a single variable t, then the composition h(t) = f(x(t), y(l)) has
derivative
h'(l) =
d
�
�
J(x(t), y(t)) = fx(x(t), y(t)) dl + fy (x(l), y(l)) dt .
dl
(2.1)
In other words,if z = f (x, y) and x and y are functions of t, then
dz
dt
8 f dx
8x d t
8 f dy
8y dl
-=--+--.
Note that, in the special case where z = f(x) is independent of y, �� = 0 and
dz
df dx
d z dx .
.
8f
df
- =-,and
the above formula reduces to - = -- = -- , 1.e.,the cham rule
dl
d x dl
dx dl
8x
dx
for functions of one variable.
Example 2.20. Let f(x, y) = x2 + y3, x(t) = t3 - 6l + 1 and y(t) =et +3t. Then
3
h(t) = J(x(t), y(l)) = (t - 6t +1)2 +(e t +3t)3
is a function of t, and
h' (t)
8 ! dy
8 ! dx
+
= 2x(t)(3t2 - 6) +3[y(t)] 2(e l +3)
8y d t
8X dt
2
t
2(t3 - 6l +1)(3t2 - 6) +3(e +3t) (i +3).
In Theorem 2.2, consider the special case where x(t) = l. In that case,
and Equation (2.1) reduces to
dx
=1,
dt
or, employing the fact that x =t,
d
dy
J(x, y(x)) = fx(x, y(x)) + fy (x, y(x)) dx ,
dx
which is the version of the chain rule required in the study of exact equations.
Exercises 2.4
1. Let f(x, y) = x3 - y2 + x2 y2 + 1.
(a) Determine J(l, 0), f(3, 1), f(l, 3) and f(2, -1).
(2.2)
28
CHAPTER 2. FIRST-ORDER EQUATIONS
(b) Find fx (x,y) and fy (x,y), and the values of these functions at the points
(1, 0), (3, 1), (1, 3) and (2, -1).
(c) Find fxx (x,y), fxy (x,y), fyx (x,y) and fyy (x,y), and the values of these
functions at the points (1, 0), (3, 1), (1, 3) and (2, -1).
2. Let J(x,y) = x2 ln(y) - y3 ex + x3 - y2 + 1. Find fx , fy , fxx , fxy , fyx and fyy ·
3. Let f(x,y) = x4 + 3x2 y2 + y4 , x(t) = t 2 + 2t and y(t) = t3 - 2t. Determine
d
f ( X (t) ' y ( t)).
dt
4. Let f(x,y) = x 2 - cos(y) + x sin(y), x(t) = 2t 2 + 3t - 1 and y(t) =et + 2t.
Determine : f(x(t),y(t)).
t
el + l and y(t)
5. Let f(x,y) = sin(xy) + y!XY, x(t)
d
f ( X (t)' y (t)).
dt
ln(t). Determine
6. Let f(x,y) = x4 + 3x2 y2 + y4 and y(x) = x3 + 6x2 - x. Determine ..!!:.._J(x,y(x)).
dx
7. Let f(x,y) = eY - x2 + y3 ln(x) and y(x) = sin(x). Determine :J(x,y(x)).
d
8. Let f(x,y) =
X2
l
d
+ xe- y and y(x) = sin(2x). Determine xf(x,y(x)).
+y
d
2
9. Let f(x,y) = x4 - y4 - l. The relation J(x,y) = 0 defines y as a function of x.
dy
Determine - m
. terms of x and y.
dx
10. Let f(x,y) = x3 + y2 - eY . The relation f(x,y) = 0 defines y as a function of
dy
x. D eterminc - m
. terms of x and y.
dx
2.5
Exact Equations
Consider a first-order equation expressed in the form
P(x,y) + Q(x,y)
dy
= 0,
dx
(2.3)
where P and Q are functions of two variables.
Suppose that there exists a function J of two variables such that fx = P and
fy = Q. Then Equation (2.3) takes the form
fx + fy
dy
=
dx 0
29
2.5. EXACT EQUATIONS
which, by Equation (2.2), is !I:_ J(x,y(x)) = 0. Integration with respect to x then
dx
yields f(x,y(x)) = c, an arbitrary constant. Thus, the general solution of Equation
(2.3) is J(x,y) = c. The problem, then, is the determination of the function f, called
a potential function.
