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

  • Thread starter Ry122
  • Start date
But the second one is more common, because it's easier to solve equations if the dy/dx is on one side alone.In summary, the two given equalities are actually the same, with the only difference being that the second is the negative of the first. The second equality is more commonly used because it is easier to solve equations with the dy/dx term on one side alone.
  • #1
Ry122
565
2
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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  • #2
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
 
  • #3
Yeah, scottie_000 is right. Multiply the first by -1 and you'll get the second.
 
  • #4
So either answer would be correct in an exam?
 
  • #5
Of course, if they were strictly equivalent and your teachers don't mind.
 

1. Why is there a prime symbol after the second y in the equation?

The prime symbol in this equation represents the derivative of y with respect to x. It is commonly used in calculus to indicate the rate of change of a variable. However, in this equation, the use of the prime symbol is incorrect and leads to an invalid equation.

2. Is there a missing operator between the two y terms?

Yes, there should be a multiplication operator between the two y terms. Without it, the equation becomes invalid and cannot be solved.

3. Can you explain the purpose of the x term on the left side of the equation?

The x term is used to indicate that the equation is a function of x. It is important to indicate the independent variable in an equation to avoid confusion and to properly solve the equation.

4. Why is there only one solution to this equation?

Even though this equation is incorrect, it is possible to find a solution by assuming that y' is a constant. However, in a valid equation, there would typically be multiple solutions as it represents a relationship between two variables.

5. How can this equation be corrected?

To correct this equation, the prime symbol should be removed and the missing multiplication operator between the two y terms should be added. The equation should also be rearranged to properly solve for y.

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