I was hoping that someone could explain to me the major steps involved in the following derivation, as I am fairly new to differential equations:
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1$\begingroup$ It is most helpful (and strongly encouraged here) to post the equation found via your link directly in your post. I clicked on your link, and it certainly wouldn't take much effort to type & include the actual equation right in your post. It also would help if you could let us know what you've tried, or how you might approach this problem. That said, Welcome! $\endgroup$– amWhyCommented Sep 2, 2012 at 23:42
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1 Answer
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There must be some extra information given beyond the quoted lines. Assuming that $$n=n(y)$$ $$y=y(z)$$ formal application of the product and chain rules gives: $$\frac{dn}{dz}\frac{dy}{dz}+n\frac{d^2 y}{dz^2}=\frac{dn}{dy}$$ $$\frac{dn}{dy}\frac{dy}{dz}\frac{dy}{dz}+n\frac{d^2 y}{dz^2}=\frac{dn}{dy}$$ $$\frac{dn}{dy}\left(\frac{dy}{dz}\right)^2+n\frac{d^2 y}{dz^2}=\frac{dn}{dy}$$ So unless $\frac{dy}{dz}\ne 0$ the given conclusion does not seem to hold.
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1$\begingroup$ Thank you for the answer. Are you sure, however, you don't mean that unless dy/dz = 0, the given conclusion does not seem to hold? $\endgroup$ Commented Sep 3, 2012 at 2:37
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$\begingroup$ Actually what you need is $$ \dfrac{dn}{dy} \left(\dfrac{dy}{dz} \right)^2 = 0$$ so either $dn/dy = 0$ or $dy/dz = 0$. $\endgroup$ Commented Sep 3, 2012 at 4:43