Unformatted text preview: 3 Fall 2010 1. (a) a2 (x) = sin(x), a1 (x) = 0, a0 (cx) .A. Campbell 2010 = 0. Since a2 (x0 ) = 0, the conditions
x0
S = cos(x),
of the Existence and Uniqueness Theorem are not satisﬁed and the theorem gives no
1. (a) a2 (x) = sin(x), a1 (x) = 0, a0 (x) = cos(x), x0 = 0. Since a2 (x0 ) = 0, the conditions
conclusion.
of the Existence and Uniqueness Theorem are not satisﬁed and the theorem gives no
x
3
0
(b) ax(x) = x2 1xy (x 5 = 0 a
x=
32
b) conclusion.000 + , a02+ ) y = ,ex ,1 (x) = (0) a0 (x) = y5, F (x)0= e y,00 (0)0= 10. All functions
1y
y x, = 0,
(0) = ,
are continuous for all x 2 IR, and a3 (x) 6= 0 for x 2 ( 1, 1). Thus, the Existence and
x
(b) a3 (x) = x2 Theorem )predicts 1 (x) =there 0will = 5a unique solution0 to 0. All functions
Uniqueness 1, a2 (x = 0, a that x, a (x) be , F (x) = e , x = the initial value
are continuous 1for all x 2 IR, and a3 (x) 6= 0 for x 2 ( 1, 1). Thus, the Existence and
problem on ( , 1).
Uniqueness Theorem predicts that there will be a unique solution to the initial value
2. (a) problem on ( x) 1).2e x .
y 00 y =...
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 Winter '14
 Heterogeneity, Homogeneity, yp

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