6.25 Well-defined region

The solution we found in the previous slide is not defined at $x=1$, since the denominator of $\frac{x+1}{x-1}$ blows up as $x$ approaches $1$. So we should only think of the solution as being defined on one of the two regions $x>1$ or $x<1$. Since we specified the initial condition $y(0)=1$, in this case we want to restrict to the region $x<1$.

There is a similar problem at the point $x=-1$, since $\log|x+1|$ is not defined when $x=-1$. We might try to solve this by observing that $\frac{x+1}{x-1}\log|x+1|$ converges to $0$ as $x$ tends to $-1$. So we could try to define $y$ at this special point as: $y(-1)=0$.

Question: why does this not give us a solution to the differential equation which is valid on the whole range $x<1$?