Daniel Ward

Posted on Wednesday, 09 April, 2003 - 08:37 pm:

Let

y

be any solution of the differential equation

$dy / dx - (x e^{6 x^{2}}) (y + 3)^{1 - k} = 0$

Find a value of $k$ such that, as $x$ tends to infinity, $y e^{- 3 x^{2}}$ tends to a finite non-zero limit, which you should determine.

I divided through by (y+3) to the $1 - k$ , multiplied by $dx$ , and rearranged to get :-

$\int (y + 3)^{k - 1} dy = \int x e^{6 x^{2}}$

Doing the integration, you get

$[(y + 3)^{k}] / k = [e^{6 x^{2}}] / 12 + C$

Rearranging, you get

$y = (k e^{6 x^{2}} / 12 + C)^{1 / k}$

thus

$y e^{- 3 x^{2}} = (e^{- 3 x^{2}}) ((k e^{6 x^{2}} / 12 + D)^{1 / k})$

Putting the $e^{- 3 x^{2}}$ under the $1 / k$ , we get

$[k e^{(6 x - 3 k) x^{2}} / 12 + D e^{- 3 x^{2}}]^{1 / k}$

If $k = 2$ , then as $x$ tends to infinity, $y$ tends to

$6^{- 1 / 2}$

Is this correct anyone?

On a random note, I'm pretty pleased that I negotiated the maths notation typing this up :-)

Demetres Christofides

Posted on Wednesday, 09 April, 2003 - 09:22 pm:

The final answer is correct, however there is a small mistake (other than typos) in your working. You changed

y + 3

y

but since

3 e^{- x^{2}}

tends to zero as

x

tends to infinity this did not affect your answer.

By the way never say that you multiply by $dx$ .

Going from $g (y) dy / dx = f (x)$ to $\int g (y) dy = \int f (x) dx$ is fine but there is no meaning at all to $g (y) dy = f (x) dx$

Demetres

Daniel Ward

Posted on Wednesday, 09 April, 2003 - 10:11 pm:

Thanks for answering so quickly. Yay, I got it right !

Out of interest, why can't you multiply by dx ?

Also, would the answer as I wrote it be enough for an alpha in the real thing, do you reckon ?

Kerwin Hui

Posted on Wednesday, 09 April, 2003 - 10:34 pm:

Well, tell me what you mean by 'dx'. Any explanation involving 'infinitesimals' and suchlikes are not accepted. Afterall, rigour is the first test and terms such as 'infinitesimals' are not rigourously defined.

Yes, 'dx' can be made rigorous, and usually done during final year undergraduate in some differential geometry course. Multiplying by 'dx' does not have the same meaning as what you were taught in school.

Kerwin

Demetres Christofides

Posted on Thursday, 10 April, 2003 - 08:59 am:

If an alpha is given for taking > 3/4 of the marks then I suppose that you would be given the alpha.
The most serious error in your whole answer is the multiplication by dx. The reason that you can do what you did is the fundamental theorem of calculus which rouglhy says that integration and differentiation are inverse processes (yes that's a theorem it's not the definition of integration !)

One more thing that you need to be aware so that you don't lose more marks where you shouldn't:
The integration is only valid for k =/= 0 (and you know this). In your A' level exams the examiner would probably not bother but in Step you would lose mark(s). Also does the question say find a k or all k ? If it says all k then you need to check what happens for k=0, and also show that the other values of k give either an infinite limit or zero.

Demetres

Michael Doré

Posted on Thursday, 10 April, 2003 - 04:23 pm:

Yes, but as this is a STEP problem f(x)dx = g(y)dy would definitely be acceptable, since this is how people are taught to solve seperable differential equations at A-Level (and no proper definition of the integral is given at A-Level anyway). Also it would be acceptable in any branch of applied mathematics, and really A-Levels and STEP are closer to applied mathematics in spirit than pure mathematics.

Demetres Christofides

Posted on Friday, 11 April, 2003 - 09:27 am:

Well, I don't know too much about STEP and how the exam is marked so Michael might be right.
Hoewever to be safe I would advise you to write f(x)dx = g(y)dy only on a separate piece of paper and let nobody to see it.

Demetres

Mark Durkee

Posted on Friday, 11 April, 2003 - 04:32 pm:

I've always been taught to take the dx to the other side and then add integral signs on the same line (although I am aware that this is equivalent to integrating both sides wrt x) - this means that no technically incorrect steps are written down.

Daniel Ward

Posted on Friday, 11 April, 2003 - 05:48 pm:

That sounds like a good idea. Thanks all for your help.