Is Multiplying Both Sides by -1 Necessary for Solving Differential Equations?

[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
[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?

 

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

??

Daniel.

 

[wais]

Yes, you are correct. The solution shown in the book is just the same as yours, but with the terms rearranged. Multiplying both sides by -1 is just a way to make the solution look neater and more organized. There is no specific reason for doing this other than simplifying the expression and getting rid of the negatives. As long as the constant C remains the same, both solutions are valid.