$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$; find my mistake solve :$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$
My try:
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}\\\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\\left(t\frac{1-x}{1+x}\right)'=\left(x-1\right)\frac{1-x}{1+x}\\\left(t\frac{1-x}{1+x}\right)=\frac{1}{2}(7+6x-x^{2}-8log(1+x))+c\\t=\frac{(1+x)(c+1/2(7+6x-x^{2}-8log(1+x))))}{1-x}\\y=-\frac{1-x}{(1+x)(c+1/2(7+6x-x^{2}-8log(1+x))))}\\\\\\\\\\$
 A: The first mistake comes in your substitution of
$$t=y^{1-2}=\frac{1}{y}\implies t'= -\frac{1}{y^2}y'.$$
The next step after
$$\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)$$
should be
$$t'-\frac{2}{x^2-1}t=1-x.$$
Then the integrating factor is
$$\mu(x)=\exp\left(\int-\frac{2}{x^2-1}\,dx\right)=\frac{1+x}{1-x},$$
so the equation becomes
$$\left(t~\frac{1+x}{1-x}\right)'=1+x$$
$$t~\frac{1+x}{1-x}=x+\frac{x^2}{2}+c=\frac{x(x+2)}{2}+c$$
$$t=\frac{(1-x)(x(x+2)+c_1)}{2(1+x)}.$$
Therefore since $t=\dfrac{1}{y}$ we find
$$y=\frac{2(1+x)}{(1-x)(x(x+2)+c_1)}=-\frac{2(x+1)}{(x-1)(x(x+2)+c_2)},$$
which matches the answer provided by Wolfram.

Note: As suggested in the comment by Aryadeva, your substitution is correct. However, it is typically the case that you substitute $t=y^{1-m}$ for a first order Bernoulli equation. The exponent of the right-hand side is $2$, so I made the substitution $t=y^{1-2}=\frac{1}{y}$.
A: You made a mistake for the integrating factor. Since $\int-\frac{2}{x^2-1}dx=\ln(\frac{x+1}{1-x})+C$, your $4$th line should be:
$$(t\frac{1+x}{1-x})'=(x-1)\frac{1+x}{1-x}=-(1+x)$$
$$(t\frac{1+x}{1-x})=-x-\frac{x^2}{2}+c$$
$$t=\frac{(1-x)(c-x-\frac{x^2}{2})}{1+x}=\frac{(x-1)(c_{1}+x(x+2))}{2(1+x)}$$
where $c_{1}=-2c.$ Thus:
$$y=-\frac{2(1+x)}{(x-1)(c_{1}+x(x+2))}.$$
A: Your 5th and 4th lines are not equivalent. You're missing a -ve sign in the second term on LHS.
