Answer:
given equation,
[tex](9-y^2)\dfrac{dy}{dx}= x^4\\(9-y^2)dy = x^4 dx\\\int (9-y^2)dy =\int x^4 dx\\9y -\dfrac{y^3}{3} = \dfrac{x^5}{5}+C[/tex]
so,
the given point from where the region is passing through is (x₀, y₀)
hence, the unique equation comes out to be
[tex]9y_0 -\dfrac{y_0^3}{3} = \dfrac{x_0^5}{5}+C[/tex]
[tex]C = 9y_0 -\dfrac{y_0^3}{3}-\dfrac{x_0^5}{5}[/tex]
hence unique equation,
[tex]9y -\dfrac{y^3}{3} = \dfrac{x^5}{5}+ 9y_0 -\dfrac{y_0^3}{3}-\dfrac{x_0^5}{5}[/tex]
