.4
Differences Between Linear and Nonlinear Equations
75
fying regions of the ty-plane where solutions exhibit interesting features that merit
more detailed analytical or numerical investigation. Graphical methods for first order equations are discussed further in Section 2.5. An introduction to numerical
methods for first order equations is given in Section 2.7, and a systematic discussion
of numerical methods appears in Chapter 8. However, it is not necessary to study
the numerical algorithms themselves in order to use effectively one of the many
software packages that generate and plot numerical approximations to solutions of
initial value problems.
Summary. The linear equation y'
+ p(t)y = g(t)
has several nice properties that can
be summarized in the following statements:
1. Assuming
that the coefficients are continuous, there is a general solution, containing an
arbitrary constant, that includes all solutions of the differential equation.
A particular
solution that satisfies a given initial condition can be picked out by choosing the proper
value for the arbitrary constant.
2.
There is an expression for the solution, namely, Eq. (7) or Eq. (8). Moreover, although it
involves two integrations, the expression is an explicit one for the solution y = cjJ(t) rather
than an equation that defines cjJ implicitly.
3.
The possible points of discontinuity, or singularities, of the solution can be identified (without solving the problem) merely by finding the points of discontinuity of the coefficients.
Thus, if the coefficients are continuous for all t, then the solution also exists and is differentiable for all t.
None of these statements is true, in general, of nonlinear equations. Although a
non linear equation may well have a solution involving an arbitrary constant, there
may also be other solutions. There is no general formula for solutions of nonlinear
equations. If you are able to integrate a nonlinear equation, you are likely to obtain an
equation defining solutions implicitly rather than explicitly. Finally, the singularities
of solutions of nonlinear equations can usually be found only by solving the equation
and examining the solution. It is likely that the singularities will depend on the initial
condition as well as the differential equation.
PROBLEMS
In each of Problems 1 through 6 determine (without solving the problem)
the solution of the given initial value problem is certain to exist.
T
l-t
+ (In t)y = 2t,
y(l) = 2
2. t(t - 5)y' + y = 0,
y(2) = 1
3. y' + (tan t)y = sin t,
y(rr) =
5. (4 - t2)y' + 2ty = 3t2,
y(1) = -3
an interval in which
1. (t - 3)y'
°
In each of Problems
are satisfied.
T
7. y' = 2~~~y
H
9. y'
= (t2
+ i)3/2
+ 2ty = 3t2,
+ y = cot t,
4. (4 - t2)y'
6. (In t)y'
7 through 12 state where in the ty-plane the hypotheses
10. y' =
In Ityl
1 - t2 + y2
y(-3)=1
y(2)
=3
of Theorem
2.4.2
Chapter 2. First Order Differential Equations
76
11. dy _
dt -
1+ t
3y _ y2
2
12. dy
(cott)y
dt
l+y
In each of Problems 13 through 16 solve the given initial value problem
the interval in which the solution exists depends on the initial value Yo.
13. y'
15.
#t
#t
y'
= -4tly,
+ i = 0,
= Yo
= Yo
y(O)
y(O)
14. y'
16. y'
how
= 2ty2,
y(O) = Yo
= t2ly(1 + t3),
y(O) = Yo
In each of Problems 17 through 20 draw a direction field and plot (or sketch) several solutions
of the given differential equation. Describe how solutions appear to behave as t increases and
how their behavior depends on the initial value Yo when t = O.
17. y'
= 1y(3
19. y' = -y(2
21. Consider
- y)
- ty)
the initial value problem
(a) Is there a solution
= y(2
~
18. y'
#t
20. y' = I - 1 - y2
y'
- ty)
= ylf3,y(O) = 0 from
Example
3 in the text.
that passes through
the point (1, I)? If so, find it.
(b) Is there a solution that passes through
the point (2, I)? If so, find it.
(c) Consider all possible solutions of the given initial value problem.
of values that these solutions have at t = 2.
T
and determine
22. (a) Verify that both Yl (t)
problem
y'
=1-
t and Y2(t)
=
+ (t2 + 4y)I/2
-t
=
the set
of the initial value
y(2) =-1.
2
Where are these solutions
are solutions
-(214
Determine
valid?
(b) Explain why the existence of two solutions
the uniqueness part of Tbeorem 2.4.2.
of the given problem
does not contradict
(c) Show that y = et + c2, where c is an arbitrary constant, satisfies the differential equation in part (a) for t ::::-2c.
