If you take the derivative of your equation, you get:
2′″−″−′+′=0
2
y
′
y
″
−
x
y
″
−
y
′
+
y
′
=
0
or
″(2′−)=0.
y
″
(
2
y
′
−
x
)
=
0.
Let =′
v
=
y
′
and we have ′(2−)=0,
v
′
(
2
v
−
x
)
=
0
,
so either ′=0
v
′
=
0
and =
v
=
c
or =/2
v
=
x
/
2
.
Then ′=
y
′
=
c
and so =+
y
=
c
x
+
d
or ′=/2
y
′
=
x
/
2
and =2/4.
y
=
x
2
/
4.
Plugging the first into the original equation gives =−2
d
=
−
c
2
. So there are two solutions =−2
y
=
c
x
−
c
2
for some constant
c
and =2/4
y
=
x
2
/
4
. I don't know if this is all the solutions
