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differential equation

Hi, I need help on this differential equation!!!! I don't understand how to do part c of coming up with a new differential equation by combining the previous and second differential equation (bacteria multiplying + bacteria being killed) Please help!!
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(edited 5 years ago)
Reply 1
Original post by faaaaa
Hi, I need help on this differential equation!!!! I don't understand how to do part c of coming up with a new differential equation by combining the previous and second differential equation (bacteria multiplying + bacteria being killed) Please help!!
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Can you show your solution to parts a) & b) pls?
You'd have to create a "new" differential equation which kicks in at the intiial time of 1.5 hrs using the solution to b) at 1.5 hrs as its iniital condition.
The new differential equation would be a) with a term which models the bacteria being killed.
Reply 2
Original post by mqb2766
Can you show your solution to parts a) & b) pls?
You'd have to create a "new" differential equation which kicks in at the intiial time of 1.5 hrs using the solution to b) at 1.5 hrs as its iniital condition.
The new differential equation would be a) with a term which models the bacteria being killed.



Hi, thank you for having a look!! i would be so grateful if you could help me

Yes, so far part a I got:
dN/dt = Nk
which after integration gives: Noe^kt=Nt
part b :
5ekt=25
k=ln5/30 =0.054
hence using the equation plugging in ln5/30 as k, I got 197.5 minutes for when Nt is 200,000

c)

new differential equation dN/dt = Nk -2500
integral [1/(Nk - 2500)] = 1 x dt
which after integrating gives: ln(0.054N- 2500)= t + c

but I can't get c because the values for N that I got from the previous differential equation at t=1.5 is 625.

so ln(0.054 x 625 - 2500) can't work as you can't calculate ln of a negative number and so I am stuck
Reply 3
Original post by faaaaa
Hi,

Yes, so far part a I got:
dN/dt = Nk
which after integration gives: Noe^kt=Nt
part b :
5ekt=25
k=ln5/30 =0.054
hence using the equation plugging in ln5/30 as k, I got 197.5 minutes for when Nt is 200,000

c)

new differential equation dN/dt = Nk -2500
integral [1/(Nk - 2500)] = 1 x dt
which after integrating gives: ln(0.054N- 2500)= t + c

but I can't get c because the values for N that I got from the previous differential equation at t=1.5 is 625.

so ln(0.054 x 625 - 2500) can't work as you can't calculate ln of a negative number and so I am stuck


Ok, you get the initial values for the new differential equation at t = 1.5 hrs from the simple equation in b).
The new differential equation which you plug them into is similar to what you give
dN/dt = Nk -2500
There are a few ways you could solve this (not sure what you've covered), but perhaps the easiest is to transform the variables
M = N -2500/k
then
dM/dt = Mk
solve as before and sub for M ...

Then use the initial values at 1.5 hrs. Note that in the new differential equation, 1.5 hrs is equivalent to t = 0.
.
(edited 5 years ago)
Reply 4
Original post by mqb2766
Ok, you get the initial values for the new differential equation at t = 1.5 hrs from the simple equation in b).
The new differential equation which you plug them into is similar to what you give
dN/dt = Nk -2500
There are a few ways you could solve this (not sure what you've covered), but perhaps the easiest is to transform the variables
M = N -2500/k
then
dM/dt = Mk
solve as before and sub for M ...

Then use the initial values at 1.5 hrs. Note that in the new differential equation, 1.5 hrs is equivalent to t = 0.
.

Hey, I had a couple of tries doing what you said but I still end up with the problem where I get ln( negative number) which is not possible to work out.
Reply 5
Original post by faaaaa
Hey, I had a couple of tries doing what you said but I still end up with the problem where I get ln( negative number) which is not possible to work out.


Are you using a definite integral where the initial conditions are (90 , 625)?
Also are you using minutes or hours consistently, i.e. 1.5 hrs = 90 min?
(edited 5 years ago)
Reply 6
Original post by mqb2766
Are you using a definite integral where the initial conditions are (90 , 625)?
Also are you using minutes or hours consistently, i.e. 1.5 hrs = 90 min?


Yes I used 90 minutes in the differential equation, and it came out with 625, which I used as t=0 for the new differential equation :frown:
Reply 7
Original post by faaaaa
Yes I used 90 minutes in the differential equation, and it came out with 625, which I used as t=0 for the new differential equation :frown:


Would you mind putting your working up, it sounds about right.
Reply 8
Original post by mqb2766
Would you mind putting your working up, it sounds about right.


dN/dt = Nk
which after integration gives: Noe^kt=Nt
part b :
5ekt=25
k=ln5/30 =0.054

hence 5e^(0.054 x 90) = 625
Reply 9
Original post by faaaaa
dN/dt = Nk
which after integration gives: Noe^kt=Nt
part b :
5ekt=25
k=ln5/30 =0.054

hence 5e^(0.054 x 90) = 625


That is obviously correct as the population increases by a factor of 5 every 30 mins. So the time goes
0 5
30 25
60 125
90 625
120 3025
What about the modified differential equation, can you show your working.
Reply 10
Original post by mqb2766
That is obviously correct as the population increases by a factor of 5 every 30 mins. So the time goes
0 5
30 25
60 125
90 625
120 3025
What about the modified differential equation, can you show your working.




new differential equation dN/dt = Nk -2500
integral [1/(Nk - 2500)] = 1 x dt
which after integrating gives: ln(0.054N- 2500)= t + c

but I can't get c because the values for N that I got from the previous differential equation at t=1.5 is 625.

so ln(0.054 x 625 - 2500) can't work as you can't calculate ln of a negative number and so I am stuck
Original post by faaaaa
new differential equation dN/dt = Nk -2500
integral [1/(Nk - 2500)] = 1 x dt
which after integrating gives: ln(0.054N- 2500)= t + c

but I can't get c because the values for N that I got from the previous differential equation at t=1.5 is 625.

so ln(0.054 x 625 - 2500) can't work as you can't calculate ln of a negative number and so I am stuck


2500 is in hours, it needs to be in minutes?

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