Exercises 2.5
Since
µ
(
x
)=
x
−
2
we must check to see if the solution
x
≡
0 was either gained or lost. The
function
x
≡
0 is a solution to the original equation, but is not given by the above implicit
solution for any choice of
C
. Hence,
y
2
2
−
y
x
+3
x
=
C
and
x
≡
0
are solutions.
9.
We compute partial derivatives of
M
(
x, y
)=2
y
2
+2
y
+4
x
2
and
N
(
x, y
)=2
xy
+
x
.
∂M
∂y
=
∂
∂y
(
2
y
2
+2
y
+4
x
2
)
=4
y
+2
,
∂N
∂x
=
∂
∂x
(2
xy
+
x
)=2
y
+1
.
Although the equation is not exact (
∂M/∂y
6
=
∂N/∂x
), the quotient
∂M/∂y
−
∂N/∂x
N
=
(4
y
+2)
−
(2
y
+1)
2
xy
+
x
=
2
y
+1
x
(2
y
+1)
=
1
x
depends on
x
only, and so the equation has an integrating factor, which can be found by
applying formula (8) on page 70 of the text. Namely,
µ
(
x
)=exp
±Z
1
x
dx
²
=exp(
ln

x

)=

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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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