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## Homework Statement

Use the "mixed partials" check to see if the following differential equation is exact.

If it is exact find a function [itex]F(x,y)[/itex] whose differential, [itex]dF(x,y)[/itex] is the left hand side of the differential equation. That is, level curves [itex]F(x,y) = C[/itex] are solutions to the differential equation:

[itex](-4xy^2+y)dx+(-4x^y+x)dy = 0[/itex]

Find:

[itex]M_{y}(x,y) = ?[/itex]

[itex]N_{x}(x,y) = ?[/itex]

[itex]F(x,y) = ?[/itex]

**2. The attempt at a solution**

I first partially differentiated each part (M and N) with respect to y and x, respectively. This got me this and they are exact:

[itex]M(x,y) = -4xy^2+y[/itex]

[itex]\frac{∂M}{∂y} = -8xy+1[/itex]

[itex]N(x,y) = 4x^2y+x[/itex]

[itex]\frac{∂N}{∂y} = -8xy+1[/itex]

Then I integrated [itex]M(x,y)[/itex] with respect to x which got me:

[itex]-2x^2y^2+xy+h(y)[/itex]

Then I partially differentiated that equation with respect to y:

[itex]\frac{∂M}{∂y} = -4x^2y+x+h'(y)[/itex]

Next, I set that equation equal to [itex]N(x,y)[/itex] and solved for [itex]h'(y)[/itex]:

[itex]-4x^2y+x+h'(y) = 4x^2y+x[/itex]

[itex]h'(y) = 0[/itex]

I then integrated with respect to y:

[itex]∫ h'(y) dy = ∫ 0 dy[/itex]

[itex]h(y) = 0 + C[/itex]

Finally I plug it back into the original equation:

[itex]-2x^2y^2+xy+h(y)[/itex]

[itex]h(y) = 0 + C[/itex]

[itex]-2x^2y^2 + xy + C = -2x^2y^2 + C[/itex]

But because the website (WeBWorK) does not allow the input of "C", I just left it as:

[itex]-2x^2y^2 + xy[/itex]

Thus, my answers were:

[itex]M_{y}(x,y) = -8xy+1[/itex]

[itex]N_{x}(x,y) = -8xy+1[/itex]

[itex]F(x,y) = -2x^2y^2 + xy[/itex]

I submitted the answers but the site said they were wrong, and I don't know which ones are wrong because the question is one of those, "get all right for a point" kind of thing.