Exercises 2.2
⇒
x
(
1+
Ce
t
)
=
t
⇒
x
=
t
t
.
Note: When
C
is replaced by
−
K
, this answer can also be written as
x
=
Ke
t
/
(
t
−
1).
Further we observe that since we divide by
x
−
x
2
=
x
(1
−
x
), then
x
≡
0and
x
≡
1area
lso
solutions. Allowing
K
to be zero gives
x
≡
0, but no choice for
K
will give
x
≡
1, so we list
this as a separate solution.
15.
To separate variables, we move the term containin
dx
to the righthand side of the equation
and divide both sides of the result by
y
. This yields
y
−
1
dy
=
−
ye
cos
x
sin
xdx
⇒
y
−
2
dy
=
−
e
cos
x
sin
xdx.
Integrating the last equation, we obtain
Z
y
−
2
dy
=
Z
(
−
e
cos
x
sin
x
)
dx
⇒−
y
−
1
+
C
=
Z
e
u
du
(
u
=cos
x
)
1
y
+
C
=
e
u
=
e
cos
x
⇒
y
=
1
C
−
e
cos
x
,
where
C
is an arbitrary constant.
17.
First we ±nd a general solution to the equation. Separating variables and integrating, we get
dy
dx
=
x
3
(1
−
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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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