Existence and uniqueness of $\frac{dy}{dx} = \frac{y^2}{x-2}$.

Question

This is from "Introduction to Ordinary Differential Equations" by Shepley Ross. Exercise 1.3.6.b.

The problem wants us to apply existence and uniqueness theorem to the initial value problem in below.

$$\frac{dy}{dx} = \frac{y^2}{x-2},$$ $$y(1)=0.$$

Here's my problem:

When I apply existence and uniqueness theorem I find both $f(x,y)$ and $\frac{\partial f}{\partial y}$ continuous at $y(1) = 0$. So by the theorem there must be one unique solution to the differential equation at that point but when I solve the differential equation I get:

$$y(x) = \frac{1}{c_1-log(x-2)},$$

which is not defined on the real domain when $x=1$.

So I guess the differential equation above does not have any solution at that point. Why am I getting this contradictory result? Where is my mistake?

Thank you in advance.

Answer 1

$$ \int \frac{1}{x-2}\mathrm dx=\ln |x-2|+c $$ and not $$ \int \frac{1}{x-2}\mathrm dx=\ln (x-2)+c $$