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# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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y2 - 2y = x3 + 2x2 + 2x + c.
Further, if x = 0 and y = 1, then c =-1. Finally, by solving for y, we obtain
y = 1 ± vx3 + 2x2 + 2x.
(14)
68
Chapter 2. First Order Differential Equations
Equation (14) provides two functions that satisfy the given differential equation for x > 0 and also satisfy the initial condition y(0) = 1.
Consider the initial value problem
EXAMPLE
3 / = y1/3, y(0) = 0 (15)
for t > 0. Apply Theorem 2.4.2 to this initial value problem, and then solve the problem.
The function f (t, y) = y1/3 is continuous everywhere, but d f/dy = y—2/3/3 is not continuous when y = 0. Thus Theorem 2.4.2 does not apply to this problem and no conclusion can be drawn from it. However, by the remark following Theorem 2.4.2 the continuity of f does assure the existence of solutions, but not their uniqueness.
To understand the situation more clearly, we must actually solve the problem, which is easy to do since the differential equation is separable. Thus we have
y—l/3dy = dt,
so
3 y2 73 = t + c
2
and
y = [ 2 (t + c)f2 .
The initial condition is satisfied ifc = 0, so
y = \$() = (21)3/2 , t > 0 (16)
satisfies both of Eqs. (15). On the other hand, the function
y = \$() = — (21)3/2 , t > 0 (17)
is also a solution of the initial value problem. Moreover, the function
y = fi(t) = 0, t > 0 (18)
is yet another solution. Indeed, it is not hard to show that, for an arbitrary positive t0,
the functions
, 0, if 0 < t < t0,
y=x(t) = i ±[2(t — yf2, if(> t„ (19)
are continuous, differentiable (in particular at t = t0), and are solutions of the initial value problem (15). Hence this problem has an infinite family of solutions; see Figure 2.4.1, where a few of these solutions are shown.
As already noted, the nonuniqueness of the solutions of the problem (15) does not contradict the existence and uniqueness theorem, since the theorem is not applicable if the initial point lies on the t-axis. If (t0, y0) is any point not on the t-axis, however, then the theorem guarantees that there is a unique solution of the differential equation / = y1/3 passing through (t0, y0).
2.4 Differences Between Linear and Nonlinear Equations
69
EXAMPLE
4
FIGURE 2.4.1 Several solutions of the initial value problem / = y1/3, y(0) = 0.
Interval of Definition. According to Theorem 2.4.1, the solution of a linear equation (1),
/ + p(t )y = g(t ),
subject to the initial condition y(t0) = y0, exists throughout any interval about t = t0 in which the functions p and g are continuous. Thus, vertical asymptotes or other discontinuities in the solution can occur only at points of discontinuity of p or g. For instance, the solutions in Example 1 (with one exception) are asymptotic to the y-axis, corresponding to the discontinuity at t = 0 in the coefficient p(t) = 2/1, but none of the solutions has any other point where it fails to exist and to be differentiable. The one exceptional solution shows that solutions may sometimes remain continuous even at points of discontinuity of the coefficients.
On the other hand, for a nonlinear initial value problem satisfying the hypotheses of Theorem 2.4.2, the interval in which a solution exists may be difficult to determine. The solution y = \$(t) is certain to exist as long as the point [t, fi(t)] remains within a region in which the hypotheses of Theorem 2.4.2 are satisfied. This is what determines the value of h in that theorem. However, since fi(t) is usually not known, it may be impossible to locate the point [t, fi(t)] with respect to this region. In any case, the interval in which a solution exists may have no simple relationship to the function f in the differential equation / = f (t, y). This is illustrated by the following example.
Solve the initial value problem
y = y2, y(0) = 1, (20)
and determine the interval in which the solution exists.
Theorem 2.4.2 guarantees that this problem has a unique solution since f (t, y) = y2 and d f/dy = 2y are continuous everywhere. To find the solution we separate the variables and integrate with the result that
y 2 dy = dt
(21)
70
Chapter 2. First Order Differential Equations
and
Then, solving for y, we have
-y 1 = t + c.
y =-~^- . (22)
t + c
To satisfy the initial condition we must choose c = —1, so
y = 1—7 (23)
is the solution of the given initial value problem. Clearly, the solution becomes unbounded as t ^ 1; therefore, the solution exists only in the interval —? < t < 1.
There is no indication from the differential equation itself, however, that the point t = 1 is in any way remarkable. Moreover, if the initial condition is replaced by
y(0) = y0, (24)
then the constant c in Eq. (22) must be chosen to be c = — 1/y0, and it follows that
y = (25)
1 — y0t
is the solution of the initial value problem with the initial condition (24). Observe that the solution (25) becomes unbounded as t ^ 1/y0, so the interval of existence of the solution is —? < t < 1/y0if y0 > 0 and is 1/y0 < t < ? if y0 < 0. This example illustrates another feature of initial value problems for nonlinear equations; namely, the singularities of the solution may depend in an essential way on the initial conditions as well as on the differential equation.
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