1 - X
IPL 14 YIf y =
prove that (1 – ~? )
+ y =0.
dx
1 + x
can anyone give me answer of this by chain rule method
Answers
Answered by
0
y=e
asin
−1
x
∴y
1
=
dx
d(e
asin
−1
x
)
=e
asin
−1
x
dx
d(asin
−1
x)
=ae
asin
−1
x
1−x
2
1
=a
(1−x
2
)
0.5
y
∴y
2
=
dx
y
1
=
dx
(ae
asin
−1
x
1−x
2
1
)
∴y
2
=a
⎣
⎢
⎢
⎢
⎡
e
asin
−1
x
dx
d
1−x
2
1
+
1−x
2
1
dx
de
asin
−1
x
⎦
⎥
⎥
⎥
⎤
∴y
2
=a[e
asin
−1
x
2(
1−x
2
)
3
−1
(−2x)+
1−x
2
y
1
]
∴y
2
=a[
(1−x
2
)
1.5
xy
+
1−x
2
y
1
]
Now, multiply by (1−x
2
) on both sides
∴(1−x
2
)y
2
=xy
1
+ay
1
1−x
2
(substituting y
1
)
∴(1−x
2
)y
2
=xy
1
+a
2
(1−x
2
)
0.5
y
1−x
2
∴(1−x
2
)y
2
=xy
1
+a
2
y
∴(1−x
2
)y
2
−xy
1
−a
2
y=0
Hence, proved.
Similar questions