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Mathematics
Mathematics
Problem
If
x
+
b
a
=
a
tan
−
1
(
a
log
y
)
\frac{x+b}{a} = a \tan^{-1}(a \log y)
a
x
+
b
=
a
tan
−
1
(
a
lo
g
y
)
,
a
>
0
a > 0
a
>
0
, then prove that
y
y
′
′
−
y
y
′
log
y
=
(
y
′
)
2
yy'' - yy' \log y = (y')^2
y
y
′′
−
y
y
′
lo
g
y
=
(
y
′
)
2
.
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