Answer:
4. A unique solution exists in the region consisting of all points in the xy-plane except the origin.
Step-by-step explanation:
Given:
[tex] (x^2 + y^2)y' = y^2[/tex]
Solving the differential equation, we have:
[tex] \frac{dy}{dx} = \frac{y^2}{x^2 + y^2}[/tex]
Thus, except at (0,0), for all real values of x and y, the function[tex] \frac{y^2}{x^2 + y^2}[/tex] is defined.
The (0,0) values of x&y causes the denominator to be 0, so the function is not defined at this (0,0) condition.
Therefore,
[tex] \frac{d}{dy} \frac{y^2}{x^2 + y^2} = \frac{x^2 + y^2 (2y) - y^2 (2y)}{(x^2 + y^2)^2} [/tex]
[tex] = \frac{2x^2y + 2y^3 - 2y^3}{(x^2 + y^2)^2} [/tex]
[tex] = \frac{2x^2y}{(x^2 + y^2)^2} [/tex].
Apart from the point of origin (0,0), this is continuous.
This means a unique solution exists in the region consisting of all points in the xy-plane except the origin.
