0
d
dy
x
dx
dx


−
=




or
(
)
2
2
2
2
(1
)
(1
)
0
d y
dy
d
x
x
dx dx
dx
−
⋅
+
⋅
−
=
or
2
2
2
2
2
(1
)
0
2 1
d y
dy
x
x
dx
dx
x
−
⋅
−
⋅
=
−
Hence
2
2
2
(1
)
0
d y
dy
x
x dx
dx
−
−
=
Alternatively, Given that y = sin–1 x, we have
1
2
1
1
y
x
=
−
, i e , (
2)
12
1
1
x
y
−
=
So
2
2
1
2
1
(1
) 2
(0
2 )
0
x
y y
y
x
−
+
−
=
Hence
(1 – x2) y2 – xy1 = 0
EXERCISE 5