Since question gave that rate of decrease of body temperature is proportional to the difference between the body and room temperature,

We can say that :

dθ/dt = k(θ - 25), where k is a constant to be determined.

1/(θ - 25) dθ/dt = k

Integrate both sides with respect to t,

ln(θ - 25) = kt + c , where c is an arbitrary constant.

When it was 12am,
t = 0 (since t is the number of hours after midnight)
θ = 32

Sub these into the equation,

ln(32 - 25) = k(0) + c
c = ln7

When it was 1am,
t = 1 , θ = 30

Sub these and c = ln7 into the equation,


ln(30 - 25) = k(1) + ln7
ln5 = k + ln7
k = ln5 - ln7
k = ln(5/7)

Therefore,

ln(θ - 25) = ln(5/7)t + ln7

θ - 25 = eˡⁿ⁽⁵/⁷⁾ᵗ ⁺ ˡⁿ⁷
θ - 25 = (eˡⁿ⁽⁵/⁷⁾)ᵗ · eˡⁿ⁷
θ - 25 = (5/7)ᵗ · 7

θ = 7(5/7)ᵗ + 25

(Shown)

Note: we do not need to put the modulus sign for the ln as the body temperature is expected to fall to 25°C eventually and be in equilibrium with the room temperature, since the body would lose heat to the surroundings.

It would not go lower than 25°C

Alternatively,


ln(θ - 25) = kt + c , where c is an arbitrary constant.

θ - 25 = eᵏᵗ⁺ᶜ
θ - 25 = eᵏᵗ · eᶜ

When it was 12am,
t = 0 (since t is the number of hours after midnight), θ = 32

Sub these into the equation,

32 - 25 = eᵏ⁽⁰⁾ · eᶜ
7 = e⁰ · eᶜ
eᶜ = 7

When it was 1am,
t = 1 , θ = 30

Sub these and eᶜ = 7 into the equation,

30 - 25 = eᵏ⁽¹⁾ · 7
5 = 7eᵏ
eᵏ = 5/7

Therefore,

θ - 25 = (5/7)ᵗ · 7
θ = 7(5/7)ᵗ + 25

Note: we do not need to put the modulus sign for the ln as the body temperature is expected to fall to 25°C eventually...

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Modulus sign is still, after all, the norm technique for students to answer differential equations in general, so most students would include the modulus sign anyway.

Modulus sign is used when the term inside has a possibility of being negative. This prevents the ln from becoming undefined.

For this question, it is not needed. Students are expected to be discerning and know when to use and when not to use, instead of just blanket usage.

If a student had not realised that the body temperature will not fall below 25°C, and applied the modulus, the following will result :

ln|θ - 25|= kt + c , where c is an arbitrary constant.

|θ - 25| = eᵏᵗ⁺ᶜ

θ - 25 = eᵏᵗ · eᶜ or θ - 25 = -eᵏᵗ · eᶜ

θ = 25 + eᵏᵗ · eᶜ or θ = 25 - eᵏᵗ · eᶜ

When t = 0, θ = 32,

32 = 25 + eᶜ or 32 = 25 - eᶜ
eᶜ = 7 or eᶜ = -7 (rejected as eᶜ > 0 for all real c)

He or she will still get the correct result, but this incurs extra working, which could have been avoided if extra attention was paid to the value of θ - 25.


And after obtaining eᶜ = 7 and eᵏ = (5/7),

|θ - 25| = 7(5/7)ᵗ

θ - 25 = 7(5/7)ᵗ or θ - 25 = -7(5/7)ᵗ

θ = 7(5/7)ᵗ + 25 or θ = 25 - 7(5/7)ᵗ

The student will still need to reject the second result since 7(5/7)ᵗ > 0 for all real t, which would imply θ < 25 (contradicts θ = 32 when t = 0 and θ = 30 when t = 1, and θ = 37 when the person just died)

Normally I would do this:

ln |x - 25| = kt + c
|x - 25| = e^(kt + c)
x - 25 = +- e^(kt + c)
x - 25 = +- e^c times e^kt
x - 25 = Ae^kt where A = +- e^c is another constant

effectively eliminating your second case.

You haven't actually eliminated anything.

Your constant A now has a ±.

You'll still need to reject one of them for the final expression to be shown (and also to give the reason why)

Substituting values gives :

32 - 25 = A
A = 7

But since A = ±eᶜ ,

7 = eᶜ
or
7 = -eᶜ (still to be rejected)

So A = eᶜ = 7 and not -eᶜ


Likewise,

30 - 25 = 7eᵏ
5 = 7 eᵏ
eᵏ = 5/7


Technically since the negative sign has been rejected, the student can immediately write :

θ - 25 = 7(5/7)^t
θ = 7(5/7)^t + 25

But if he/she did not realise this and just substituted the values back into the modulus, the negative case will appear again (and will still need to be rejected)


Update : using the A = ±eᶜ technique can allow one to show the required result per se (effectively circumventing any rejection of sign), but it also means that the student did not demonstrate understanding that eᶜ > 0 for all real c and the rejection of 7 = -eᶜ

Strictly speaking, the constant A can't have two signs for this case since the negative is not possible for real c, t and k in the context of the question.

Alternatively with your approach,

|θ - 25| = eᵏᵗ⁺ᶜ

θ - 25 = ±eᵏᵗ⁺ᶜ

θ - 25 = Bᵗ · eᶜ , where B = ±eᵏ

Sub θ = 32 and t = 0,

32 - 25 = eᶜ
eᶜ = 7

Sub θ = 30 and t = 1 ,

30 - 25 = 7B
7B = 5
B = 5/7

Then θ - 25 = 7(5/7)ᵗ
θ = 7(5/7)ᵗ + 25

But this ignores :

5/7 = ±eᵏ
eᵏ = ± 5/7 (negative is rejected for all real k)


Should a question ask to solve the values of k and c, rejection is still needed.

It is better to write out the rejection (who knows, marks deduction may ensue for 'lack of understanding')

Lastly, this could all have been avoided since the modulus is unnecessary for this question (if the context was correctly understood)

In this case we are actually solving for a “particular” solution, rather than general solution, of a differential equation. As such, there is no need to reject the “negative exponential case”.

And what if you were asked to solve for c and k? You'll still need to reject the negative.

Besides, if solving for a general solution :

θ = Aeᵏᵗ + 25 , where A = ±eᶜ

For a typical contextless question, this is alright.

But in the context of this question, the negative is clearly not possible for real c, k and t (the body temperature cannot go lower than the system it is in) so you'll still need to reject it.

Or, simply not use modulus in the first place.

(Be it for particular or general solution. Not rejecting the negative will get you the required expression, but ignores the context and logic.

For a 8m question, I'm pretty sure some demonstration of that is required. Such questions are usually worth only 5-6m)