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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
.Thu
s
,theso
lu
t
ioni
s
y
=
1
√
1
−
x
2
.
Similarly we Fnd 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
p
(1
/
4)
−
x
2
.
(c)
±or the solution to the Frst 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
6
= 0
(for existence of the quotient)
⇒
1
−
x
2
>
0
.
Solving for
x
,weget
x
2
<
1
⇒
x

<
1o
r
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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