Separation of variables now shows that
±
e
y
dy/
√
c
+ 4
e
y
=
dt
and then
±
1
2
(
c
+ 4
e
y
)
1
/
2
=
t
+
d
. Algebraic manipulations then yield the implicitly defined
solution
e
y
= (
t
+
c
2
)
2
+
c
1
.
48.
Suppose that
y
0
=
v
(
y
) and then
y
00
=
v
0
(
y
)
v
(
y
).
The equation is
v
2
v
0
= 2,
90
Chapter 2. First Order Differential Equations
which gives us
v
3
/
3 = 2
y
+
c
.
Now plugging 0 in place of
t
gives that 2
3
/
3 =
2
·
1 +
c
and we get that
c
= 2
/
3. This turns into
v
3
= 6
y
+ 2, i.e.
y
0
= (6
y
+ 2)
1
/
3
.
This separable equation gives us (6
y
+ 2)

1
/
3
dy
=
dt
, and integration shows that
1
6
3
2
(6
y
+ 2)
2
/
3
=
t
+
d
. Again, plugging in
t
= 0 gives us
d
= 1 and the solution is
(6
y
+ 2)
2
/
3
= 4(
t
+ 1). Solving for
y
here yields
y
=
4
3
(
t
+ 1)
3
/
2

1
3
.
49.
Set
y
0
=
v
(
y
).
Then
y
00
=
v
0
(
y
)(
dy/dt
) =
v
0
(
y
)
v
(
y
).
We obtain the equation
v
0
v

3
y
2
= 0, where the differentiation is with respect to
y
. Separation of variables
gives
vdv
= 3
y
2
dy
, and after integration this turns into
v
2
/
2 =
y
3
+
c
. The initial
conditions imply that
c
= 0 here, so (
y
0
)
2
=
v
2
= 2
y
3
. This implies that
y
0
=
√
2
y
3
/
2
(the sign is determined by the initial conditions again), and this separable equation
now turns into
y

3
/
2
dy
=
√
2
dt
.
Integration yields

2
y

1
/
2
=
√
2
t
+
d
, and the
initial conditions at this point give that
d
=

√
2.
Algebraic manipulations find
that
y
= 2(1

t
)

2
.
50. Set
v
=
y
0
, then
v
0
=
y
00
. The equation with this substitution is
(1 +
t
2
)
v
0
+ 2
tv
= ((1 +
t
2
)
v
)
0
=

3
t

2
.
Integrating this we get that (1 +
t
2
)
v
= 3
t

1
+
c
, and
c
=

5 from the initial con
ditions. This means that
y
0
=
v
= 3
/
(
t
(1 +
t
2
))

5
/
(1 +
t
2
)
.
The partial fraction decomposition of the first expression shows that
y
0
= 3
/t

3
t/
(1 +
t
2
)

5
/
(1 +
t
2
) and then another integration here gives us that
y
= 3 ln
t

3
2
ln(1 +
t
2
)

5 arctan
t
+
d
.
The initial conditions identify
d
= 2 +
3
2
ln 2 + 5
π/
4,
and we obtained the solution.
51. Set
v
=
y
0
, then
v
0
=
y
00
. The equation with this substitution is
vv
0
=
t
. Inte
grating this separable differential equation we get that
v
2
/
2 =
t
2
/
2 +
c
, and
c
= 0
from the initial conditions. This implies that
y
0
=
v
=
t
, so
y
=
t
2
/
2 +
d
, and the
initial conditions again imply that the solution is
y
=
t
2
/
2 + 3
/
2.
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 Spring '13
 YOYL
 Boundary value problem, Stability theory