Problem 04 | Integrating Factors Found by Inspection

$2t \, ds + 2s \, dt + s^3t \, dt = 0$

$(2t \, ds + 2s \, dt) + s^3t \, dt = 0$

$2(t \, ds + s \, dt) + s^3t \, dt = 0$

$2 \, d(st) + s^3t \, dt = 0$

$\dfrac{2 \, d(st)}{s^3t^3} + \dfrac{s^3t \, dt}{s^3t^3} = 0$

$\dfrac{2 \, d(st)}{(st)^3} + \dfrac{dt}{t^2} = 0$

$2 (st)^{-3} \, d(st) + t^{-2} \, dt = 0$

$\displaystyle 2\int (st)^{-3} \, d(st) + \int t^{-2} \, dt = 0$

$-(st)^{-2} - t^{-1} = -c$

$(st)^{-2} + t^{-1} = c$

$\dfrac{1}{(st)^2} + \dfrac{1}{t} = c$

$\dfrac{1}{s^2t^2} + \dfrac{1}{t} = c$
 

Multiply by s²t²
$1 + s^2t = cs^2t^2$           answer