Solving Exact Differential: x(dy/dx) = 2xe^x - y + 6x^2

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Homework Statement



x(dy/dx) = 2xe^x - y + 6x^2

Homework Equations





The Attempt at a Solution



I am trying to show this is an exact differential (book has an answer, it must be )

I rearranged the problem to be :

2xe^x - y + 6x^2 dx = x dy

Partial derivatives (using & for that symbol that represents partial)
&m/&y = -1
&m/&x = 1

-1 doesn't equal 1

Any idea where my mistake is?
 
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illidari said:

Homework Statement



x(dy/dx) = 2xe^x - y + 6x^2

Homework Equations





The Attempt at a Solution



I am trying to show this is an exact differential (book has an answer, it must be )

I rearranged the problem to be :

2xe^x - y + 6x^2 dx = x dy

Partial derivatives (using & for that symbol that represents partial)
&m/&y = -1
&m/&x = 1

-1 doesn't equal 1

Any idea where my mistake is?
Write the equation as Mdx + Ndy = 0, and you'll see that this equation is exact, since My = Nx.
 
*face palm*

Thank you lol :P

That would flip the sign of y and make it 1=1.
 
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