MAC 2313, Fall 2010 — Midterm 1 Review Solutions
4
We then integrate both sides with respect to
x
,
integraldisplay
1
y
dy
dx
dx
=
integraldisplay
x
sin
x dx,
integraldisplay
1
y
dy
=
sin
x
−
x
cos
x
+
C
(obtained by integration by parts),
ln

y

=
sin
x
−
x
cos
x
+
C,
y
(
x
)
=
±
e
C
e
sin
x

x
cos
x
,
y
(
x
)
=
Ce
sin
x

x
cos
x
.
Note that in going from the secondtolast line to the last line above, we have done our
standard abuse of constants, replacing
±
e
C
by
C
. It is worth noting that
C
cannot be 0.
c.
dy
dx
= sin(
x
+
y
2
).
Solution:
The equation is not separable
2
.
The equation is also not linear, because we
cannot express it in the form
dy
dx
+
P
(
x
)
y
=
Q
(
x
). Finally, the equation is not exact; to see
this, we express it in the form
M
(
x, y
) +
N
(
x, y
)
dy
dx
= 0, where we see that
M
(
x, y
) =
−
sin(
x
+
y
2
) and
N
(
x, y
) = 1
,
and the partials do not match:
∂M
∂y
= 2
y
cos(
x
+
y
2
)
negationslash
= 0 =
∂N
∂x
.
Since none of our methods apply to this equation, we cannot solve it (yet).
d.
dy
dx
=
x

y
2
x
.
2
Again, if we wanted to, we could verify this by considering the three points (
x,y
)=(0
,
0)
,
(
π
/
2
,
0)
,
(0
,
radicalbig
π
/
2
), but
again, that won’t be required on the midterm.