# differential equation

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#1
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!!
Attachment 770670
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2 years ago
#2
(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!!
Attachment 770670
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.
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#3
(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
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2 years ago
#4
(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.
.
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#5
(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.
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2 years ago
#6
(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?
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#7
(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
0
2 years ago
#8
(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
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#9
(Original post by mqb2766)
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
0
2 years ago
#10
(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
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#11
(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

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
0
2 years ago
#12
(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?
0
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