Exercises 2.2
Solving for
y
yields
y
=
±
1
√
1
−
x
2
.
(2.6)
Since, at the initial point,
x
= 0,
y
(0) = 1
>
1, we choose the positive sign in the above
expression for
y
. Thus, the solution is
y
=
1
√
1
−
x
2
.
Similarly we find solutions for the other two initial conditions:
y
(0) =
1
2
⇒
C
= 4
⇒
y
=
1
√
4
−
x
2
;
y
(0) = 2
⇒
C
=
1
4
⇒
y
=
1
(1
/
4)
−
x
2
.
(c)
For the solution to the first initial problem in (b),
y
(0) = 1, the domain is the set of all
values of
x
satisfying two conditions
1
−
x
2
≥
0
(for existence of the square root)
1
−
x
2
= 0
(for existence of the quotient)
⇒
1
−
x
2
>
0
.
Solving for
x
, we get
x
2
<
1
⇒

x

<
1
or
−
1
< x <
1
.
In the same manner, we find domains for solutions to the other two initial value problems:
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Boundary value problem, Initial value problems, ﬁrst initial problem

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