Question

G find the domain for the particular solution to the differential equation dy dx equals the quotient of negative 1 times x and y, with initial condition y(2) = 2.

Answer 1

$\frac{dy}{dx} = \frac{-x}{y}$
by separating the variables
∴ y dy = -x dx
integrating both sides with respect to x
∴ ∫y dy = ∫-x dx
∴ $\frac{ y^{2} }{2} = - \frac{ x^{2} }{2} +c$
finding c using initial condition y(2) = 2.
∴ $\frac{ 2^{2} }{2} = - \frac{ 2^{2} }{2} +c$
∴ c = 4
 ∴ $\frac{ y^{2} }{2} = - \frac{ x^{2} }{2} +4$
multiplying all the equation by 2
∴ y² = -x² + 8
∴ x² + y² = 8
The resultant equation represents the equation of the circle
with center (0,0)  and radius = √8 = 2√2
so, the domain of this function is [-2√2 , 2√2]
See the attached figure