Definition 2.11. An equation of the form P(x,y) + Q(x,y) ;� = 0, with P and Q
continuous in an open region R of the (x, y)-plane, is exact in R if there exists a
potential function J defined for all (x, y) in R.
Example 2.21. Consider the equation y + xy' = 0. Evidently, J(x,y) = xy is a
potential function because fx = y = P and fy = x = Q. Hence, the equation is exact
in IR 2 , with the general solution f(x,y) = c, i.e., xy = c.
Theurern 2.3. Suppose that P and Q have continuous first-order partial derivatives
in an open region R of the plane. If the equation P(x,y) + Q(x,y)y' = 0 is exact in
R, then Py (x,y) = Qx (x,y) for all (x,y) in R.
Proof. Since the equation is exact, there exists a potential function f. Then fx = P
fxy = Py and fyx = Qx , Since Py and Qx are continuous in R, fxy and
and fy = Q
are
continuous
in R, hence equal by Theorem 2.1. Thus, Py = Qx in R.
fyx
*
Theorem 2.3 is equivalent to the following:
Theorem 2.4. If Py =I= Qx in R, then the equation P(x,y) + Q(x,y)y' = 0 is not
exact in R.
Proof. Suppose that Py =I= Qx in R. If the equation is exact, then Py = Qx by
Theorem 2.3, which contradicts Py =I= Qx . Hence, the equation is not exact.
Example 2.22. Consider the equation x 2 + y2 + xy
dy
= 0.
dx
P(x,y) = x 2 + y2 , Q(x,y) = xy, Py = 2y, Qx = y, Py =I= Qx, hence, the equation is
not exact, by Theorem 2.4.
It is not true, in general, that if Py = Qx, then the equation P(x, y)+Q(x,y)y' = 0
is exact. In order to state a converse of Theorem 2.3, i.e., a theorem which gives the
conditions under which exactness follows from Py = Qx , the following definition is
required.
30
CHAPTER 2. FIRST-ORDER EQUATIONS
Definition 2.12. A region R in the (x, y)-plane is connected if it consists of a single
"piece." It is simply connected if it is connected and has no "holes" in it.
Definition 2.12 is, of necessity, merely informal, since the mathematically rigorous
definitions of connected and simply connected sets are beyond the scope of this book.
For example, the region R enclosed by a non-self-intersecting closed curve (e.g.,
a circle) is simply connected. The entire plane is simply connected. However, the
plane with the origin excluded is not simply connected due to the "hole" at the origin.
Theorem 2.5. Suppose that P and Q have continuous first-order partial derivatives
in an open, simply connected region R of the plane. If Py = Qx in R, then the
equation P(x, y) + Q(x, y)y' = 0 is exact in R.
Note that, if the region R is not simply connected, then the equation may not be
exact, even if Py = Qx Example 2.23. Consider the equation x 2 + y2 + 2xyy' = 0.
2
P(x, y) = x + y 2 , Q(x, y) = 2xy, Py = 2y, Qx = 2y, Px = 2x and Qy = 2x
are continuous in the plane, which is simply connected, and Py = Qx . Hence, the
equation is exact by Theorem 2.5. A potential function f exists, and satisfies the two
conditions fx = P and fy = Q.
fx = P
⇒
f(x, y) =
J
P(x, y) dx =
J
3
x 2 + y2 dx = � + xy2 + g(y),
where g(y) is an arbitrary function of y. Note that, in the above integration with
respect to x, y is regarded as a constant, and g(y) has the role of the arbitrary
d
"constant" of integration, being the most general expression such that
g(y) = .
0
dx
Then fy = Q ⇒
2xy + g'(y) = 2xy
⇒ g'(y) = 0 ⇒ g(y) = c1
and the general solution of the equation is
f ( x, y) =
x3
3 + xy + c1 = c2, or
2
=?
f ( X, Y) =
x3
3
+ xy2 + C1,
2.5. EXACT EQUATIONS
31
Example 2.24. Consider the equation 4x3 - 2x + 3y + (3x + 2y)y' = 0.
P(x,y) = 4x3 - 2x + 3y, Q(x,y) = 3x + 2y, all first-order partial derivatives of both
P and Q are continuous in the plane, which is simply connected, and Py = 3 = Qx .