If c = -1, the initial condition is also satisfied, and the
solution y = Yl (t) is obtained.
Show that there is no choice of c that gives the second
solution y = yz(t).
23. (a) Show that cjJ(t) = eZI is a solution of y' - 2y
of this equation for any value of the constant c.
= 0 and
that y
= ccjJ(t)
is also a solution
(b) Show that cjJ(t) = lit is a solution of y' + y2 = 0 for t > 0 but that y = ccjJ(t) is not
a solution of this equation unless c = 0 or c = 1. Note that the equation of part (b) is
non linear, while that of part (a) is linear.
24. Show that if y = cjJ(t) is a solution
any value of the constant c.
25. Let y
= Yl (I) be
a solution
of y'
+ p(t)y = 0,
then y
= ccjJ(t)
is also a solution
for
of
y'
+ p(t)y = 0,
(i)
and let y = Y2(t) be a solution of
y'
Show that y = Yl (t)
+ Y2(t)
26. (a) Show that the solution
form
is also a solution
=
constant.
get)·
(ii)
of Eq. (ii).
(7) of the general
y
where c is an arbitrary
+ p(t)y
linear equation
(1) can be written
= CYI (t) + yz(t),
Identify the functions Yl and Y2.
in the
(i)
77
ences Between Linear and Nonlinear Equations
(b) Show that y, is a solution of the differential equation
y'
+ p(t)y = 0,
(ii)
corresponding to get) = O.
(c) Show that Yz is a solution of the fulJ linear equation (1). We see later (for example,
in Section 3.5) that solutions of higher order linear equations have a pattern similar to
Eq. (i).
Bernoulli Equations.
Sometimes it is possible to solve a nonlinear equation by making a
change of the dependent variable that converts it into a linear equation. The most important
such equation has the form
y'
+ p(t)y = q(t)y",
and is called a Bernoulli equation after Jakob Bemoulli. Problems 27 through 31 deal with
equations of this type.
rl
27. (a) Solve Bernoulli's equation when n = 0; when n = 1.
(b) Show that if n 1= 0,1, then the substitution v = yl-II reduces Bernoulli's equation to a
linear equation. This method of solution was found by Leibniz in 1696.
In each of Problems 28 through 31 the given equation is a Bernoulli equation. In each case
solve it by using the substitution mentioned in Problem 27(b).
28. tZy' + 2ty -l = 0,
t »0
29. y' = ry - kyZ, r > 0 and k > O.This equation is important in population dynamics and is
discussed in detail in Section 2.5.
30. y' = Ey - ai, E > 0 and a> o. This equation occurs in the study of the stability of fluid
flow.
31. dyjdt = (F cos t + T)y - i, where rand T are constants. This equation also occurs in
the study of the stability of fluid flow.
Coefficients.
Linear differential equations sometimes occur in which one or
both of the functions p and g have jump discontinuities. If to is such a point of discontinuity,
tben it is necessary to solve the equation separately for t < to and t > to. Afterward, the two
solutions are matched so that y is continuous at to; this is accomplished by a proper choice of
the arbitrary constants. The following two problems illustrate this situation. Note in each case
that it is impossible also to make y' continuous at to.
32. Solve the initial value problem
Discontinuous
y'
+ 2y = get),
y(O)
= 0,
where
get)
=
I,
{0,
0:::: t ::::1,
t » 1.
33. Solve the initial value problem
y'
+ p(t)y = 0,
y(O)
= 1,
where
pet)
4,
0:::: t ::::1,
2,
t :» 1.
={
Autonomous Equations and Population Dynamics
T
M
89
as asymptotically stable, unstable, or semis table (see Problem 7). Draw the phase line, and
sketch several graphs of solutions in the ty-plane.
8. dyf dt = -key - 2)2,
k > 0, -00 < Yo < 00
9. dyf dt = y2(y2 -1),
< Yo <
< Yo <
10. dyt dt = y(4 - y2),
-00
11. dyf dt = ay - b.,fY,
a> 0,
12. dyf dt
13. dy t dt
= l(1- y?,
= y2(4 -l),
00
00
b > 0,
-00 <
Yo <
00
-00 <
Yo <
00
Yo::: 0
14. Consider the equation dy t dt = fey) and suppose that y, is a critical point, that is,f(yl) = O.
Show that the constant equilibrium solution fjJ (t) = Yl is asymptotically stable if j' (YI) < 0
and unstable ifj'(YI) > O.
