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Unformatted text preview: Chapter 1
diﬀerentiable function on an interval cannot have two relative minima on an interval without having a point of relative maximum, we conclude that x∗ is the only point where φ (x) = 0. Continuity of φ (x) implies that it has the same sign for all x > x∗ and, therefore, it is positive there since it is positive for x > x∗ and close to x∗ (φ (x∗ ) = 0 and φ (x∗ ) > 0). By Monotonicity Test, φ(x) increases for x > x∗ . (e) For y = x − 1, dy/dx = 1 and x − y = x − (x − 1) = 1. Thus the given diﬀerential equation is satisﬁed, and y = x − 1 is indeed a solution. To show that the curve y = φ(x) always stays above the line y = x − 1, we note that the initial value problem dy = x − y, dx y (x0 ) = y0 (1.5) has a unique solution for any x0 and y0 . Indeed, functions f (x, y ) = x−y and ∂f /∂y ≡ −1 are continuous on the whole xy -plane, and Theorem 1, Section 1.2, applies. This implies that the curve y = φ(x) always stays above the line y = x − 1: φ(0) = 1 > −1 = (x − 1)
x=0 , and the existence of a point x with φ (x) ≤ (x − 1) would imply, by intermediate value theorem, the existence of a point x0 , 0 < x0 ≤ x, satisfying y0 := φ(x0 ) = x0 − 1 and, therefore, there would be two solutions to the initial value problem (1.5). Since, from part (a), φ (x) = 1−φ (x) = 1−x+φ(x) = φ(x)−(x−1) > 0, we also conclude that φ (x) is an increasing function and φ (x) < 1. Thus there exists limx→∞ φ (x) ≤ 1. The strict inequality would imply that the values of the function y = φ(x), for x large enough, become smaller than those of y = x − 1. Therefore,
x→∞ lim φ (x) = 1 ⇔ x→∞ lim [x − φ(x)] = 1, and so the line y = x − 1 is a slant asymptote for φ(x). (f), (g) The direction ﬁeld for given diﬀerential equation and the curve y = φ(x) are shown in Figure B.6 in the answers of the text. 14 ...
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