Shadow Posted June 19, 2009 Share Posted June 19, 2009 Hey all, I've been trying to learn second order differential equations with x and y' missing. I've been doing so with some material my math teacher gave me, and they give the following example: [math]y'' = -\frac {1}{y^3}, \ y \neq 0[/math] They go on to multiply both sides of the equation by [math]2y'[/math]: [math]2y'y'' = -2 \frac {y'}{y^3}[/math] Now here comes my problem. They rearrange the left side like so: [math](y'^{\ 2})'[/math] That is correct, when we use the chain rule the result is [math]2y'y''[/math]. However, it's the right side rearrangement that confuses me. From [math]-2 \frac {y'}{y^3}[/math] they somehow get [math]\left(\frac {1}{y^2}\right)'[/math] How is this possible? As far as I can see, [math]\left(\frac {1}{y^2}\right)'=\left(y^{-2}\right)'=-2y^{-3}=-\frac{2}{y^3}[/math]. Is there something I'm missing, or is the book in error? Cheers, Gabe Link to comment Share on other sites More sharing options...
Severian Posted June 19, 2009 Share Posted June 19, 2009 No - the book is correct. You need to remember that the prime is a differentiation with respect to [math]x[/math], not [math]y[/math]. So [math]\left(\frac {1}{y^2}\right)' \equiv \frac{d}{dx} \left(y^{-2}\right) = \frac{dy}{dx} \frac{d}{dy}\left(y^{-2}\right) = - y^\prime 2y^{-3}=-\frac{2y^\prime}{y^3}[/math]. Link to comment Share on other sites More sharing options...
Shadow Posted June 19, 2009 Author Share Posted June 19, 2009 (edited) Okay, me feeling stupid... Thanks man EDIT: Just a question, why does [math]\frac{d}{dx} \left(y^{-2}\right) = \frac{dy}{dx} \frac{d}{dy}\left(y^{-2}\right)[/math]? Edited June 19, 2009 by Shadow Link to comment Share on other sites More sharing options...
D H Posted June 19, 2009 Share Posted June 19, 2009 Chain rule. Link to comment Share on other sites More sharing options...
Shadow Posted June 20, 2009 Author Share Posted June 20, 2009 Thanks again Link to comment Share on other sites More sharing options...
GetBG Posted June 20, 2009 Share Posted June 20, 2009 maybe because dy/dy=1 Link to comment Share on other sites More sharing options...
Bignose Posted June 20, 2009 Share Posted June 20, 2009 maybe because dy/dy=1 No, that has nothing to do with it. You can't just manipulate "dy" like you can an x in x/x or an h in h/h and the like. The differential there obeys certain extras rules. Sometime you can manipulate it like an algebraic term and sometimes not. In this case, you can't. The expansion works because of the chain rule, not because of algebra. http://en.wikipedia.org/wiki/Chain_rule Link to comment Share on other sites More sharing options...
the tree Posted June 22, 2009 Share Posted June 22, 2009 Although to be fair, for most sensible interpretations of [imath]y[/imath] it does hold that [imath]\frac{d y}{d y} = 1[/imath]. Not that that is at all relevant. Link to comment Share on other sites More sharing options...
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