The Road to Exams [Week 20, Differential equations]

Added: 2026-05-25

[00:00:01]Okay, we are on week 20 of 24. They are getting really close right now. We've got this question that is a differential equation, bit of partial fractions and then a bit of modeling stuff to do with this as well. Okay, so we're going to write this in partial fraction form that we have got here. Pretty easy to begin with. So that's 1 over 1 over h - 5 and then we have h + 3. Is that right? Yes, it is. So, we know this is going to go to an a over hus 5 plus a b over h + 3.

[00:00:34]I wonder why they're asking us to do partial fractions. I'm sure you've got an idea. It's probably going to be useful somehow in the next parts of the question. And there's lots of different ways of doing this. I'm probably just going to do substitution. I think substitution seems to be the easiest here. So, if I let h be equal to 5, I get that one, excuse me, h is equal to 5. 1 is equal to 8 a. And so a is 1/8.

[00:00:57]And if I let h= to minus3, I get that 1 is = -8 b. And so b is minus an 8. And it's not enough just to do this. You do actually have to finish off and write it out. So this means that 1 / h - 5h + 3 is equal to an 8 over h - 5 - an 8 over h + 3. Now they would accept it in this form here. This is the form I prefer it in for when I do the integration part.

[00:01:30]Bit of a spoiler there, but you might more commonly see it written like with the 8 that's up here going down into the denominator. So you have it as a 1 over 8 8 h - 5 - 1 over 8 h + 3. There's no need to expand the brackets, but they would accept it if they were expanded brackets as well. So it then says the depth of the water in a storage tank is being monitored and the depth of the water in the tank h meters is modeled by this differential equation. Know that there's a minus here and obviously it's related to what we've got earlier on. T is the time in days from when the monitoring began. And given that the initial depth of the water was 13. As soon as I see that, I think initial the depth of the water was 13. Now, this is a differential equation. This is one we're going to separate the variables and put everything to do with h onto the left hand side. And then everything to do with the t, which is going to then be on the right hand side instead. Why is it not letting me change color? There we go. So, there's going to be the minus one over 40 that remains there. So to get this H minus 5 H + 3 to the other side, I can only get it over there as a division. So what we have I'm definitely going to need another page. Might as well add it in. What we have for part B, it would be a 1 / H - 5 H + 3, and then we have the DH DT, but it was a capital DH dt, which would mean if we've divided by that part that's there, we're left with min -1 over 40. Now, sometimes people want to put the 40 over here as well. The reason I didn't put the 40 up to the top was because I just want to use exactly what it looks like on this side. If you do it with the 40, it might make a difference in the long run, but I just thought, let's make it look exactly like the thing that we've done the partial fractions on. So, that leaves us with the minus1 over 40. And what's actually happening here now that we've put the hes onto the left, we're going to put this dt to this side. And then we're just going to simply integrate both sides to solve this differential equation. Now, the whole point of doing partial fractions was so that we could then integrate this. You can't integrate it like this, but you can integrate it in the partial fraction form. And this is the one where I prefer the eth in the numerator. It just is a bit easier for us to process what needs to go on in this question. So, we've got these two parts with respect to h. Apologies, I've got a bit of a sniff here coming at the end of a cold.

[00:03:48]Okay, so let's actually do the integration. I think the left hand side is going to be pretty straightforward, right? It's going to be a 1/8 ln of h minus 5. Put the modulus signs around it for when you do an ln integration. The eth is just because that's the factor.

[00:04:03]Nothing else really needs to change because of the fact there's no coefficient on the h. And then we will have - 1/8 ln of h + 3. And obviously when we integrate this we just get - 1 over 40t and plus some constant. People often say why don't we need a constant on both sides? If you had a constant on both sides, you could subtract them and it would combine into one single constant. So now that we've got to that stage, what does it even want us to do?

[00:04:28]It wants us to make h the subject. But I'm going to need to use the information that when t is zero, h is 13. So when t is zero, h is 13. Let's sub then. So h is 13 here would be the ln an 8 ln of 13 - 5 which is 8 minus an 8 ln of 16.

[00:04:52]The t is a zero and we just get the c part here. So if I was going to combine these as to one single thing remember with a subtraction as long as you've got the same factor outside the front we can do it as a division. So it is 1/8 of the ln of 8 / 16 is equal to c. Well, I know that 8 over 16 is just a half like this.

[00:05:13]So, I have the 8 ln a half. Meaning, I've now got what the equation is, right? The equation is 1/8 ln of h - 5.

[00:05:22]I'm actually going to start combining this into one single thing because we've got them both with the 1/8, right? With the subtract means it's going to be a division. So I have an h - 5 / an h + 3 and that's equal to -1 over 40 t + c is 1/8 ln of a half. Okay. So let's try and make h the subject. I think it's going to be better to combine the ln terms to begin with. So I have 1/8 ln of h - 5 / h + 3us 18 ln of a half. Don't really need those brackets but I've got them now anyway.

