Final answer:
Any separable equation presented as m(x) n(y) y' = 0 can be shown to be exact by rewriting it in the form m(x) dx + n(y) dy = 0, which meets the definition of an exact differential equation.
Step-by-step explanation:
To show that any separable equation is also exact when presented in the form m(x) n(y) y' = 0, one must first understand that a separable equation can be written as the product of two functions, one involving only x and the other involving only y. When the equation is rearranged, it can be written as m(x) dx + n(y) dy = 0, which is the definition of an exact differential equation.
Let's consider an equation of the form h(x)g(y)y'=0 where h(x) is a function of x only and g(y) is a function of y only. When we have y' = dy/dx,
the equation can be rewritten as h(x)g(y)(dy/dx) = 0 or,
after multiplying through by dx, h(x)g(y) dy + 0 dx = 0.
This can then be recognized as the differential M dx + N dy = 0,
where M=0 and N=h(x)g(y).
If both M and N are continuously differentiable, and ∂M/∂y = ∂N/∂x holds true,
which it does in this case (since ∂M/∂y = 0 and ∂N/∂x = 0), then by definition the differential equation is exact.