Unformatted text preview: 24. 25. 26. 27. 28. 29. 30. 31. 32. 35. 36. 12 f(x, y) = ym is continuous in a neighborhood of (0, 0), but 81‘fo = ('1/3)y'm is
not, so the theorem guarantees existence but not uniqueness in some neighborhood of
x = 0. f(x, y) = (x — y)”2 is notcontinuous at (2, 2) because it is not even defined if y > 3;.
Hence the'theorem guarantees neither existence nor uniqueness in any neighborhood of
the point x = 2. f(x,y) = (x y)”2 and affay = —(1/2)(x— 30'”? are continuous in aneighborhood
of (2, I), so the theorem guarantees both existence and uniqueness of a solution in some
neighborhood of x = 2. Both f("x, y) = (x m 1/)! and Elf/By = —(x— l)/y2 are continuous near (0, 1), so the
theorem guarantees both existence and uniqueness of a solution in some neighborhood of x=0. Neither f(x,y) = (x— l)/y nor dfiay : —(x—1)r‘y2 is continuous near (1.0), so the
existenceuniqueness theorem guarantees nothing. Both f(x,y) = ln(1 + yz) and Bf/dy = 2y/(l + ya) are continuous near (0, 0), so
the theorem guarantees the existence of a unique solution near it = 0. Both f(x,y) = 33— y2 and Bf/By = “2y are continuous near (0. 1). so the theorem
guarantees both existence and uniqueness of a solution in some neighborhood of x = 0. If f(x,y) = —(l w 322)”2 then Bflay = y(1— ff“2 is not continuous when y = 1,
so the theorem does not guarantee uniqueness. The two solutions are y1(x) = 0 (constant) and y2(x) = x3 The isoclines of y' yfx are the straight lines y = Cx through the origin, and
y’ = C at points of y = Cx, so itappears that these same straightlines are the solution curves of xy’z 'y. Then we observe that there is (i) a unique one of these lines through any point not on the yaris;
(ii) no such line through any point on the yaxis other than the origin; and
(iii) infinitely many such lines through the origin. fbny) = Any“? and (if/By = fom are continuous if y>0, so for all a and all b>0 there existsaunique solution near x = a such that y(a) = b. If .5 = 0 then the theorem guarantees neither existence nor uniqueness. For any a, both y1(x) = 0 and
y2(x) = (x2 — (12)2 are solutions with y(a) = 0. Thus we have existence but not uniqueness near points on the x—axis. Chapter 1 ...
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