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

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?

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