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.
[tex] \frac{dy}{dx} = \frac{-x}{y} [/tex] by separating the variables ∴ y dy = -x dx integrating both sides with respect to x ∴ ∫y dy = ∫-x dx ∴ [tex] \frac{ y^{2} }{2} = - \frac{ x^{2} }{2} +c[/tex] finding c using initial condition y(2) = 2.
∴ [tex] \frac{ 2^{2} }{2} = - \frac{ 2^{2} }{2} +c[/tex] ∴ c = 4 ∴ [tex] \frac{ y^{2} }{2} = - \frac{ x^{2} }{2} +4[/tex] 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
