Solve 2x + y + y' x = 3y^2 y', why is this wrong?

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The discussion centers on the equation 2x + y + (dy/dx)x = 3y^2(dy/dx) and the perceived correctness of two forms of the equation. Participants clarify that both forms, (dy/dx)(x - 3y^2) = -2x - y and 2x + y = (3y^2 - x)(dy/dx), are equivalent, differing only by a negative sign. This equivalence suggests that both answers could be considered correct in an exam context, depending on the instructor's preferences. The conversation emphasizes the importance of recognizing equivalent expressions in mathematical equations. Ultimately, understanding the relationship between the two forms is key to resolving the confusion.
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2x+y+(dy/dx)x=3y^2(dy/dx)
This is wrong:
(dy/dx)(x-3y^2)=-2x-y
This is right:
2x+y=(3y^2-x)(dy/dx)

Can someone please explain why.
 
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Ry122 said:
This is wrong:
(dy/dx)(x-3y^2)=-2x-y
This is right:
2x+y=(3y^2-x)(dy/dx)


Those two equalities are actually the same, so I'm not sure "right" and "wrong" really apply.
The only difference is that the second is the negative of the first.
If you move each side of the first equality over the equals sign you will get the second
 
Yeah, scottie_000 is right. Multiply the first by -1 and you'll get the second.
 
So either answer would be correct in an exam?
 
Of course, if they were strictly equivalent and your teachers don't mind.
 
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