Consider the ODE \[t\,\frac{dy}{dt}=y+2t^3\] for positive \(t\). It can be rewritten in standard form \[\frac{dy}{dt}=\frac{1}{t}y+2t^2\] and it is in differential form equal to \[dy-\frac{1}{t}y\,dt=2t^2\,dt\] The natural logarithm \(\ln\) is an antiderivative of \(\frac{1}{t}\). This suggests \(e^{-\ln(t)}=\frac{1}{t}\) as an integrating factor.

But an intellectual guess of this integrating factor based on division of both sides by \(t\) in the differential form would be as good because then we get the differential form \[\frac{1}{t}\,dy-\frac{1}{t^2}y=2t\,dt\] which can be simplified to \[d\left(\frac{1}{t}y\right)= d(t^2)\] So: \[\frac{y}{t}=t^2+c\] for some constant \(c\). In other words, the general solution of the given ODE is \[y=c\,t+t^3\]