Hence, the equation is exact by Theorem 2.5. A potential function f exists, and
satisfies the two conditions fx = P and fy = Q.
fx = P
=>
f(x,y) =
J
P(x,y)dx =
f
2
4x3 -2x+3ydx=x4 -x +3xy+g(y),
where g(y) is an arbitrary function of y. Then fy = Q ⇒
3x + g'(y) = 3x + 2y
⇒ g'(y) = 2y ⇒ g(y) = y2 + c1
⇒ J(x,y) = x4 - x2 + 3xy + y2 + c1,
and the general solution of the equation is
I
'f(x,y) = x4 -x2 +3xy+y2 +c 1 = c2, or x4 -x2 +3xy+y2 = k.
Note that, if we begin with the condition fy = Q instead of fx = P, then the result
will be the same. Thus,
fy = Q =>
J(x,y) =
J
Q(x,y)dy=
f
2
3x+2ydy= 3xy+y +g(x),
where g(x) is an arbitrary function of x, being the most general expression such that
d
g(x) = 0. In the above integration with respect toy, x is regarded as a constant.
dy
Then fx = P =>
3
3y+g'(x)=4x -2x+3y
=>
⇒ g'(x)=4x3 -2x ⇒ g(x)=x4 -x2 +c 1
f(x,y) = 3xy + y2 + x4 - x 2 + c 1,
as before.
Example 2.25. Consider the equation x2 + y2 + xyy' = 0.
As shown in Example 2.22, this equation is not exact because Py =I= Qx. Thus, any
attempt to determine a function f such that fx = P and fy = Q must fail. Proceeding
as in the last two examples,
fx = P =>
and fy = Q ⇒
f(x,y) =
J
P(x,y)dx =
2xy + g'(y)= xy
J
3
x 2 + y2 dx = � + xy2 + g(y),
⇒ g'(y)= -xy,
32
CHAPTER 2. FIRST-ORDER EQUATIONS
which is a contradiction because the left-hand side is independent of x, whereas the
right-hand side is not, making their equality impossible. It follows that a function
f(x,y) with fx = P and fy = Q does not exist.
Example 2.26. Consider the initial-value problem
x
ye + sin(y) +[e x +xcos(y) + 1]:� = 0, y(0)= -3.
P(x,y) = ye x + sin(y), Q(x,y)=ex +xcos(y) + 1, all first-order partial derivatives
of both P and Q are continuous in the plane, and Py = e x +cos(y) = Qx - Hence, the
equation is exact by Theorem 2.5.
fx = P ⇒ f(x,y) =
and fy = Q ⇒
J
P(x,y)dx =
J
ye x + sin(y) dx = ye + x sin(y) + g(y),
x
ex +xcos(y) + g'(y)=e x +xcos(y) + 1
⇒ g'(y) = 1 ⇒ g(y) = y + c1
⇒ f(x,y)=yex +xsin(y)+y +c1,
and the general solution is ye x + x sin(y) + y = k. Setting x = 0 and y = -3 then
gives k = -6, and the solution of the initial-value problem is ye x + x sin(y) + y = -6.
Example 2.27. Consider the initial-value problem
2
yln(x)+2x-[x- xln(x)+3y ]y'=0, x>0, y(1)=2.
P(x,y) = y ln(x) +2x, Q(x,y) = -x+x ln(x)-3y2 , all first-order partial derivatives of
both P and Qare continuous in the right half-plane x > 0, which is simply connected,
and Py = ln(x) = Qx - Hence, the equation is exact by Theorem 2.5.
fy = Q
and fx = P ⇒
⇒ J(x,y)
J Q(x,y)dy = f-x +xln(x)- 3y dy
2
-xy + xy ln(x) - y3 + g(x),
-y + y ln(x) + y + g'(x) = y ln(x) + 2x
⇒ g'(x) = 2x ⇒ g(x) = x2 + c1
⇒ J(x,y) = -xy + xy ln(x) - y3 + x2 + c1,
and the general solution is xy ln(x) - xy + x2 - y3 = k. Setting x = 1 and y = 2 then
gives k = -9, and the solution of the initial-value problem is
xyln(x)- xy +x 2 - y3 = -9.
