<s>[INST] Differential equation: $y + t\frac{dy}{dt} = \frac{y}{y^2 - 1} $ $$y + t\frac{dy}{dt} = \frac{y}{y^2 - 1} \Rightarrow t\frac{dy}{dt} = \frac{-y^3 + 2y}{y^2 - 1} \Rightarrow \frac{1}{t} dt = \frac{y^2 - 1}{-y^3 + 2y}dy \Rightarrow$$
$$\Rightarrow \ln |t| = - \int \frac{y^2 - 1}{y^3 - 2y}dy$$
How to solve the integral in the right part? Had it been $\frac{3y^2 - 2}{y^3 - 2y}$, then the answer would be  $\ln |y^3 - 2y|$.
 [/INST] HINT : We have
$$\frac{y^2-1}{y^3-2y}=\frac 12\left(\frac{1}{y}+\frac{y}{y^2-2}\right)=\frac{1}{2y}+\frac{(y^2-2)'}{4(y^2-2)}.$$
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