Answer:
Since, [tex]\dfrac{\partial M}{\partial y }\neq \dfrac{\partial N}{\partial x};[/tex] then the given equation is not exact.
Step-by-step explanation:
Consider the differential equation
[tex](-4x^2-1y) dx + ( -2x-2y^2) \ dy = 0[/tex]
Let's recall that the differential equation must choose the form Mdx + Ndy = 0 is exact; given that it satisfies the below proposition.
[tex]\dfrac{\partial M}{\partial y }= \dfrac{\partial N}{\partial x}[/tex]
where;
M = [tex]-4x^2 -y[/tex] and N = [tex]-2x -2y^2[/tex] in the given equation.
Thus; to calculate;
[tex]\dfrac{\partial M}{\partial y }= \dfrac{\partial }{\partial y} \bigg( -4x^2-y \bigg)[/tex] and [tex]\dfrac{\partial N}{\partial x }= \dfrac{\partial }{\partial x} \bigg( -2x-2y^2 \bigg)[/tex]
[tex]M_y (x,y) =-8x-1[/tex] and [tex]N_x (x,y)= -4y - 2[/tex]
Since, [tex]\dfrac{\partial M}{\partial y }\neq \dfrac{\partial N}{\partial x};[/tex] then the given equation is not exact.
