Answer
$y=-\frac{1}{4(4t+3)}+\frac{1}{28}$
Work Step by Step
$\frac{dy}{dt}=(4t+3)^{-2};$ $y(1)=0$
$dy=(4t+3)^{-2}dt$
$\int dy=\int(4t+3)^{-2}dt$
$y+C=\frac{1}{4}\int4(4t+3)^{-2}dt$
$y=\frac{1}{4}(-(4t+3)^{-1})+C$
Substitute in initial condition to solve for $C$
$0=\frac{1}{4}(-\frac{1}{7})+C$
$C=\frac{1}{28}$
$y=-\frac{1}{4(4t+3)}+\frac{1}{28}$
