## Calculus (3rd Edition)

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