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**Unformatted text preview: **With y(0) = 1, l- l + y2 = 2e_2’2 y = ix/Ze‘z‘z — 1. We must choose the branch that contains the initial
condition y(0) = 1. Thus, y = «29-212 — 1. This
solution is deﬁned, provided that t2 < —21nl
2 t2 < ln4 ltl < Vln4. Thus, the interval of existence is (—Vln 4, Vln 4). 20. dy_ x
dx_ y
ydy=—xdx
12 l2 _ =_- C
2y 2x +
y2=2C—x2 y = :|:\/2C —)c2 With y(0) = 2, we choose the positive branch
and 2 = J2—C leads to C = 2 and the solution
y = v4 —x2 with interval of existence (—2, 2).
This solution is plotted as a solid curve in the fol-
lowing ﬁgure. With y(0) = —2, we choose the neg-
ative branch and *2 = —J2E leads to C = 2 and
the solution y — V4 — x2 with interval of existence
(—2, 2). This solution is shown as a dashed curve in
the ﬁgure. 36. x’ = [at+by+c]l = a+by’ = a+bf(at+by+c) =
a + f(x). For the equation y’ = (y + t)2 we use x :
t+y. Thenx’ =1+y’:1+(y+t)2 =1+x2.Solv—
ing this separable equation in the usual way we get
the general solution x (t) = tan(t+C). In terms of the
unknown y, we gety(t) = x(t) —t = tan(t+C) —t. ...

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