Consequently,
y
0
(
x
) = 6
parenleftBig
e
x
2
−
1
parenrightBig
1
/
2
,
and so
y
0
(1) = 6
parenleftBig
e
−
1
parenrightBig
1
/
2
.
021
10.0points
If
y
0
is the particular solution of the differ-
ential equation
dy
dx
+ 2
xy
+ 4
x
= 0
,
such that
y
(0) = 3, find the value of
y
0
(
√
ln 2).
1.
y
0
(
√
ln 2 ) =
1
2
correct
2.
y
0
(
√
ln 2 ) = 1
3.
y
0
(
√
ln 2 ) =
5
2

atchison (bma862) – Homework 5 – staron – (53940)
12
4.
y
0
(
√
ln 2 ) = 3
5.
y
0
(
√
ln 2 ) = 7
Explanation:
The differential equation
dy
dx
+ 2
xy
+ 4
x
= 0
becomes
integraldisplay
dy
y
+ 2
=
−
integraldisplay
2
x dx
after separating
variables and
integrating.
Thus the general solution of this differential
equation is
ln(
y
+ 2) =
−
x
2
+
C,
which in explicit form is given by
y
=
−
2 +
e
C
−
x
2
=
−
2 +
Ae
−
x
2
with
A
an arbitrary constant. For the partic-
ular solution
y
0
the value of
A
is determined
by the condition
y
(0) = 3 since
y
(0) = 3
=
⇒
3 =
−
2 +
A,
so
y
0
(
x
) =
−
2 + 5
e
−
x
2
.
At
x
=
√
ln 2, therefore,
y
0
(
√
ln 2 ) =
1
2
,
since
e
−
ln2
=
1
2
.
022(part1of3)10.0points
For the differential equation
2
y
dy
dx
= 49
e
x
+1
,
(i) First find the general solution.
1.
y
=
1
7
(
e
x
+1
+
A
)
2.
y
= 7(
e
x
+1
+
A
)
−
1
/
2
3.
y
= 7(
e
x
+1
+
A
)
4.
y
= 7(
e
x
+1
+
A
)
1
/
2
correct
5.
y
=
1
7
(
e
x
+1
+
A
)
1
/
2
Explanation:
The differential equation
2
y
dy
dx
= 49
e
x
+1
becomes
2
integraldisplay
y dy
= 49
e
integraldisplay
e
x
dx
after
separating variables
and
integrating.
Consequently, the general solution of this dif-
ferential equation is given by
y
2
= 49
e
x
+1
+
C
= 49(
e
x
+1
+
A
)
,
which in explicit form is given by
y
= 7
parenleftBig
e
1+
x
+
A
parenrightBig
1
/
2
with
A
an arbitrary constant.
023(part2of3)10.0points
(ii) Then find the particular solution
y
1
such
that
y
(
−
1) = 0.
1.
y
1
(
x
) =
1
7
(
e
x
+1
−
1)
1
/
2
2.
y
1
(
x
) = 7(
e
x
+1
−
1)
3.
y
1
(
x
) =
1
7
(
e
x
+1
−
1)
4.
y
1
(
x
) = 7(
e
x
+1
−
1)
1
/
2
correct
5.
y
1
(
x
) = 7(
e
x
+1
−
1)
−
1
/
2

atchison (bma862) – Homework 5 – staron – (53940)
13
Explanation:
For the particular solution
y
1
of (
†
) the
arbitrary constant
A
in
y
= 7(
e
x
+1
+
A
)
1
/
2
is determined by the condition
y
(
−
1) = 0,
since
y
(
−
1) = 0
−→
A
=
−
1
.
Hence
y
1
= 7(
e
x
+1
−
1)
1
/
2
.
024(part3of3)10.0points
(iii) For the particular solution
y
1
in (ii), deter-
mine the value of
y
1
(0).
1.
y
1
(0) =
1
7
(
e
−
1)
1
/
2
2.
y
1
(0) = 7(
e
−
1)
1
/
2
correct
3.
y
1
(0) = 7(
e
−
1)
4.
y
1
(0) =
1
7
(
e
−
1)
5.
y
1
(0) = 7(
e
−
1)
−
1
/
2
Explanation:
When
x
= 0 the value of the particular
solution
y
1
= 7(
e
x
+1
−
1)
1
/
2
is given by
y
1
(0) = 7
parenleftBig
e
−
1
parenrightBig
1
/
2
.
025(part1of3)10.0points
For the differential equation
2
xy
dy
dx
= 4 + 4 ln
x,
(i) first find its general solution.
1.
y
=
parenleftBig
4
x
+ 4(ln
x
)
2
+
C
parenrightBig
1
/
2
2.
y
=
parenleftBig
4 ln
x
−
2(ln
x
)
2
+
C
parenrightBig
1
/
2
3.
y
=
parenleftBig
4 ln
x
+ 2
x
2
+
C
parenrightBig
1
/
2
4.
y
=
parenleftBig
4 ln
x
+ 2(ln
x
)
2
+
C
parenrightBig
1
/
2
correct
5.
y
= 4 ln
x
+ 4(ln
x
)
2
+
C
Explanation:
The differential equation
2
xy
dy
dx
= 4 + 4 ln
x
becomes
integraldisplay
2
y dy
= 4
integraldisplay
1
x
dx
+ 4
integraldisplay
ln
x
x
dx
after
separating variables
and
integrating.