M341 H2
(S. Zhang) .
1. EP – Separable equations and applications.
§
1.4:
3, 12,
17, 23, 30, 33, 34, 49, 52
•
ans:
2. Find the general solution of
dy
dx
=
y
sin
x
•
ans:
§
1.4:3.
Z
dy
y
=
Z
sin
xdx
ln
y
=

cos
x
+
c
y
=
Ce

cos
x
3. Find the general solution of
y
dy
dx
=
x
(
y
2
+ 1)
•
ans:
§
1.4:12.
Z
ydy
y
2
+ 1
=
Z
2
xdx
1
2
ln(
y
2
+ 1) =
1
2
x
2
+
C
y
2
=
ce
x
2

1
4. Find the general solution of
dy
dx
= 1 +
x
+
y
+
xy
•
ans:
§
1.4:17.
Z
dy
1 +
y
=
Z
(1 +
x
)
dx
ln

1 +
y

=
x
+
1
2
x
2
+
C
5. Solve the IVP
dy
dx
+ 1 = 2
y
;
y
(1) = 1
•
ans:
§
1.4:23.
Z
dy
2
y

1
=
Z
dx
1
2
ln

2
y

1

=
x
+
c
y
=
1
2
(
1 +
Ce
2
x
)
By the initial condition,
C
=
e

2
,
y
=
1
2
(
1 +
e
2
x

2
)
6. Solve the di±erential equation
±
dy
dx
²
2
= 4
y
to ²nd the general solution and singular solution. Deter
mine the points (
a, b
) for which the initial problem
±
dy
dx
²
2
= 4
y
;
y
(
a
) =
b
has (a) no solution,
(b) in²nitely many solutions that are
de²ned for all
x
,
(c) on some neighborshood of the point
x
=
a
, only ²nitely many solutions.
•
ans:
§
1.4:30. Note that due to nonlinearity with
y
0
,
we get two slope (direction) ²elds! See the two graphs.
1
0
2
3
y(x)
0
3
1
2
3
1
1
2
x
2
In order to divide the equation by
√
y
, we need to check if
y
6
= 0 and if
y
= 0 is a solution. Yes,
y
≡
0 is a singular
solution!
±
Z
dy
2
√
y
=
Z
dx
±
√
y
=
x

C
;
y
= (
x

C
)
2
Note that
y
(
x
) is always nonnegative. Note also that the
whole curve of
y
= (
x

C
)
2
is a solution curve, not only
the right half!
(a) No solution when
b <
0.
(b) If
b
≥
0, we can combine
y
= 0 and
y
= (
x

C
)
2
(
x > C
) for any
C > a
to get in²nitely many solutions,
also we can combine
y
= (
x

C
)
2
and
y
= 0 (for
x > C
)
to get in²nitely many solutions.
(c) If
b >
0, locally, we have only two integral curves, one
increasing curve
y
= (
x

C
)
2
passing through point (
a, b
)
and one decreasing curve
y
= (
x

C
1
)
2
passing through
point (
a, b
).
1