# differential equation

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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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#2

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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

**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

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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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.

**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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#4

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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

**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

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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(Original post by

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.

.

**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.

.

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#6

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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.

**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.

Also are you using minutes or hours consistently, i.e. 1.5 hrs = 90 min?

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(Original post by

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?

**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?

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#8

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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

**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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(Original post by

Would you mind putting your working up, it sounds about right.

**mqb2766**)Would you mind putting your working up, it sounds about right.

which after integration gives: Noe^kt=Nt

part b :

5ekt=25

k=ln5/30 =0.054

hence 5e^(0.054 x 90) = 625

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#10

(Original post by

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

**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

0 5

30 25

60 125

90 625

120 3025

What about the modified differential equation, can you show your working.

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(Original post by

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.

**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

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#12

(Original post by

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

**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

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