Answer
$$p = 4{e^{1 - \frac{1}{t}}} - 1$$
Work Step by Step
$$\eqalign{
& \frac{{dp}}{{dt}} = \frac{{p + 1}}{{{t^2}}} \cr
& {\text{separate}} \cr
& \frac{{dp}}{{p + 1}} = \frac{1}{{{t^2}}}dt \cr
& {\text{integrate both sides}} \cr
& \int {\frac{1}{{p + 1}}} dp = \int {\frac{1}{{{t^2}}}dt} \cr
& \ln \left| {p + 1} \right| = - \frac{1}{t} + C \cr
& {\text{the initial condition }}p\left( 1 \right) = 3{\text{ implies that}} \cr
& \ln \left| {3 + 1} \right| = - \frac{1}{1} + C \cr
& \ln 4 + 1 = c \cr
& \ln \left| {p + 1} \right| = - \frac{1}{t} + \ln 4 + 1 \cr
& {\text{solve for }}p \cr
& p + 1 = 4{e^{1 - \frac{1}{t}}} \cr
& p = 4{e^{1 - \frac{1}{t}}} - 1 \cr} $$
