ANSWERS
to LBS 119 – Practice problems for Exam # 3.
1.
4
2
(
)
0
y
x y
dx
xdy

+
=
First, since it is in this form, check to see if it is EXACT  You should be able to see quickly that this one is not!
Therefore, it must be either Separable, Linear, or Bernoulli.
Divide through by dx
4
2
4
2
3
2
and get the dy/dx symbol isolated on one side.
Now, get
by
itself on one side
Notice that the y's will not factor out and separate from the x's, so it
.
,
(
)
0
+

+
=
= 
+
= 
dy
dx
dy
dx
dy
dy
y
so
dx
dx
x
y
x y
x
x
y
x y
x y
3
2
is not separable.
This equation has both a first power of y and a higher
power of y, therefore it is a Bernoulli equation, we want the single power of y on the side with
dy/dx
Bernoulli
:
1
+
=
dy
dx
x
y
x y
1n
12
1
2
2
2
1
3
3
standard form with n=2.
dv
dy
Let
v = y
y
= y
and
dx
dx
So, we should multiply the standard form by y
and then substitute.
Now this is a LINEAR equation in s
=
1
1



= 



= 
= 
y
dy
y
dx
x
dv
v
dx
x
y
x
x
[
]
1
1

ln 
ln 
1
x
1
2
2
2
tandard form.
=e
Multiply the standard form through by
.
The left side always becomes [
]'.
Substitute
back into the equation and integrate both
1
1
μ
μ
μ
μ
μ




=
=
=

= 
= 
dx
x
x
v
e
e
x
or
x
dv
x
v
x
dx
x
v
x
1
2
3
1
4
4
1
1
4
Solve for y
sides.
Finish solving for v.
Plug y back in.
y
=
y =
3
3
3
3



�
�
=

�
�
= 
+
= 
+

+
�
�

+
�
�
�
�
�
�
x
v
dx
x dx
x
x v
C
x
v
Cx
x
Cx
x
Cx
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2.
(1
)
0;
(0)
1
y dx
x dy
y
+
+
=
=
First, since it is in this form, check to see if it is EXACT  You should be able to see quickly that this one is not!
Therefore, it must be either Separable, Linear, or Bernoulli.
Divide through by dx
and get the dy/dx symbol isolated on one side.
Get the
on one side by itself.
Notice that this is separable, since we can factor the y's out from the x's!
Get t
(1
)
0
(1
)
1
+
+
=
+
= 

=
+
dy
dx
dy
y
x
dx
dy
x
y
dx
y
dy
dx
x
1
2
1
2
1
2
1
2
he y's on the left and x's on the right.
Integrate both sides and then solve for y.
Use initial condi
1
1
1
1
1
1
1
ln
ln 1

1
2
2
ln 1

1
ln 1

2

=
+


=
+
=
+
=
+

=
+
+
� �
� �
� �

=
+
+

=
+
+
�
�
1
dy
dx
x
y
y
dy
dx
x
u
x
du
dx
du
u
u
C
y
x
C
y
x
C
y
x
C
1
2
2
tions to find C.
1
1
ln 1
0 
2
1
Plug C back into the solution and finish solving for y.
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 Spring '08
 HanniNichols
 Derivative, #, 0 m, 2 2 m

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