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For the problems that I'm currently working on, we are just solving the problems for an unknown constant C.

I just finished one were I came up with

[tex]\frac{x^2}{2}-y^2cos x-xy^3=C[/tex]

The book shows the solution as

[tex]y^2cos x+xy^3-\frac{x^2}{2}=C[/tex]

Because C is an arbitrary, unknown constant, did they just multiply both sides by -1? Is there some reason for doing this other then getting rid of 2 negatives in the answer?