15. Suppose that a certain population obeys the logistic equation dyf dt = ry[l - (y/K)].
(a) If Yo = K /4, find the time r at which the initial population has doubled. Find the value
of r corresponding to r = 0.025 per year.
(b) IfYo/K = 0:, find the time Tat whichy(T)/K
= fJ, where 0 < o:,fJ < 1. Observe that
T -7 00 as 0: -7 0 or as fJ -7 1. Find the value of T for r = 0.025 per year, a = 0.1, and
fJ = 0.9.
16. Another equation that has been used to model population growth is the Gornpertz!"
equation
dyf dt
= ryln(K/y),
where rand K are positive constants.
(a) Sketch the graph off(y) versus y, find the critical points, and determine whether each
is asymptotically stable or unstable.
(b) For 0 :::y :::K, determine where the graph of y versus t is concave up and where it is
concave down.
(c) For each y in 0 < y ::: K, show that dy jdt as given by the Gompertz equation is never
less than dy [dt as given by the logistic equation.
17. (a) Solve the Gornpertz equation
dyf d:
subject to the initial condition
= ry
lu(K/y),
y(O) = Yo.
Hint: You may wish to let u = In(y/K).
(b) For the data given in Example 1 in the text (r = 0.71 per year, K = 80.5 X 106 kg,
= 0.25), use the Gompertz model to find the predicted value of y(2).
(c) For the same data as in part (b), use the Gornpertz model to find the time r at which
yo/ K
y(r)
= 0.75K.
18. A pond forms as water collects in a conical depression of radius a and depth h. Suppose
that water flows in at a constant rate k and is lost through evaporation at a rate proportional
to the surface area.
(a) Show that the volume V(t) of water in the pond at time t satisfies the differential
equation
where a is the coefficient of evaporation.
14Benjamin Gompertz (1779-1865) was an English actuary. He developed his model for population
growth, published in 1825, in the course of constructing mortality tables for his insurance company.
omous Equations and Population Dynamics
----r
fluid. Problems 25 through
equations of the form (i).
25. Consider
27 describe
93
three types of bifurcations
that can occur in simple
the equation
dyjdt
=a
-i.
(ii)
(a) Find all of the critical points for Eq. (ii). Observe that there are no critical points if
a < 0, one critical point if a = 0, and two critical points if a > O.
(b) Draw the phase line in each case and determine
totically stable, semistable, or unstable.
(c) In each case sketch several solutions
whether
each critical point is asymp-
of Eq. (ii) in the ty-plane.
(d) If we plot the location of the critical points as a function of a in the ay-plane, we obtain
Figure 2.5.10. This is called the bifurcation diagram for Eq. (ii). The bifurcation at a = 0
is called a saddle-node bifurcation. This name is more natural in the context of second
order systems, which are discussed in Chapter 9.
y
2
Asymptotically
1
3
2
stable
4 a
Unstable
-1
-2
FIGURE
2.5.10
Bifurcation
diagram for y' = a -yZ.
26. Consider the equation
dyjdt
= ay -
i = yea -i)·
(iii)
(a) Again consider the cases a < 0, a = 0, and a> O. In each case find the critical points,
draw the phase line, and determine whether each critical point is asymptotically stable,
semis table, or unstable.
(b) In each case sketch several solutions
of Eq. (iii) in the ty-plane.
(c) Draw the bifurcation diagram for Eq. (iii), that is, plot the location of the critical points
versus a. For Eq. (iii) the bifurcation point at a = 0 is called a pitchfork bifurcation; your
diagram may suggest why this name is appropriate.
27. Consider
the equation
dyjdt
=
ay -
i = yea -
y).
(iv)
(a) Again consider the cases a < 0, a = 0, and a> O. In each case find the critical points,
draw the phase line, and determine whether each critical point is asymptotically stable,
semistable, or unstable.
(b) In each case sketch several solutions
of Eq. (iv) in the ty-plane.
100
Chapter 2. First Order Differential Equatio
= _ ax + by
5. dy
~
6.
dy
~+~
= _ ax
~
=0
- by
~-~
7. (e+sin y - 2ysinx) dx + (er cosy + 2 cos x) dy
8. (e siny + 2y) dx - (3x - er sin y) dy = 0
9. (yerY cos 2x - 2efY sin 2x + 2x) dx + (xety cos 2r - 3) dy
f
10. (xlny+xy)dx+(ylnx+xy)dy=O;
x>O,
11. (y/x+6x)dx+(lnx-2)dy=0,
12.
x dx
=0
y>O
x>O
ydy
+ y2)3/2 + (x2 + y2)3/2 = 0
(x2
In each of Problems 13 and 14 solve the given initial value problem and determine at le
approximately where the solution is valid.