33
2.5. EXACT EQUATIONS
2.5.1
Integrating Factors
If the equation P(x, y) + Q(x, y)y' = 0 is not exact because Py =/- Qx , then it can be
made exact by multiplication by an integrating factor I(x, y), chosen such that the
resulting, equivalent, equation
T(x, y)P(x, y) + I(x, y)Q(x, y)y' = 0
(2.4)
is exact. The exactness of Equation (2.4) requires that (IP)y = (IQ)x - Employing
the product rule, this condition becomes
This partial differential equation for the unknown function I ( x, y) is more difficult to
solve than the given ordinary differential equation, and simplifying assumptions must
be made, which may or may not succeed.
Seek an integrating factor J(x) which is independent of y. Then Iy = 0, Ix = I'(x),
and the condition (2.5) becomes
I'(x)
I(x)
(2.6)
If the right-hand side of Equation (2.6) is independent of y, then I(x) is determined
by Equation (2.6). Otherwise, Equation (2.6) is a contradiction because the left-hand
side is independent of y. In that case, I ( x) does not exist.
If I(x) does not exist, then seek an integrating factor I(y) which is independent
of x. Then Ix = 0, Iy = J'(y), and the condition (2.5) becomes
I'(y)
I(y)
(2.7)
If the right-hand side of Equation (2.7) is independent of x, then I(y) is determined
by Equation (2.7). Otherwise, Equation (2.7) is a contradiction because the left-hand
side is independent of x. In that case, I (y) does not exist.
Example 2.28. Consider the equation x 2 + y 2 + xyy' = 0.
As shown in Example 2.22, this equation is not exact because Py =/- Qx - Since
Py - Qx
Q
2y -y
1
xy
X
is independent of y, I ( x) exists and is determined by
J'(x)
I(x)
1
X
(
CHAPTER 2. FIRST-ORDER EQUATIONS
34
⇒ ln III= ln /xi ⇒
I(x) = ±x,
where arbitrary constants of integration are unnecessary and have been suppressed,
because we seek only one solution I(x), and not all possible solutions. Since both
±x are integrating factors, we select the simpler one, I(x) = x. The equation then
becomes
and is exact. Then
fx = P
x
3
+ xy2 + x2 yy' = 0,
⇒ f(x,y) =
J
⇒
⇒
and fy = Q ⇒
2
2
x y+g'(y)=x y
I
g (y)=0
x
l
1
+ xy2 dx = 4 x4 + 2 x2 y2 + g(y),
3
1 4
1
f(x,y)= x + x y +c1,
⇒
g(y)=c 1
2
4
. 1
.
and the general solution 1s x 4 + 1 x2 y2 = k�, or x4 + 2x2 y2 = 4k = k1-
2 2
2
4
Example 2.29. Consider the equation
jx + y
3
+ 2xy2 y' = 0,
X > 0.
P(x,y) = ..jx + y , Q(x,y) = 2xy , Py (x,y) = 3y , Qx (x,y) = 2y , Py =/= Qx, hence,
3
2
the equation is not exact. Since
2
Py - Qx - 3y
Q
2
2 -
2y2 = �
2x
2xy2
is independent of y, I ( x) exists and is determined by
I'(x)
= �
I(x)
2x
⇒ ln II(x)I = !2 ln(x) = ln (x 112 ) ⇒
I(x) = ±x 112 = ±Jx.
With I(x) = x 112, the equation becomes
x+x f y
1 2 3
and is exact. Then
fx = p
+ 2x3f2y2 y' = o,
⇒ J(x,y) = J x + x
y dx = 1x
2
1f 2 3
+ �x3f2y3 + g(y),
and fy = Q ⇒
2x3 f 2 y2 + g'(y) = 2x3 f2y2
⇒
g(y) = c1
⇒
J(x,y) =
1
2
2
x
+
2
. . 1 2 2 3/2 3
3/ 2 3
2
- k 1and the general solut10n 1s x + x y = k�, or 3x + 4x Y _
2
3
3/2 3
3x
y + c1,
35
2.5. EXACT EQUATIONS
Example 2.30. Consider the equation
X
x2
+y+ (
2
)
+ X + XY y1 = 0.
Py= 1, Q x = x + 1 + y, Py # Q x , hence, the equation is not exact. Since
-x-y
+ X + xy
x2
is not independent of y, I(x) does not exist. Since
x+
Qx - Py
y = l
=
p
x+y
is independent of x, J(y) exists and is determined by
ThP equation then becomes
��?
