Answer
$s=(t+1)+\dfrac{\ln (t+1)}{(t+1)^2}+\dfrac{c}{(t+1)^2}$
Work Step by Step
The standard form of the given equation is:
$s'+\dfrac{2s}{(t+1)}=3+\dfrac{1}{(t+1)^3}$ ....(1)
The integrating factor is: $e^{\int (2/(t+1)) dx}=(t+1)^2$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [s(t+1)^2]' dt=\int [3(t+1)^2+\dfrac{1}{(t+1)}] dt$
or, $s(t+1)^2=\dfrac{3(t+1)^3}{3}+\ln (t+1)+c$
Thus, the general solution is as follows:
$s=(t+1)+\dfrac{\ln (t+1)}{(t+1)^2}+\dfrac{c}{(t+1)^2}$
