Section 35–7
◆
Geometric Applications of First-Order Differential Equations
3 xy
7. y 2x2
2y
8. y x x
9. xy 2y x
10. (x 1)y 2y (x 1)4
2 4x2y
11. y x x3
12. (x 1)y 2(x y 1)
13. xy x2y y 0
14. (1 x3) dy (1 3x2y) dx
15. y y ex
16. y e2x y
17. y 2y 4e2x
18. xy ex y xy 0
4 ln x 2x2y
19. y x3
With Trigonometric Expressions
20. y y sin x 3 sin x
21. y y sin x
22. y 2xy 2x cos x
23. y 2 cos x y
2
24. y sec x y cot x
Bernoulli’s Equation
y
25. y 3x2y2
x
27. y y xy2(x 2)
26. xy x2y2 y 0
28. y 2xy xex y3
2
Particular Solution
Using the given boundary condition, find the particular solution to each differential equation.
29. xy y 4x, x 1 when y 5
dy
30. 5x x xy, x 2 when y 1
dx
31. y y/x 5, x 1 when y 2
3y
32. y 2 , x 2 when y 6
x
p
33. y tan2 x y cot x, x when y 2
4
dy
34. 5y 3ex, x 1 when y 1
dx
35–7 Geometric Applications of First-Order Differential Equations
Now that we are able to solve some simple differential equations of first order, we turn to
applications. Here we not only must solve the equation but also must first set up the differential
equation. The geometric problems we do first will help to prepare us for the physical applications that follow.
Setting Up a Differential Equation
When reading the problem statement, look for the words “slope” or “rate of change.” Each of
these can be represented by the first derivative.
1051
1052
Chapter 35
◆◆◆
◆
Differential Equations
Example 30:
(a) The statement “the slope of a curve at every point is equal to twice the ordinate” is represented by the differential equation
dy
2y
dx
(b) The statement “the ratio of abscissa to ordinate at each point on a curve is proportional to
the rate of change at that point” can be written
x
y
dy
k
dx
(c) The statement “the slope of a curve at every point is inversely proportional to the square
of the ordinate at that point” can be described by the equation
dy
k
y2
dx
◆◆◆
Finding an Equation Whose Slope Is Specified
Once the equation is written, it is solved by the methods of the preceding sections.
◆◆◆
Example 31: The slope of a curve at each point is one-tenth the product of the ordinate
and the square of the abscissa, and the curve passes through the point (2, 3). Find the equation
of the curve.
Solution: The differential equation is
dy
x2y
dx
10
In solving a differential equation, we first see if the variables can be separated. In this case they
can be.
10 dy
y
x2dx
Integrating gives us
x3
10 ln | y | C
3
At (2, 3),
8
C 10 ln 3 8.32
3
Our curve thus has the equation ln | y | x3/30 0.832.
◆◆◆
Tangents and Normals to Curves
In setting up problems involving tangents and normals to curves, recall that
dy
slope of the tangent y
dx
1
slope of the normal y
◆◆◆
Example 32: A curve passes through the point (4, 2), as shown in Fig. 35–5. If from any
point P on the curve, the line OP and the tangent PT are drawn, the triangle OPT is isosceles.
Find the equation of the curve.
Section 35–7
◆
1053
Geometric Applications of First-Order Differential Equations
y
4
P(x, y)
(4, 2)
2
T
O
2
4
x
6
FIGURE 35–5
Solution: The slope of the tangent is dy/dx, and the slope of OP is y/x. Since the triangle is
isosceles, these slopes must be equal but of opposite signs.
dy
y
x
dx
We can solve this equation by separation of variables or as the integrable combination x dy y dx 0. Either way gives the hyperbola
xy C
(see Eq. 335). At the point (4, 2), C 4(2) 8. So our equation is xy 8.
◆◆◆
Orthogonal Trajectories
In Sec. 30–3 we graphed a relation that had an arbitrary constant, and we got a family of curves.
For example, the relation x2 y2 C2 represents a family of circles of radius C, whose centre
is at the origin. Another curve that cuts each curve of the family at right angles is called an
orthogonal trajectory to that family.
To find the orthogonal trajectory to a family of curves:
1. Differentiate the equation of the family to get the slope.
2. Eliminate the constant contained in the original equation.
3. Take the negative reciprocal of that slope to get the slope of the orthogonal trajectory.
4. Solve the resulting differential equation to get the equation of the orthogonal trajectory.
◆◆◆
Example 33: Find the equation of the orthogonal trajectories to the parabolas y2 px.
Solution:
1. The derivative is 2yy p, so
p
y 2y
(1)
2. The constant p, from the original equation, is y2/x. Substituting y2/x for p in Eq. (1) gives
y y/2x.
Common
Error
Be sure to eliminate the constant (p in this example) before
continuing.
