Question

(1 point) Use the "mixed partials" check to see if the following differential equation is exact. If it is exact find a function F(x,y) whose differential, dF(x,y) is the left hand side of the differential equation. That is, level curves F(x,y)=C are solutions to the differential equation: (−4x2−1y)dx+(−2x−2y2)dy=0

Answer 1

Answer:

Since, $\dfrac{\partial M}{\partial y }\neq \dfrac{\partial N}{\partial x};$ then the given equation is not exact.

Step-by-step explanation:

Consider the differential equation

$(-4x^2-1y) dx + ( -2x-2y^2) \ dy = 0$

Let's recall that the differential equation must choose the form Mdx + Ndy = 0 is exact; given that it satisfies the below proposition.

$\dfrac{\partial M}{\partial y }= \dfrac{\partial N}{\partial x}$

where;

M = $-4x^2 -y$ and N = $-2x -2y^2$ in the given equation.

Thus; to calculate;

$\dfrac{\partial M}{\partial y }= \dfrac{\partial }{\partial y} \bigg( -4x^2-y \bigg)$ and $\dfrac{\partial N}{\partial x }= \dfrac{\partial }{\partial x} \bigg( -2x-2y^2 \bigg)$

$M_y (x,y) =-8x-1$ and $N_x (x,y)= -4y - 2$

Since, $\dfrac{\partial M}{\partial y }\neq \dfrac{\partial N}{\partial x};$ then the given equation is not exact.