help separate eq: y' = (x^2 y^2 - 2y^4) / (x^3 - xy^3)

[ibanez1608]

is my way is good? cause im stuck

 

[Ishuda]

is my way is good? cause im stuck

Sorry, your images are turned 90 degrees and I can't make them out.

 

[ibanez1608]

turned it

 

[Ishuda]

Just to make sure I understand what you are asking [sometimes I'm not exactly sure of what is on the image], you have
y' = \(\displaystyle \dfrac{x^2\, y^2\, -\, 2\, y^4}{x^3\, -\, x\, y^3}\)

You want to know if substituting
y = x f
is a good way to go to get a separable equation.

Yes, it is. Perhaps one of the reasons you are having problems is that you made a mistake at the point you have
y' = \(\displaystyle f\, \dfrac{1\, -\, 2\, f^2}{1\, -\, f}\)
This should be
y' = \(\displaystyle f\, \dfrac{1\, -\, 2\, f^2}{1\, -\, f^2}\)

Can you continue from there?

 

[ibanez1608]

Just to make sure I understand what you are asking [sometimes I'm not exactly sure of what is on the image], you have
y' = \(\displaystyle \dfrac{x^2\, y^2\, -\, 2\, y^4}{x^3\, -\, x\, y^3}\)

You want to know if substituting
y = x f
is a good way to go to get a separable equation.

Yes, it is. Perhaps one of the reasons you are having problems is that you made a mistake at the point you have
y' = \(\displaystyle f\, \dfrac{1\, -\, 2\, f^2}{1\, -\, f}\)
This should be
y' = \(\displaystyle f\, \dfrac{1\, -\, 2\, f^2}{1\, -\, f^2}\)

Can you continue from there?

this is the eq:
y' = [FONT=MathJax_Math-italic]4x²-2y³/[/FONT][FONT=MathJax_Math-italic]x³-xy²[/FONT]

 

[HallsofIvy]

this is the eq:
y' = [FONT=MathJax_Math-italic]4x²-2y³/[/FONT][FONT=MathJax_Math-italic]x³-xy²[/FONT]

So it really is \(\displaystyle y'= 4x^2- \frac{2y^3}{x^3}- xy^2\)
and not \(\displaystyle y'= \frac{3x^2- 2y^3}{x^3- xy^2}\)?

 

[ibanez1608]

So it really is \(\displaystyle y'= 4x^2- \frac{2y^3}{x^3}- xy^2\)
and not \(\displaystyle y'= \frac{3x^2- 2y^3}{x^3- xy^2}\)?

y' = ([FONT=MathJax_Math-italic]4x²-2y³)/([/FONT][FONT=MathJax_Math-italic]x³-xy²)

this way
i dont know how to do line for division[/FONT]

 

[ibanez1608]

any help?