The value of y as t → ∞ for an initial value of y(1) = 0, for the differential equation \(\left( {4{t^2} + 1} \right)\frac{{dy}}{{dt}} + 8yt - t = 0\), is

The value of y as t → ∞ for an initial value of y(1) = 0, for the

1.	The value of y as t → ∞ for an initial value of y(1) = 0, for the differential equation \(\left( {4{t^2} + 1} \right)\frac{{dy}}{{dt}} + 8yt - t = 0\), is
A.	1
B.	\(\frac{1}{2}\)
C.	\(\frac{1}{4}\)
D.	\(\frac{1}{8}\)
Answer» E.