Find the rectangular regions where the initial value problem
$$ \frac{dy}{dx}=\frac{\sqrt{y^2-9}}{\sqrt{x^2-4}}, \ \ y(x_0)=y_0 $$
has a unique solution at every point.
Answer:
The given equation is $$ \frac{dy}{dx}=\frac{\sqrt{y^2-9}}{\sqrt{x^2-4}}, \ \ y(x_0)=y_0 $$
Here $ \ f(x,y)=\frac{\sqrt{y^2-9}}{\sqrt{x^2-4}} $
Now $ \ f(x,y) \ $ exists when $ \ y^2-9 \geq 0 \ $ and $ \ x^2-4>0 \ $
i.e., $ f(x,y) \ $ exists when $ \ \ y \in (-\infty,-3] \cup [3,\infty) \ $ and $ \ \ x \in (-\infty,-2) \cup (2,\infty) \ $
Now,
$ \frac{\partial f}{\partial y}=\frac{y}{\sqrt{(y^2-9)(x^2-4)}} \ $
Now $ \frac{\partial f}{\partial y} \ $ exists and continuous when $ \ \sqrt{(y^2-9)(x^2-4)} \neq 0 \ $
i.e., either $ \ (x,y) \in (-2,2) \times (-3,3) \ $ or $ \ (x,y) \in (-\infty,-2) \cup (2, \infty) \times (-\infty,-3) \cup(3, \infty) \ $
So the largest rectangle is $ \ (-2,2) \times (-3,3) \ $ , where the unique solution exists .
Am I right?