Extraneous solution to differential equation - modulus sign causing problems

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I see what you mean and I can't see why you wouldn't get two answers, but is it actually the modulus that's causing the problems for you? It seems to me if you ignore the modulus you would still have this problem, maybe im being stupid tho

However in the textbooks; when solving a differential equation, they just quietly ignore this technicality and proceed as if the modulus sign had never been discovered. I suspect they did this here in the official mark scheme.

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I cannot find any reason to refute k=65/9 as a valid solution. Am I missing something? My exams are in 3 weeks so how do I proceed with this type of question. Do I simply forget about the modulus sign? That leaves a bad taste in my mouth.

So, there's a detail that is often omitted or glossed over when people talk about integrating f'(x)/f(x) to get ln |f(x)|. If f(x) is always > 0 (in the region you are considering), then yes, you get ln(f(x)), and if (f(x) < 0) then you get ln(-f(x)). But if f(x) actually changes sign, then the integral is not well defined. (And similarly if you are solving a differential equation and get ln |f(x)| in your solution, then this solution isn't defined when f(x) = 0).

Explicitly, your 2nd solution is undefined when x^3 = 65 / 9 (which is < 8 = 2^3), and since you know x covers the range 1 to 2, there will be a point where this happens. (Conversely, k = 9 is outside the range of values of x^3 we need to consider).

In "practical terms", when solving a differential equation (with initial conditions), I don't think you're likely to go wrong by ignoring the modulus signs, and then if you find you want to take the log of something negative, putting them in. If you want to be careful, think about what the *actual* sign is of what you're taking the log of, and in particular, whether that sign might change in the interval of consideration.

I actually think it's you who are unwittingly glossing over a technical point on the difference between definite and indefinite integrals.

Forgetting about the limits and then adding them on later is where it's muddying the waters. Just go straight to the actually true equation

$\int_1^2 { 2 x^2 \over {k - x^3}} dx = \int_1^4 {dt \over t}$

Now it's clear the left-hand integrand's denominator can't change sign over the range [1,2] as that would mean a blow-up at zero

Nothing unwitting about it, I assure you My recollection was that at A-level (not FM A-level), differential equations were always solved by forming the 2 indefinite integrals, and I didn't want to go too far from the route the OP had taken. I did emphasize the need to consider the behaviour over the entire range. I also thought showing that explicitly "your actual solution is undefined at x^3 = k" was a bit more direct than merely discussing behaviour of the integrand, when (I assume) A-level students will have encountered integrals where the integrand is undefined at a point but the integral still converges.

I disagree that the limits are where the waters are muddied. To my mind, the real problem is that (arguably), it's not clear whether k - x^3 should be treated as +ve or -ve, or indeed, whether both are valid possibilities. I don't see using definite integrals as resolving this. It's not knowing this sign that causes a potential ambiguity; if not unresolved, you're going to have to consider both possibilities, find the two consequent potential values for k and only then will you be able to tell "yeah, that second value doesn't work".

If I was doing it personally, having noted that k-x^3 can't change sign, I'd note that x is an increasing function of t (and so k-x^3 must be +ve) to get rid of the modulus signs early. Again, this is something I did try to hint to the OP, although it doesn't look like he took it on board.

And don't forget, in this midst of all this, that from the OP's description of the mark scheme, it doesn't seem like the examiners considered this at all. I do think "don't worry about modulus until you find out you need them" is actually pretty reasonable advice at this level.

Thanks for all the detailed responses. The reason I sounded so arrogant was that none of what was mentioned in this discussion is even addressed in any textbook or mark scheme by CIE. I understand now that the term inside the modulus can be any sign but cannot change.

Also isn't x a decreasing function of t?

Just to be clear: Does the manual discarding of k=65/9 HAVE to be done for rigor? The mark scheme does not even mention this which is why I think they just took the positive root of (1). Are CIE not expecting us to go this far or is their something I'm missing that allowed taking only the positive root.

I've thought about this some more ... The anti-derivative function can be ln(f(x)) or ln(|f(x)|) but not both.

You are evaluating one anti-derivative function at two points, not two different functions. Since the numerator is obviously +ve the denominator must be +ve over the range, or -ve over the range but it can't cross zero.