Answer
$$y = 4{e^{{t^4} + t}}$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = y\left( {4{t^3} + 1} \right),{\text{ }}y\left( 0 \right) = 4 \cr
& {\text{Write }}y'\left( t \right){\text{ as }}\frac{{dy}}{{dt}} \cr
& \frac{{dy}}{{dt}} = y\left( {4{t^3} + 1} \right) \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{y} = \left( {4{t^3} + 1} \right)dt \cr
& {\text{Integrate both sides}} \cr
& \ln \left| y \right| = {t^4} + t + C \cr
& {\text{Solve for }}y \cr
& {e^{\ln \left| y \right|}} = {e^{{t^4} + t + C}} \cr
& y = {e^{{t^4} + t}}{e^C} \cr
& y = C{e^{{t^4} + t}} \cr
& {\text{Use the initial condition }}y\left( 0 \right) = 4 \cr
& 4 = C{e^{{{\left( 0 \right)}^4} + \left( 0 \right)}} \cr
& 4 = C \cr
& {\text{Substitute }}C{\text{ into }}y = C{e^{{t^4} + t}} \cr
& y = 4{e^{{t^4} + t}} \cr} $$
