This preview shows pages 1–3. Sign up to view the full content.
GE 207K
Homogenous Equations – Problem Set
September 14, 2010
Solve the following ﬁrstorder diﬀerential equations.
1.
(
x
2
+
y
2
)
dx

2
xydy
.
Rewrite the equation as
2
xy
dy
dx

(
x
2
+
y
2
)
= 0
.
Divide through by
x
2
:
2
±
y
x
²
y
0

³
1 +
±
y
x
²
2
´
= 0
(1)
Perform the following substititions
u
=
y
x
y
=
ux
⇒
dy
dx
=
u
+
u
0
x
Therefore,
2
u
(
u
+
u
0
x
)

(
1 +
u
2
)
= 0
Simplifying, we have
u
+
u
0
x
=
1
2
³
1
u
+
u
´
u
0
x
=

u
2

1
2
u
Now, we can clearly see we have a separable equation. Solve by separating the variables on
either sides of the equation and integrating:

Z
2
u
u
2

1
du
=
Z
1
x
dx
Therefore,

ln

u
2

1

= ln
x
+
C
1
⇒
1
u
2

1
=
C
2
x
Don’t forget the backsubstitute
u
=
y
x
. The problem statement was in terms of
y
and
x
,
and so should be the solution!
1
(
y
x
)
2

1
=
C
2
x
Beatifying the solution, we get
∴
y
2

x
2
=
C
3
x
.
(2)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentGE 207K
Homogenous Equations – Problem Set
September 14, 2010
2.
Solve the diﬀerential equation
(3
x
+
y
)
dx
+
xdy
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '10
 None

Click to edit the document details