13. (2x - y) dx
14. (9x2
y -
+
+ (2)' 1) dx
-
x) d)' = 0,
=3
y(l)
= 0,
(4y - x) dy
y(l)
=0
In each of Problems 15 and 16 find the value of b for which the given equation is exact,
then solve it using that value of b.
15. (xl + bx2y) dx + (x + Y)X2 dy = 0
J 6. (ye2xy + 4x3) dx + bxe2xy dy = 0
17. Assume that Eq. (6) meets the requirements of Theorem 2.6.1 in a rectangle Rand
therefore exact. Show that a possible function 1jJ(x,y) is
1Y
1
1jJ(x,y)=
xM(s,yo)ds+
where
(xo,Yo)
N(x.,t)dt,
Yu
XO
is a point in R.
h 18. Show that any separable equation
M (x)
+ N(y)y' = 0
is also exact.
~
In each of Problems 19 through 22 show that the given equation is not exact but becomes ex;
when multiplied by the given integrating factor. Then solve the equation.
19. x2y3 +x(l + y2)y' = 0,
f.k(x,y) = l/xy3
20. y dx
+ (2'( -
y
21. (Sin--
-
y
yeY) dy
2e -".)
'smx
= 0,
dx
x
+ (COSY + 2e-
+2) siny dx +xcosy
dy
23. Show that if (Nx - My)/M
equation
22. (x
T
=y
f.k(x,y)
dy = 0,
COSX)
Y
=
=
0,
f.k(x,y) = xe'
Q, where Q is a function of
M+Ny'
f.k(x,y)
y
= yer
only, then the differen
=0
has an integrating factor of the form
f.k(y) =
T
24. Show that if (Nf - My)/(xM
the differential equation
- yN)
exp
J
= R, where
Q(y) dy.
R
depends on the quantity
xy
only, the
M+Ny'=O
has an integrating factor of tbe form
factor.
f.k(xy).
Find a general formula for this integratia;
101
erical Approximations: Euler's Method
1-1
In each of Problems 25 through 31 find an integrating factor and solve the given equation.
25. (3x2y + 2xy + i) dx + (x2 + y2) dy = 0 26. y' = e2x + y - 1
27. dx + (x/y - siny) dy = 0
28. ydx + (2xy - e-2y) dy = 0
29. e' dx + (e' coty + 2ycscy) dy = 0
30.
[4(X3/y2)
+
31.
(3X+~)+(~
(3/y)] dx
+
[3(x/i)
+
4y] dy
=0
+3n~=0
Hint: See Problem 24.
32. Solve the differential equation
(3xy
+ l) +
(x2
+ xy)y' = 0
»r'.
using the integrating factor j.t(x,y) = [xy(2x +
Verify that the solution is the same
as that obtained in Example 4 with a different integrating factor.
erical Approximations: Euler's Method
Recall two important facts about the first order initial value problem
dy
dt
= f(t,y),
y(to)
= Yo·
(1)
First, if f and BflBy are continuous, then the initial value problem (1) has a unique
solution y = tP(t) in some interval surrounding the initial point t = to. Second, it is
usually not possible to find the solution tP by symbolic manipulations of the differential equation. Up to now we have considered the main exceptions to the latter
statement: differential equations that are linear, separable, or exact, or that can be
transformed into one of these types. Nevertheless, it remains true that solutions of
the vast majority of first order initial value problems cannot be' found by analytical
means, such as those considered in the first part of this chapter.
Therefore it is important to be able to approach the problem in other ways. As we
have already seen, one of these ways is to draw a direction field for the differential
equation (which does not involve solving the equation) and then to visualize the
behavior of solutions from the direction field. This has the advantage of being a
relatively simple process, even for complicated differential equations. However, it
does not lend itself to quantitative computations or comparisons, and this is often a
critical shortcoming.
For example, Figure 2.7.1 shows a direction field for the differential equation
dy
dt
=3
1
- 2t - 2:Y-
(2)
From the direction field you can visualize the behavior of solutions on the rectangle
shown in the figure. A solution starting at a point on the y-axis initially increases with
t, but it soon reaches a maximum value and then begins to decrease as t increases
-further.