= 1, which gives J(y) = eY .
2
(x+y)e Y + (� +x+xy)eY y'=O,
and is exact. Then
and fu = Q ==>
)e Y
(�x2 +xy+x)e Y +g'(y)= (�x2 +xy+x
1
==>
==>
g(y)=c1
f (x, y) = ( � x + xy)eY + c1,
2
and the general solution is (�x2 + xy)e Y = k, or ( x2 + 2xy) e Y = k 1.
Example 2.31. Consider the initial-value problem
4xy + (x 2 +
v'Y) y' = 0, y > 0, y(l) = 4.
Py = 4x, Q x = 2x, Py -I Qx , hence, the equation is not exact. An integrating factor
I (y) exists and is determined by
J'(y)
J(y)
-2x
4xy
1
2y
==>
1
ln II(y)I=-- ln(y) ==> I(y) =y- 112.
2
36
CHAPTER 2. FIRST-ORDER EQUATIONS
The equation then becomes
and is exact. Then
fx = 4xy112 ⇒ f(x, y) = J 4xy112 dx = 2x 2 y 112 + g(y),
and fv = Q ⇒
x 2 y-l/ 2 + g'(y) = x 2y -l/ 2 + 1
⇒ g'(y) = 1 ⇒ g(y) = y + C1
⇒ J(x, y) = 2x2y 112 + y + c1,
and the general solution is 2x 2 y 112 + y = k. Setting x = 1 and y = 4 gives k = 8, and
the solution of the initial-value problem is 2x 2 y112 + y = 8.
Exercises 2. 5
1. Find the general solution of 3x2 + y + 1 - (3y2 - x + l)y' = 0.
2. Find the general solution of xy + x + 1 + ( �x 2 + y + 1)
!� =
0.
3. Find the general solution of 3x2y2 - 2x + (2x3 y + 3y2)y' = 0.
4. Solve the initial-value problem y3 - 3x2y2 + 4x3 + (3xy2 - 2x 3 y - 4y3 ) y' = 0,
y(0) = 1.
5. Solve the initial-value problem y' =
3x 2y2 - y - 2xy3
, y(2) = 1.
3x 2y2 - 2x 3 y+x
6. Solve the initial-value problem 21 + y + ( � + x)
2
y(l) = 4.
��
= 0, x > 0, y > 0,
7. Find the general solution of (6xy + eY)y' + xex + 3y2 = 0.
.
.
2+ Y2
8. Fmd the general solution of 2xe x
d
+ 4x3 + ( 2yex2+Y2 - 2y dy = 0.
x
)
9. Find the general solution of ex+y + x + y + cos(x) + (ex+y + x - y) y' = 0.
x
y
y
- 2x + ( 2
10. Solve the initial-value problem 2
+ 2y) dd = 0,
2
2
x
x+y+l
x+y+l
y(0) = 0.
37
2.5. EXACT EQUATIONS
11. Find the general solution of
dy = .
2y - 3
2x + 2
-------O
+ 1 + -------2
2
2
2
x + y + 2x - 3y + 3
x + y + 2x - 3y + 3 dx
12. Consider the equation
✓x x+ y + 1 + ( ✓x y+ y + 1) dxdy = 0.
2
2
2
2
(a) Show that Py = Qx in the region R = {(x,y): (x,y) =/= (0,0)}.
(b) Solve the equation.
(c) Is the equation exact in R?
13. Consider the equation
X2
y
+y
X
X2
2
+y 2
yI = O.
(a) Show that Py = Qx in the region R = {(x, y): (x, y) =/= (0, 0)}.
(b) Solve the equation.
(c) Is the equation exact in R?
14. Solve the initial-value problem x + 6y2 + 4xyy' = 0, y(2) = 0, and express y in
terms of x.
15. Find the general solution of x3 + 2y2 + xyy' = 0.
16. Find the general solution of e- x - cos(y) + sin(y )y' = 0.
17. Find the general solution of xex + xy2 + yy' = 0.
2
18. Find the general solution of ft+ y2 + 4xyy' = 0, x > 0.
19. Find the general solution of x + y2 + 3xyy' = 0.
20. Solve the initial-value problem 4xy + (6x 2 + y)y' = 0, y(l) = 1.
21. Find the general solution of sin(x) + [e-Y - cos(x)] y' = 0.
22. Find the general solution of y + (1 + 2x + xy)y' = 0.
23. Find the general solution of
dy
(y)
= ;r�
, y > 0.
dx
x n y + 2x
\
38
CHAPTER 2. FIRST-ORDER EQUATIONS
Chapter 2 Exercises
1. Consider the equation xy + y2 - x2 y' = O.
(a) Solve it as a homogeneous equation.
