Math modelling diffrential equation question?

[Eveflutter]

So there's this question I've to do. I got through until a certain point and now I'm stuck. >->

1&2The question:
The rate of deterioration of a product in a container is proportional to the amount of product present. At time t, the amount of product is x.
(i) State the diffrential equation relating x and t and solve the general solution to show that x=Ae^-kt where A is an arbitrary contant and k is the contant of proportionality.
(ii) Before t=0, no product was present in the container. At t=0, P amount of the product was added to the container. When t=1, the amount of product in the container was Pq, where q is a constant such that 0<q<1. Show that x=Pq^t.
(iii) When t=1 and again when t=2, another P amount of x was added to the container. Show that the amount of product in the container immediately after t=2 was P(q²+q+1)

3I got through with the first two parts:
(i) dx/dt=-kx
-> x=e^-kt+c
-> x=Ae^-kt , A=e^c
(ii)when t=0, x=P
-> A=P
-> x= Pe^-kt
when t=1, x=Pq
-> Pq=Pe^-kt
-> q=e^-kt
-> x=Pq^t

I'm stuck at part (iii)
I tried reasoning it out but I don't know. Would the equation x=Pq^t change since the intials are now changing twice (at t=1 and t=2)? But if so, how would I get the new ones from tha information?

Alright so as t approahes 1, x tends to Pq. At t=1 the amount of product changes to Pq+P. Then as t approaches 2, x deteriorates from Pq+P. But by how much?? Let's say it deteriorates till an amount B? Now at t=2, the amount of product goes to B+P.
I guess the P part of the required result (P(q²+q+1)) is that P in B+P. How do i get B?

I would really appreciate any help on this question, thank you!

 

[Dick]

So there's this question I've to do. I got through until a certain point and now I'm stuck. >->

1&2The question:
The rate of deterioration of a product in a container is proportional to the amount of product present. At time t, the amount of product is x.
(i) State the diffrential equation relating x and t and solve the general solution to show that x=Ae^-kt where A is an arbitrary contant and k is the contant of proportionality.
(ii) Before t=0, no product was present in the container. At t=0, P amount of the product was added to the container. When t=1, the amount of product in the container was Pq, where q is a constant such that 0<q<1. Show that x=Pq^t.
(iii) When t=1 and again when t=2, another P amount of x was added to the container. Show that the amount of product in the container immediately after t=2 was P(q²+q+1)

3I got through with the first two parts:
(i) dx/dt=-kx
-> x=e^-kt+c
-> x=Ae^-kt , A=e^c
(ii)when t=0, x=P
-> A=P
-> x= Pe^-kt
when t=1, x=Pq
-> Pq=Pe^-kt
-> q=e^-kt
-> x=Pq^t

I'm stuck at part (iii)
I tried reasoning it out but I don't know. Would the equation x=Pq^t change since the intials are now changing twice (at t=1 and t=2)? But if so, how would I get the new ones from tha information?

Alright so as t approahes 1, x tends to Pq. At t=1 the amount of product changes to Pq+P. Then as t approaches 2, x deteriorates from Pq+P. But by how much?? Let's say it deteriorates till an amount B? Now at t=2, the amount of product goes to B+P.
I guess the P part of the required result (P(q²+q+1)) is that P in B+P. How do i get B?

I would really appreciate any help on this question, thank you!

Think of it this way. The amount $P$ at $t=0$ decreases according to $Pq^t$. The amount $P$ added at $t=1$ decreases like $Pq^{t-1}$ (it's just 1 second delayed from the first). Then at $t=2$ you add $Pq^{t-2}$. So the total after $t=2$ is $Pq^t+Pq^{t-1}+Pq^{t-2}$. Correct?

 

[Eveflutter]

Think of it this way. The amount $P$ at $t=0$ decreases according to $Pq^t$. The amount $P$ added at $t=1$ decreases like $Pq^{t-1}$ (it's just 1 second delayed from the first). Then at $t=2$ you add $Pq^{t-2}$. So the total after $t=2$ is $Pq^t+Pq^{t-1}+Pq^{t-2}$. Correct?

Ooooh! I understood what you were doing but not why you were doing it but it took me a couple minutes to completely get it. Thank you so much! I really appreicate your answer :)