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

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].

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, 200916 yr by Shadow

Chain rule.

Thanks again

maybe because dy/dy=1

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

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.