(b) Solve it as a Bernoulli equation.
2. Consider the equation
y2
X2
- y'
= 0.
(a) Solve it as a separable equation.
(b) Solve it as a homogeneous equation.
(c) Solve it as a Bernoulli equation.
(d) Find an integrating factor which makes the equation exact and solve it.
(e) Find the orthogonal trajectories of the one-parameter family of curves
defined by the general solution.
3. Consider the equation x + y - xy' = 0.
(a) Solve it as a homogeneous equation.
(b) Solve it as a linear equation.
(c) Find an integrating factor which makes the equation exact and solve it.
( d) Find the orthogonal trajectories of the one-parameter family of curves
defined by the general solution.
4. Consider the equation 3xy' + 2y = 0.
(a) Solve it as a separable equation.
(b) Solve it as a homogeneous equation.
(c) Solve it as a linear equation.
(d) Find an integrating factor which makes the equation exact and solve it.
(e) Find the orthogonal trajectories of the one-parameter family of curves
defined by the general solution.
5. Find the orthogonal trajectories of the one-parameter family of curves defined
by the equation 2x - 2e-x - y2 = k.
6. Find the orthogonal trajectories of the one-parameter family of curves defined
by the equation y = 2x ln Jxl + ex.
7. The relation f(x, y) = 0 defines y as a function of x, and x as a function of y.
dy
dx
dy
dx
Determine - and - in terms of Jx and fy, and express - in terms of -.
dy
dX
dX
dy
CHAPTER 2 EXERCISES
39
8. Consider the equation 3x - 2y + (y - 2x)y' = 0.
(a) Solve it as an exact equation and express yin terms of x.
(b) Solve it as a homogeneous equation.
9. Consider the equation xex - x ln(x) + y + xy' = 0, x > 0.
(a) Solve it as an exact equation and express yin terms of x.
(b) Solve it as a linear equation.
10.
(a) Show that any separable equation can be expressed as an exact equation.
(b) Solve the equation y + x ln(x) ln(y)y' = 0, x > 0, y > 0, as a separable
equation.
(c) Solve the separable equation in part (b) as an exact equation.
11. Consider an equation of the form y' = J(ax +by+ c), b-=/- 0.
(a) Show that u( x) = ax + by + c transforms the equation into one which is
separable.
(b) Solve y' = (x + y + 1)2.
1
(c) Solve y' =
, x + y > 2.
Jx + y- 2
(d) Solve y' = (x + y) ln(x + y) - 1, x + y > 0.
12. Consider an equation of the form a(y) + [b(y)x - c(y)]y' = 0.
(a) Show that an integrating factor I (y) exists.
(b) Regard y as the independent variable and x as the unknown function of
y. Show that the resulting equation for x(y) is linear, employing the fact
1
that y' = - by the chain rule (see Exercise 7).
x'
13. Consider the equation y + [(2y2 + l)x - 2y]y' = 0.
(a) Find an integrating factor which makes the equation exact and solve it.
(b) Proceed as in Exercise 12(b) and solve the linear equation.
14. Consider the equation y2 + (3xy + 2)y' = 0.
(a) Find an integrating factor which makes the equation exact and solve it.
(b) Proceed as in Exercise 12(b) and solve the linear equation.
15. Solve the equation 1 + (xy - x- 3 y3 )y' = 0 by transforming it into an equation
for x as a function of y (see Exercise 12(b)).
40
CHAPTER 2. FIRST-ORDER EQUATIONS
16. Consider the equation y' = 2-jy=1".
(a) Find the general solution.
(b) By inspection of the equation, find a solution which is not a particular
solution, i.e., it does not correspond to any particular value of the arbitrary
constant c. Such a solution is called a singular solution, and may occur
only as the solution of a nonlinear equation.
