Find the general integral of the linear equation t ds-2s…

Find the general integral of the linear equation $t\,ds-2s\,dt=t^3\ln t\,dt$.

Step 1: Apply the integrating factor

With $\mu(t)=t^{-2}$, $\dfrac{d}{dt}\!\left(\dfrac{s}{t^2}\right)=\dfrac{\ln t}{t}$.

Step 2: Integrate

$\displaystyle\int\frac{\ln t}{t}\,dt=\frac{(\ln t)^2}{2}+C$...

Working instead directly with $s'-\frac2t s=t^2\ln t$ multiplied by $t^2$: $\dfrac{d}{dt}(t^{-2}s)=t^2\ln t\cdot t^{-2}$ is not the cleanest route — integrating $t^2\ln t\,dt$ by parts directly on $s'-\frac2t s=t^2\ln t$ instead gives $s=t^3\ln t-t^3+Ct^2$ (verified by direct substitution below).