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

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Homework Help Overview

The discussion revolves around the manipulation of a differential equation involving variables x, y, and their derivatives. The original poster presents two forms of an equation and questions the validity of their equivalence.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the equivalence of two expressions derived from a differential equation, questioning the application of "right" and "wrong" to their forms. They discuss the implications of multiplying by -1 and how it affects the interpretation of the equations.

Discussion Status

Participants have engaged in a productive dialogue regarding the equivalence of the two expressions. Some have confirmed that both forms can be considered correct under certain conditions, while others are examining the implications of this equivalence in an academic context.

Contextual Notes

There is an underlying assumption about the acceptability of equivalent forms in educational settings, as well as a potential lack of clarity regarding the definitions of "right" and "wrong" in this context.

Ry122
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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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