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Quick diff eq question

  • Thread starter kdinser
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  • #1
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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
[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?
 

Answers and Replies

  • #2
dextercioby
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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:
[tex] x+y=4 [/tex]
would u like it more than
[tex] -x-y=-4 [/tex]

??

Daniel.
 

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