[00:06:03]Now when you have this you're subtracting two logarithms which means you're doing this divided by a half.

[00:06:10]When you divide by a half think of what dividing by a half is the same as it's the same as multiplying by two. So actually this thing is going to get multiplied by two because of the subtracting with the half. So it's going to be the 8th ln of 2 h - 5 / h + 3 is equal to -1 over 40 t. Now what I think I'm going to do now is get rid of this eth. So I'm going to multiply both sides by 8. Yes, I am going to do minus 1 over 40 times by 8 on my calculator because I feel like it it is -0.2 which is actually quite a good thing because we end up with that -0.2.

[00:06:55]I think that's actually featured in the questionus0.2 t. Is that what we've got? Yeah, look -0.2t. So that's looking good. We now just need to make h the subject of this.

[00:07:06]And this is where people can find this quite overwhelming because there's just so much manipulating that needs to be done. So the ln that I've got, I'm going to get rid of that and I'm going to change it to an e onto the right hand side. So I get two lots of h + 5 h + 3 is equal to e to the -0.2t.

[00:07:23]I'm trying to make h the subject here.

[00:07:26]So I get my 2h - 5 which is my 2h - 10 is equal to this thing multiplied by this. So that's e -0.2 h + 3. So 2 h - 10 is h e -0.2 t + 3 e -0.2t. Remember the goal is you are trying to find out what h is. So I'm going to collect this h onto this side.

[00:07:53]So I have 2 h - h e -0.2t. And I'll put the 10 onto the other side. So I have 10 + 3 eus 0.2t.

[00:08:04]I will factoriize like this.

[00:08:08]And then I think we've nearly got what they wanted, but I'm going to double check and see have we got it written exactly as they want. So they've got it as a 10 plus this. It's going to be divided by this. Yeah, they want it 10 + 3 / that, which is what we've got. So h is equal to 10 + 3 e -0.2t / 2 - e -0.2t. Quite a lot of algebraic manipulation. Wow, look how much algebraic manipulation there is. loads and loads and loads for this. Okay. It then says, hence find the time taken for the depth of the tank, excuse me, the depth of the water in the tank to fall to 8 m. In other words, they've now told us that h is equal to 8 and we want to find out what the time is. Now, if you couldn't do part B or even part A, you could still get these marks for part C and part D because you can just solve this equation where h is equal to 8. So, we're going to continue with this for part I think it's part C. So we get 8 is equal to 10 + 3 e -0.2t / 2 - e -0.2t.

[00:09:14]I'll multiply this by this. So I'll have the 8 * 2 which is a 16 and a - 8 e -0.2t is equal to 10 + 3 e -0.2t.

[00:09:27]So I will add this 8 onto this side to get 11 eus 0.2t.

[00:09:35]and I will do the 16 - the 10 which is 6. So 6 / 11 is eus 0.2t.

[00:09:44]I will then ln both sides like this. So when you do the ln of the right hand side you get that. So my final thing is to divide it by -0.2 and then put that on the calculator. So the ln of 6 over 11 / -0.2 two and you get that t is 3.03.

[00:10:06]Was it hours or days? It is days and it wants it doesn't say what it wants it to. So do three significant figures. So this is 3.03 days to three significant figures. Now I can probably fit part D of the question in here as well. Let's see what D says. According to the model, the depth of the water will eventually fall to K meters. State the value of the constant K. state just means you can just give an answer straight up really.

[00:10:35]So this is saying what's going to happen to it as it eventually gets to a particular level as t becomes a very very big value. Well the key thing to notice here is that when t becomes a very big value if you have e the minus0.2t think about what that becomes e to theus0.2 to is a graph an exponential graph an exponential decay graph that looks like this. And as t gets very very big this line from the graph goes towards 0. So this means that h is going to go towards 10 + 3 * 0 over 2 - 0. I'm just saying this part's going to become zero and this part is going to become zero. In other words, the top part is 10, the bottom part is two. So it is going to go to five. In other words, K is going to go to five. It will go and fall to five meters, but according to the model, it won't be able to go any lower than this. So, we knew that this part was well, we didn't know this part was correct. Notice the different ways you can present it. I probably would most commonly accept it like this, but you could have it like this, too. That was the show that kind of thing, which we knew we could do. We got the 3.03 days, and we got the K was equal to 5.

[00:11:44]So for next week you have this question that is here obviously a probability question and I hope all of your revision is really going well. Um if you check out the link in the description um you can see what kind of events and stuff I've got going on. I've got an event that's going on between the first and second pure paper that you might find helpful. See you in another one soon guys. by