Homework Help: Quick diff eq question

1. Jan 22, 2005

kdinser

I'm pretty sure I know the answer to this, just want to double check.

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
$$\frac{x^2}{2}-y^2cos x-xy^3=C$$

The book shows the solution as
$$y^2cos x+xy^3-\frac{x^2}{2}=C$$

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?

2. Jan 22, 2005

dextercioby

It's irrelevant which solution u chose,u can verify that both satisfy the same diff.eq.,the one which u were supposed to solve.In general,a smaller number of minuses is preferable.Par éxample:
$$x+y=4$$
would u like it more than
$$-x-y=-4$$

??

Daniel.

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