1054
Chapter 35
◆
Differential Equations
3. The slope of the orthogonal trajectory is, by Eq. 295, the negative reciprocal of the
slope of the given family.
2x
y y
4. Separating variables yields
y
O
y dy 2x dx
x
Integrating gives the solution
2x2
2
2
which is a family of ellipses, 2x2 y2 C (Fig. 35–6).
y2
C1
FIGURE 35–6 Orthogonal
ellipses and parabolas. Each
intersection is at 90°.
◆◆◆
Exercise 7 ◆ Geometric Applications of
First-Order Differential Equations
Slope of Curves
y
P(x, y)
(1, 2)
C
O
x
FIGURE 35–7
y
B
1. Find the equation of the curve that passes through the point (2, 9) and whose slope is
y x 1/x y/x.
2. The slope of a certain curve at any point is equal to the reciprocal of the ordinate at that
point. Write the equation of the curve if it passes through the point (1, 3).
3. Find the equation of a curve whose slope at any point is equal to the abscissa of that point
divided by the ordinate and which passes through the point (3, 4).
4. Find the equation of a curve that passes through (1, 1) and whose slope at any point is
equal to the product of the ordinate and abscissa.
5. A curve passes through the point (2, 3) and has a slope equal to the sum of the abscissa and
ordinate at each point. Find its equation.
P(x, y)
Tangents and Normals
(4, 1)
x
A
O
FIGURE 35–8
y
T
P(x, y)
x
O
FIGURE 35–9
y
P(x, y)
PC y 1 (y)2
(0, 1)
O
FIGURE 35–10
6. The distance to the x intercept C of the normal (Fig. 35–7) is given by OC x yy. Write
the equation of the curve passing through (1, 2) for which OC is equal to three times the
abscissa of P.
7. A certain first-quadrant curve (Fig. 35–8) passes through the point (4, 1). If a tangent is
drawn through any point P, the portion AB of the tangent that lies between the coordinate
axes is bisected by P. Find the equation of the curve, given that
y
AP p q 1 (y)2 and BP x 1 (y)2
y
8. A tangent PT is drawn to a curve at a point P (Fig. 35–9). The distance OT from the origin
to a tangent through P is given by
xy y
OT 1 (y)2
Find the equation of the curve passing through the point (2, 4) so that OT is equal to the
abscissa of P.
9. Find the equation of the curve passing through (0, 1) for which the length PC of a normal
through P equals the square of the ordinate of P (Fig. 35–10), where
C
x
10. Find the equation of the curve passing through (4, 4) such that the distance OT (Fig. 35–9)
is equal to the ordinate of P.
Section 35–8
◆
1055
Exponential Growth and Decay
Orthogonal Trajectories
The equations for the various
distances associated with
tangents and normals were given
in Chapter 22 Review Problems.
Write the equation of the orthogonal trajectories to each family of curves.
11. the circles, x2 y2 r2
12. the parabolas, x2 ay
13. the hyperbolas, x2 y2 Cy
35–8 Exponential Growth and Decay
In Chapter 20 we derived Eqs. 199, 201, and 202 for exponential growth and decay and exponential growth to an upper limit by means of compound interest formulas. Here we derive
the equation for exponential growth to an upper limit by solving the differential equation that
describes such growth. The derivations of equations for exponential growth and decay are left
as an exercise.
◆◆◆ Example 34: A quantity starts from zero and grows with time such that its rate of growth
is proportional to the difference between the final amount a and the present amount y. Find an
equation for y as a function of time.
Solution: The amount present at time t is y, and the rate of growth of y we write as dy/dt. Since
the rate of growth is proportional to (a y), we write the differential equation
dy
n(a y)
dt
where n is a constant of proportionality. Separating variables, we obtain
dy
n dt
ay
Integrating gives us
ln(a y) nt C
Going to exponential form and simplifying yields
a y entC enteC C1ent
where C1 eC. Applying the initial condition that y 0 when t 0 gives C1 a, so our equation becomes a y aent, which can be rewritten as follows:
Exponential
Growth to an
Upper Limit
y a (1 ent)
202
◆◆◆
Motion in a Resisting Fluid
Here we continue our study of motion. In Chapter 29 we showed that the instantaneous velocity
is given by the derivative of the displacement and that the instantaneous acceleration is given
by the derivative of the velocity (or by the second derivative of the displacement). In Chapter 31
we solved simple differential equations to find displacement given the velocity or acceleration.
Here we do a type of problem that we were not able to solve in Chapter 31.
In this type of problem, an object falls through a fluid (usually air or water) which exerts
a resisting force that is proportional to the velocity of the object and in the opposite direction.
We set up these problems using Newton’s second law, F ma, and we’ll see that the motion
follows the law for exponential growth described by Eq. 202.