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exponentials maths question help!

I'm not sure where to start with this question :/
Screenshot 2021-02-21 at 13.02.17.png189.5KB

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Reply 1
Avatar for mqb2766
mqb2766
21
Original post by dnejfn
I'm not sure where to start with this question :/


Shift the -P to the left hand side and solve/integrate the first order ode.
What have you covered about solving them?
It should be a straightforward case/solution.
(edited 3 years ago)
Reply 2
Avatar for mqb2766
mqb2766
21
Original post by dnejfn
dP/dt + P = 100
so integrate 100? that's 100t

Do you know how to solve first order odes?
Original post by mqb2766
Shift the -P to the left hand side and solve/integrate the first order ode.
What have you covered about solving them?
It should be a straightforward case/solution.

dP/dt + P = 100

Integrate 100? that's 100t

I'm not sure if I have actually covered this in the lesson
https://www.mathsisfun.com/calculus/differential-equations-first-order-linear.html
I found this, is this what you mean by integrating the first order ode?
(edited 3 years ago)
Reply 4
Avatar for mqb2766
mqb2766
21
Original post by dnejfn
dP/dt + P = 100

Integrate 100? that's 100t

I'm not sure if I have actually covered this in the lesson
https://www.mathsisfun.com/calculus/differential-equations-first-order-linear.html
I found this, is this what you mean by integrating the first order ode?

I'm presuming you have a textbook. But if you've not covered it, why try questions on it?
Original post by mqb2766
I'm presuming you have a textbook. But if you've not covered it, why try questions on it?

What's this sort of topic called? First Order Linear Differential Equations?
Original post by mqb2766
Shift the -P to the left hand side and solve/integrate the first order ode.
What have you covered about solving them?
It should be a straightforward case/solution.

Would have thought this was more appropriately a "separation of variables" problem (which at least in my time, was covered in the normal A-level, while "dP/dt + kP = C" type equations would only be in FM).
Original post by DFranklin
Would have thought this was more appropriately a "separation of variables" problem (which at least in my time, was covered in the normal A-level, while "dP/dt + kP = C" type equations would only be in FM).

I attempted it and got up to here. Put then the 100t cancelled out so think I've gone wrong somewhere
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Reply 8
Avatar for mqb2766
mqb2766
21
The answer is an exponential+ ... What have you covered "at school".
Original post by dnejfn
I attempted it and got up to here. Put then the 100t cancelled out so think I've gone wrong somewhere

So, what I'm saying is to solve dPdt=(100−P)\dfrac{dP}{dt} = (100-P) as a separable differential equation. Do not move the P over (i.e. don't add P to both sides). Solve it directly. If you don't know how to do that, say so.

But if you don't know how to solve a separable DE, and you don't know the method for solving a first order linear DE (which is what mqb2766 was leading you to), then you won't be able to solve this.

At which point "why are you trying to solve questions you haven't covered the material for?" definitely requires an answer.
(edited 3 years ago)
Original post by DFranklin
So, what I'm saying is to solve dPdt=(100−P)\dfrac{dP}{dt} = (100-P) as a separable differential equation. Do not move the P over (i.e. don't add P to both sides). Solve it directly. If you don't know how to do that, say so.

But if you don't know how to solve a separable DE, and you don't know the method for solving a first order linear DE (which is what mqb2766 was leading you to), then you won't be able to solve this.

At which point "why are you trying to solve questions you haven't covered the material for?" definitely requires an answer.

well, I've finished all the content for a level maths lol I just probably wasn't there when my class went over this. I'm gonna do some practise questions before doing this one then so then I can get the method right 👍
I got this but not sure if it's right
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No. Using separation of variables, divide by the right hand side

Int 1/(100-P) dP = Int 1 dt

Can you see that and now to proceed?
Original post by mqb2766
No. Using separation of variables, divide by the right hand side

Int 1/(100-P) dP = Int 1 dt

Can you see that and now to proceed?

do i then integrate both sides, then find c ?
Reply 14
Avatar for TSRTD
TSRTD
11
Original post by dnejfn
well, I've finished all the content for a level maths lol I just probably wasn't there when my class went over this. I'm gonna do some practise questions before doing this one then so then I can get the method right 👍

In A level mathematics, to solve a differential equation like this, the P term must be i) on the same side as the dP/dt and ii) must be multiplied by dP/dt. You should not be adding P to both sides as then the P term will not be being multiplied by dP/dt, so you cannot solve using any method learned in the standard A level course. Dividing by (100 - P), however, will give you the equation in a solvable form. Remember that when you take the integral of both sides, you take the integral of the whole of the side and not just one of the terms. Also remember that you must be integrating with respect to a variable - in this case to integrate both sides it would be ∫ dt - integrate with respect to t. Finally, can you spot why the ∫ dt becomes a ∫ dP on the left?
Original post by dnejfn
do i then integrate both sides, then find c ?

Yes. You'd expect an exponential solution, so logs may play a part. You could use the given initial conditions and treat it as a definite integral, or just do an indefinite integral, then find C using the same info.

Do you understand why/how you get it in that form?
(edited 3 years ago)
Original post by mqb2766
Yes. You'd expect an exponential solution, so logs may play a part. You could use the given initial conditions and treat it as a definite integral, or just do an indefinite integral, then find C using the same info.

Do you understand why/how you get it in that form?

i assume its one of the natural log rules but not sure which one. I have dP=(100-P)dt so have dt on the left and then you divide by 100-P but not sure where ln comes from
Original post by dnejfn
i assume its one of the natural log rules but not sure which one. I have dP=(100-P)dt so have dt on the left and then you divide by 100-P but not sure where ln comes from

You've changed the integral from the previous post :banghead:. Don't.

On the left, you want to integrate
1/(100-P)
It's roughly 1/P. Can you integrate that wrt P?

Do you understand why I divided by 100-P?


Spoiler

Original post by mqb2766
You've changed the integral from the previous post :banghead:. Don't.

On the left, you want to integrate
1/(100-P)
It's roughly 1/P. Can you integrate that wrt P?

Do you understand why I divided by 100-P?


Spoiler



I integrated 1/100-P and got -ln|100-P| + C. You want to get P on the left with dP so you can multiply them so can then integrate which is why you divide by 100-P
Original post by dnejfn
I integrated 1/100-P and got -ln|100-P| + C. You want to get P on the left with dP so you can multiply them so can then integrate which is why you divide by 100-P

Ok, so what is on the right when you integrate and how do you get P = ...

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