Absolute value signs in differential equations

I solved a separable differential equation to get

\ln |sec y| = \tan x

.
The official textbook solutions go on to "simplify" it and say

y = \arccos(e^{-\tan x})

.
Unless I'm missing something this is invalid for two reasons:
Ignoring the absolute value signs loses a lot of information about the function. For example, the range of

y

in my answer includes

\pi

, but the answer they give doesn't allow any

y

such that

\cos y < 0

.

Secondly,

\cos y = x \not\Rightarrow y = \arccos x

I don't know what I should do in, say, the real exam. These sorts of things always seemed to be missed by the official solutions to the A level textbook.

EDIT: Also, another related question is that when your solving for the general solution, the constant often deals with the absolute value problem right?

(edited 5 years ago)

Anyone?

Bump

You are basically right but you're being too clever for the exam. Just do what they expect you to do even if it isn't strictly true.

Reply 4

ghostwalker

Think I'd like to see the original question and textbook solution before commenting.

Reply 5

RuneFreeze

Original post by ghostwalker

Think I'd like to see the original question and textbook solution before commenting.

https://activeteach-prod.resource.pearson-intl.com/r00/r0069/r006953/r00695390/current/alevelsb_p2_ex11j.pdf

Question 2d. The differential equation is at the top of the solution. boundary condition is it goes through the origin.

Reply 6

RuneFreeze

Original post by B_9710

You are basically right but you're being too clever for the exam. Just do what they expect you to do even if it isn't strictly true.

I'm not trying to be clever it just really confused me. I've already spent ages trying to really understand the solutions to the differential equations, like how it's justified to remove the absolute value signs when there general solutions etc... But this just threw a spanner at that and confused me more.

Reply 7

B_9710

Original post by RuneFreeze

So what is it specifically that is confusing you?

Reply 8

DFranklin

Original post by RuneFreeze

When you're solving a DE and end up with ln|stuff| on one side, it's virtually never valid for stuff to change sign. So if the initial condition indicates that stuff is +ve, it's reasonable to assume it stays so.

Reply 9

RuneFreeze

Original post by DFranklin

What do you mean by it's not valid for it to change sign? Because I input both graphs to Geogebra and they were the same except the non simplified one had a much broader range of values.

Reply 10

DFranklin

Original post by RuneFreeze

What do you mean by it's not valid for it to change sign? Because I input both graphs to Geogebra and they were the same except the non simplified one had a much broader range of values.

Can you post a link to your graph?

If you know about complex numbers, you can think of taking the modulus as adding a (complex) arbitrary constant.

That is, you do the integral in the normal way (without the mysterious modulus appearing from no where), and get ln(f(x)) + C, where f(x) < 0.
But

\ln f(x) = \ln(-f(x)) + i \pi

, so you can rewrite

\ln(f(x)) + C

\ln (-f(x)) + K

(where K = C + i pi).

But of course, by definition, the arbitrary constant is *constant*. So it can't keep changing to flip between -f(x) and f(x).

It's also worth noting that (in general), you can't meaningfully take a solution across the point where you need to find ln(0) (and would get infinity).

e.g. if you have the differential equation

\dfrac{dy}{dx} = 1/x

, then for x > 0 you have a solution y = C + ln x, and for y < 0 you have a solution y = C + ln(-x), but the most general solution (*) for

x \neq 0

is *not* y = C + ln |x|. You can in fact have y = C + ln x for x > 0, y = D + ln |x| for x < 0, where C and D are *different* constants.

(*) Really, assuming you can have *one* solution when you cross a point where y is undefined is what's causing the problem.