(c) The envelope of a one-parameter family of curves is a curve C with the
property that, at every point of C, it is tangent to the graph of a member
of the family. The envelope of the one-parameter family of curves defined
by the general solution, if it exists, is a singular solution. However, a
singular solution is not necessarily an envelope. Confirm that the graph of
the singular solution found in part (b) is the envelope of the one-parameter
family of curves defined by the general solution found in part (a).
17. Given the general solution J(x, y, c) = 0 of a nonlinear equation, where f is
a (single-valued) function with continuous first-order partial derivatives, if a
singular solution exists, then it is obtained by the elimination of c from the two
of
equations f(x, y, c) = 0 and Be (x, y, c) = 0.
1
y2
(a) Find the general solution of the equation (y') 2 = �­
y
(b) By inspection of the equation, find two singular solutions.
(c) Express the general solution as f (x, y, c) = 0 and determine the singular
solutions by elimination of c from J(x, y, c) = 0 and fc (x, y, c) = 0.
(d) Confirm that each singular solution is an envelope of the general solution.
18. Consider the equation
(y') 2 + 1
= 1.
(y-xy')2
(a) Show that y = mx ± ✓m2 + 1 is the general solution, where m is an
arbitrary constant.
(b) Express the general solution in part (a) as f(x, y, m) = 0 and determine x2
as a function of m 2 from the equations f(x, y, m) = 0 and fm (x, y, m) = 0.
(c) Employ the result of part (b) to determine y2 as a function of m2.
(d) Employ the results of parts (b) and (c) to eliminate m, thereby obtaining
a relation g(x, y) = 0 between x and y, which defines a singular solution.
(e) The general solution y = mx± ✓m 2 + l of part (a) defines a one-parameter
family of straight lines of slope m and y-intercept ± ✓m2 + 1. Identify
the graph of the singular solution obtained in part (d) as a well-known
geometrical object, and confirm that it is the envelope of the family of
straight lines.
CHAPTER 2 EXERCISES
41
19. Peano 's existence theorem states that, if f is continuous on a rectangular region
R with sides parallel to the axes, and if (x0 , y0) is an interior point of R, then
the initial-value problem y' = f(x, y), y(x0) = y0, has a solution.
(a) Find the general solution of the equation y' = '#._ and express y as a function
of x.
X
(b) Show that the initial-value problem with the initial condition y(0) = 1 has
no solution, and give the reason.
20. Picard 's existence and uniqueness theorem states that, if f and fy are continuous
on a rectangular region R with sides parallel to the axes, and if (x0, y0) is an
interior point of R, then the initial-value problem y' = J(x, y), y(x0) = y0, has
a unique solution.
(a) Find the general solution of the equation y' = 3xy 1 1 3.
(b) Solve the initial-value problem with the initial condition y(0) = 0.
(c) Find a singular solution (which is not an envelope).
(d) Show that the solution of the initial-value problem is not unique, and give
the reason.
L
Chapter 3
Second-Order Equations
3.1
Basic Definitions
Definition 3.1. The general solution of a second-order equation is the solution which
contains two arbitrary constants which cannot be combined. It is a two-parameter
family of solutions, i.e., a set of solutions such that every solution corresponds to
particular values of the arbitrary constants. A solution in which the constants are
assigned particular values is called a particular solution.
Example 3.1. The second-order equation y" = x can be solved by two integrations:
yII = X
⇒
yI = 1 x 2 + C1
2
is the general solution since it contains two arbitrary constants c1 and c2 which cannot
be replaced by a single arbitrary constant. Particular solutions are
y=
1 3
1
1 3
x , y = x 3 + 2x + 1, and y = x - 4x - 3,
6
6
6
corresponding to c1 = c2 = 0, c1 = 2, c2 = 1, and c1 = -4, c2 = -3, respectively.
Definition 3.2. An initial-value problem for a second-order equation consists of the
equation, together with two initial conditions, which specify the values of y and y' at
a single point x and, thereby, determine a particular solution.
In Example 3.1, if the initial conditions y(0) = 2 and y'(0) = 3 are imposed, then
1 3
12 + C1,
0 = 2 ⇒ C2 = 2 ⇒ y = 61x3 + C1X + 2 ⇒ yI = 2x
y = 6x + C1X + C2 and y ()
1
and y' (0) = 3 ⇒ c 1 = 3 ⇒ y = -x 3 + 3x + 2
6
43