Final answer:
The equation √(-10y-y²) dx + (6+7x-x²)dy = 0 is not classified as separable, linear, or exact as it does not satisfy the conditions for these categories; hence, it is none of these types.
Step-by-step explanation:
To classify the equation √(-10y-y²) dx + (6+7x-x²)dy = 0, let's examine it according to the definitions of separable, linear, and exact equations:
- Separable equations can be written in the form g(y)dy = f(x)dx, where each side of the equation depends on only one variable. However, this equation involves a square root function with variable y embedded within it, affecting the dx term, and as such, it cannot be separated into a form where each side depends only on one variable.
- Linear equations have the form dy/dx + P(x)*y = Q(x). This equation does not fit this form, as there are nonlinear terms (such as the square root and the x squared term). Therefore, it is not linear.
- Exact equations have the form M(x, y)dx + N(x, y)dy = 0, where ∂M/∂y = ∂N/∂x. In this case, ∂M/∂y is not continuous everywhere because of the square root requiring negative y values, and ∂N/∂x is merely the derivative of a polynomial, which does pose some potential but is not assured without the exactness condition being satisfied verifiably.
Based on these considerations, the equation is none of these types because it does not satisfy the conditions of being separable, linear, or exact.