Answer:
given equation,
![(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](https://tex.z-dn.net/?f=%289-y%5E2%29%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%20x%5E4%5C%5C%289-y%5E2%29dy%20%3D%20x%5E4%20dx%5C%5C%5Cint%20%289-y%5E2%29dy%20%3D%5Cint%20x%5E4%20dx%5C%5C9y%20-%5Cdfrac%7By%5E3%7D%7B3%7D%20%3D%20%5Cdfrac%7Bx%5E5%7D%7B5%7D%2BC)
so,
the given point from where the region is passing through is (x₀, y₀)
hence, the unique equation comes out to be
![9y_0 -\dfrac{y_0^3}{3} = \dfrac{x_0^5}{5}+C](https://tex.z-dn.net/?f=9y_0%20-%5Cdfrac%7By_0%5E3%7D%7B3%7D%20%3D%20%5Cdfrac%7Bx_0%5E5%7D%7B5%7D%2BC)
![C = 9y_0 -\dfrac{y_0^3}{3}-\dfrac{x_0^5}{5}](https://tex.z-dn.net/?f=C%20%3D%209y_0%20-%5Cdfrac%7By_0%5E3%7D%7B3%7D-%5Cdfrac%7Bx_0%5E5%7D%7B5%7D)
hence unique equation,
![9y -\dfrac{y^3}{3} = \dfrac{x^5}{5}+ 9y_0 -\dfrac{y_0^3}{3}-\dfrac{x_0^5}{5}](https://tex.z-dn.net/?f=9y%20-%5Cdfrac%7By%5E3%7D%7B3%7D%20%3D%20%5Cdfrac%7Bx%5E5%7D%7B5%7D%2B%209y_0%20-%5Cdfrac%7By_0%5E3%7D%7B3%7D-%5Cdfrac%7Bx_0%5E5%7D%7B5%7D)
Add 2+x
9 = x
If you're wanting the value of x, the answer is ...
x = 9
_____
You can always solve an equation like this by subtracting one side, then dividing by the coefficient of x, then adding the opposite of the constant. If you choose to subtract the right side, these steps will give you ...
9 -x = 0 . . . . . . subtract -2 from both sides
-9 +x = 0 . . . . . divide by the coefficient of x, which was -1
x = 9 . . . . . . . . .add the opposite of the constant, so add 9
Whatever you do to one side of the equation, you must also do to the other side. When we say "subtract -2", we mean "subtract -2 from both sides of the equation."
Answer:
free answer
Step-by-step explanation:
it is free