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STEP III question



Part i) fine.
Part ii) fine in the most part, but I have an almost trivial question with the solutions booklet.

After applying a similar substitution to that in part i) in part ii), I ended up with d2udx2+x1−xdudx−11−xu=0\displaystyle \frac{d^{2}u}{dx^{2}} + \frac{x}{1-x} \frac{du}{dx} - \frac{1}{1-x} u = 0. Now, the solutions booklet notes that this is the original equation but with x replaced with -x, but surely the dudx\displaystyle \frac{du}{dx} term would be −x1−xdudx\displaystyle - \frac{x}{1-x} \frac{du}{dx} rather than a +. I did consider that the x in dudx\displaystyle \frac{du}{dx} might also change sign, and that would yield the right answer, though I've never seen that done before, so I'm a bit confused at the moment.

As always, any help would be appreciated. Thanks!
Reply 1
Avatar for DFranklin
DFranklin
18
If x′=−xx' = -x, then dudx′=−dudx\frac{du}{dx'} = - \frac{du}{dx} (which I think is what you were saying). I'm pretty certain you need to factor this in.
Reply 2
Avatar for Zuzuzu
Zuzuzu
OP
12
Yep, that's what I was getting at, albeit not very eloquently.

Cheers, DF.
This is the 07 paper?
Reply 4
Avatar for Slowbro93
Slowbro93
20
Yeah it is, doing the question this evening :smile: de :love:

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