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# Ordinary differential equation watch

1. Hey, I've been struggling to find a solution for this equation.

If anyone could help me or give me a clue as to how to get started? Thanks

y' = 3y/(3y^(2/3) - x)
2. I think it goes
dy/dx=((vdu/dx)-(udv/dx))/v^2
((3y^(2/3)-x)(3dy/dx)-(3y(2y^(-1/3)-1))/(3y^(2/3)-1)
3. (Original post by jo62)
I think it goes
dy/dx=((vdu/dx)-(udv/dx))/v^2
((3y^(2/3)-x)(3dy/dx)-(3y(2y^(-1/3)-1))/(3y^(2/3)-1)
Thanks, but I meant the differential equation, not the derivative
4. Flip it upside down to get

, which can be solved using integrating factor methods.
5. (Original post by Glutamic Acid)
Flip it upside down to get

, which can be solved using integrating factor methods.
Oh thanks, I'd completely forgotten to try flipping it.

Brilliant thanks )

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