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Chapter 2
(d)
By dividing both sides by
y
2
we lost the solution
y
≡
0.
31.
Following the proof of Theorem 2, we come to the expression (10) on page 63 of the text for
g
0
(
y
), that is
g
0
(
y
)=
N
(
x, y
)
−
∂
∂y
x
Z
x
0
M
(
s, y
)
ds
(2.15)
(where we have replaced the integration variable
t
by
s
). In other words,
g
(
y
) is an antideriva
tive of the righthand side in (2.15).
Since an antiderivative is de±ned up to an additive
constant and, in Theorem 2, such a constant can be chosen arbitrarily (that is,
g
(
y
)canbe
any
antiderivative), we choose
g
(
y
) that vanishes at
y
0
. According to fundamental theorem
of calculus, this function can be written in the form
g
(
y
y
Z
y
0
g
0
(
t
)
dt
=
y
Z
y
0
N
(
x, t
)
−
∂
∂t
x
Z
x
0
M
(
s, t
)
ds
dt
=
y
Z
y
0
N
(
x, t
)
dt
−
y
Z
y
0
∂
x
Z
x
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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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