6.24 Example continued further

Integrating both sides, we obtain

(Note that the absolute value sign now matters. It is only in the integrating factor that we can ignore it.) Dividing through by $\frac{x-1}{x+1}$, we obtain the general solution

We now want the particular solution satisfying $y(0)=1$. To obtain this, we evaluate $y(0)={\frac{1}{-2}\left(\log 1+c\right)=-\frac{c}{2}.}$ Setting this equal to 1, we obtain $c={-2}$ and therefore the solution is $y={\frac{x+1}{x-1}\left(\log|x+1|-2\right).}$