Question

Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point (x0, y0) in the region. (x2 + y2)y' = y2
1. A unique solution exists in the region y ≥ x.
2. A unique solution exists in the entire xy-plane.
3. A unique solution exists in the region y ≤ x.
4. A unique solution exists in the region consisting of all points in the xy-plane except the origin.
5. A unique solution exists in the region x2 + y2 < 1.

Answer 1

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:

$(x^2 + y^2)y' = y^2$

Solving the differential equation, we have:

$\frac{dy}{dx} = \frac{y^2}{x^2 + y^2}$

Thus, except at (0,0), for all real values of x and y, the function$\frac{y^2}{x^2 + y^2}$ 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,

$\frac{d}{dy} \frac{y^2}{x^2 + y^2} = \frac{x^2 + y^2 (2y) - y^2 (2y)}{(x^2 + y^2)^2}$

$= \frac{2x^2y + 2y^3 - 2y^3}{(x^2 + y^2)^2}$

$= \frac{2x^2y}{(x^2 + y^2)^2}$.

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.