Please someone please find the second differential for this equation

(x^{3}) + (y^{3}) = 1

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- Jun 17th 2012, 04:57 PMrighteous818Urgent second differential problem
Please someone please find the second differential for this equation

(x^{3}) + (y^{3}) = 1 - Jun 17th 2012, 05:03 PMProve ItRe: Urgent second differential problem
- Jun 17th 2012, 05:06 PMrighteous818Re: Urgent second differential problem
i did but i am not the gettin the answer the book said the answer is -2x/(y

^{5}) - Jun 17th 2012, 05:07 PMProve ItRe: Urgent second differential problem
- Jun 17th 2012, 05:07 PMrighteous818Re: Urgent second differential problem
how do i get fractions

- Jun 17th 2012, 05:11 PMProve ItRe: Urgent second differential problem
If you are unsure how to use LaTeX, just use a / for the vinculum and use brackets where they are necessary. Otherwise inside tex tags, use the command \frac{}{}, with the numerator in the first set of brackets and the denominator in the second.

- Jun 17th 2012, 05:15 PMrighteous818Re: Urgent second differential problem
well for te first differential i got

3x^{2}+ 3y^{2}(dy/dx)=0

dy/dx = (-x^{2})/(y^{2})

Then seoncd differntial using quotient rule

[(y^{2})(-2x)-(-x^{2})(2y(dy/dx))]/(y^{4})

then finally the second differntial came down to for me

[-2xy^{3}+ 2x^{4}]/y^{5}

Please help - Jun 17th 2012, 05:27 PMProve ItRe: Urgent second differential problem
You are correct up to where you have used the Quotient Rule to find the second derivative. So you have

$\displaystyle \displaystyle \begin{align*} \frac{d^2y}{dx^2} &= \frac{-2xy^2 + 2x^2y\,\frac{dy}{dx}}{y^4} \\ &= \frac{-2xy + 2x^2\frac{dy}{dx}}{y^3} \\ &= \frac{-2xy + 2x^2\left(-\frac{x^2}{y^2}\right)}{y^3} \\ &= \frac{-2xy - \frac{2x^4}{y^2}}{y^3} \\ &= \frac{-2xy^3 - 2x^4}{y^5} \end{align*}$

Your final answer has a sign error but is otherwise correct. - Jun 17th 2012, 05:31 PMrighteous818Re: Urgent second differential problem
thanks