I was studying Existence and uniqueness theorem from here. Now I got the following question.
Test the uniqueness of the initial value problem $(x-2y+1)dy - (3x-6y+2)dx = 0; y(0) = 0$ over the rectangle $|x| \le 1/4 ; |y| \le 1/4$ . Determine the interval for $x$ over which the solution is guaranteed.
Now from the theorem 1, I have to show that $|f(x,y)| \le K \text{ } \forall (x,y) \in R$,i.e.,
$$ \left| \frac{3x-6y+2}{x-2y+1} \right| \le K \text{ } \forall (x,y) \in R$$ where, $$R = \{(x, y) : |x − x_0 | ≤ 1/4, |y − y_0| ≤ 1/4\}, (a, b > 0). $$ So, my question is how to calculate the upper bound of the function ?
In order to find AN upper bound (not the minimal one) note that for all $(x,y)\in R=[-1/4,1/4]^2$, $$\begin{align}\left| \frac{3x-6y+2}{x-2y+1} \right| &\leq \frac{\max_R|3x-6y+2|}{\min_R|x-2y+1|}\\&\leq \frac{2+3\max_R|x|+6\max_R|y|}{1-\max_R|x|-2\max_R|y|}\\&\leq \frac{2+3/4+6/4}{1-1/4-2/4}=17. \end{align}$$ P.S. Note that $|f|$ is continuous in the compact set $R$ ($f$ is a rational function such that its denominator is never zero in $R$). Therefore $|f|$ is bounded in $R$.