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Hard differential equation problem?

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APersonYo
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#1
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#1
In a revised model, it is assumed also that fish are removed from the pond, by anglers and natural wastage, at the constant rate of p per unit time so that:

dn/dt= kn-p

Given that k=2 p =100

and that there were initially 500 fish in the pond, solve the differential equation, expressing n in terms of t. Give a reason why this revised model is not satisfactory for large values of t.

answer... n=450 e^(2t) + 50

So far I have...

(1/kn-p ) dn= 1 dt

Intergrate both sides...

ln (kn-p)= t+c

And now Im stuck lol.
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Pangol
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#2
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#2
That's nearly right - you need a constant outside of you natural log term.

You now need to turn the statement "there were initially 500 fish in the pond" into something in terms of n and t so that you can find c. After that, it should just be algebra all the way.
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RDKGames
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#3
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(Original post by APersonYo)
...
Firstly; \displaystyle \int \frac{dn}{kn-p} \neq \ln\vert kn-p \lvert+A - you need to adjust the integral slightly.

Secondly, after you done that, you need to use the fact that there were 500 fish in the pond initially and solve for the constant of integration.
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Erasman
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#4
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#4
(Original post by APersonYo)
In a revised model, it is assumed also that fish are removed from the pond, by anglers and natural wastage, at the constant rate of p per unit time so that:

dn/dt= kn-p

Given that k=2 p =100

and that there were initially 500 fish in the pond, solve the differential equation, expressing n in terms of t. Give a reason why this revised model is not satisfactory for large values of t.

answer... n=450 e^(2t) + 50

So far I have...

(1/kn-p ) dn= 1 dt

Intergrate both sides...

ln (kn-p)= t+c

And now Im stuck lol.
It should be (1/k)* ln |kn-p| = t + ln c
which is kn - p = A*e^(kt)
or n = p/k + B*e^(kt)

then as n(t=0) = 500 it means that 500 = p/k + B so you can get B from that.

sorry if I'm wrong, I did it really quickly
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APersonYo
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#5
(Original post by RDKGames)
Firstly; \displaystyle \int \frac{dn}{kn-p} \neq \ln\vert kn-p \lvert+A - you need to adjust the integral slightly.

Secondly, after you done that, you need to use the fact that there were 500 fish in the pond initially and solve for the constant of integration.
1) Firstly, I didnt write that at all.

I wrote:

ln(kn-p) = t+ c

How am I going to use that with the information given?
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RDKGames
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#6
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#6
(Original post by APersonYo)
1) Firstly, I didnt write that at all.

I wrote:

ln(kn-p) = t+ c

How am I going to use that with the information given?
That's what your working implied.

To use the information, what is the variable represented by the amount of fish? What is the numerical value of time referred to as "initially"?

Then you have your initial conditions which you use to find the particular solution.
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APersonYo
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#7
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#7
(Original post by RDKGames)
That's what your working implied.

To use the information, what is the variable represented by the amount of fish? What is the numerical value of time referred to as "initially"?

Then you have your initial conditions which you use to find the particular solution.
This is what I've done afterwards...

ln(kn-p)= t+ lnA

as lnA is a constant

ln(kn-p)-ln A= t

ln (kn-p/A) = t

... Honestly. it just leads to a dead answer. Im totally lost
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APersonYo
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#8
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#8
(Original post by RDKGames)
That's what your working implied.

To use the information, what is the variable represented by the amount of fish? What is the numerical value of time referred to as "initially"?

Then you have your initial conditions which you use to find the particular solution.
This is what I've done afterwards...

ln(kn-p)= t+ lnA

as lnA is a constant

ln(kn-p)-ln A= t

ln (kn-p/A) = t


... Honestly. it just leads to a dead answer. Im totally lost
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RDKGames
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#9
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#9
(Original post by APersonYo)
This is what I've done afterwards...

ln(kn-p)= t+ lnA

as lnA is a constant

ln(kn-p)-ln A= t

ln (kn-p/A) = t

... Honestly. it just leads to a dead answer. Im totally lost
You should have \frac{1}{k}\ln\vert kn-p \vert = t+c \Rightarrow kn-p=Ae^{kt}

You're given that k=2 and p=100

So you have 2n-100=Ae^{2t}

Initially conditions give n(0)=500

Solve for A.

Write down the particular solution. Job done.
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APersonYo
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#10
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#10
(Original post by RDKGames)
You should have \frac{1}{k}\ln\vert kn-p \vert = t+c \Rightarrow kn-p=Ae^{kt}

You're given that k=2 and p=100

So you have 2n-100=Ae^{2t}

Initially conditions give n(0)=500

Solve for A.

Write down the particular solution. Job done.
How do you get 1/k ln(kn-p)

because when I initially did it I divided both sides by kn-p

so I would get (1/kn-p) dn/dt = 1

(1/(kn-p) dn = t dt

And so if you were to i eventually integrate both sides I would get ln(kn-p) = t +c
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RDKGames
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#11
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#11
(Original post by APersonYo)
How do you get 1/k ln(kn-p)

because when I initially did it I divided both sides by kn-p

so I would get (1/kn-p) dn/dt = 1

(1/(kn-p) dn = t dt

And so if you were to i eventually integrate both sides I would get ln(kn-p) = t +c
That is precisely why I said \displaystyle \int \frac{dn}{kn-p} \neq \ln\vert kn-p \lvert+A

It doesn't integrate to that because the numerator is not the derivative of the denominator with respect to n, hence why I said you need to adjust the integral slightly.

\displaystyle \int \frac{dn}{kn-p} = \frac{1}{k} \int \frac{k}{kn-p} .dn

Now you're allowed to deal with the integral in a way you know how.
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APersonYo
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#12
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#12
(Original post by APersonYo)
How do you get 1/k ln(kn-p)

because when I initially did it I divided both sides by kn-p

so I would get (1/kn-p) dn/dt = 1

(1/(kn-p) dn = t dt

And so if you were to i eventually integrate both sides I would get ln(kn-p) = t +c
its okay I understand its because if you intergrate 1/(ax+b) you get 1/a ln(ax+b)
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APersonYo
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#13
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#13
(Original post by RDKGames)
That is precisely why I said \displaystyle \int \frac{dn}{kn-p} \neq \ln\vert kn-p \lvert+A

It doesn't integrate to that because the numerator is not the derivative of the denominator with respect to n, hence why I said you need to adjust the integral slightly.

\displaystyle \int \frac{dn}{kn-p} = \frac{1}{k} \int \frac{k}{kn-p} .dn

Now you're allowed to deal with the integral in a way you know how.
Im sorry, Im finding this question severely complex.

So. I now understand why you would get:

1/k ln |kn-p|= t +c

If I plug in my values...

1/2 ln |100-100|= c

so ln 30 = c

Right so now if I do

ln|kn-p|= kt + kc

e^(kt+kc)= kn-p

therefore n= (e^(kt+kc) +p)/k

And if I plug in the values we are given... I still dont get the right answer
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RDKGames
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#14
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#14
(Original post by APersonYo)
Im sorry, Im finding this question severely complex.

So. I now understand why you would get:

1/k ln |kn-p|= t +c

If I plug in my values...

1/2 ln |100-100|= c
Firstly, you should be aware than \ln(0) is undefined so you shouldn't had been able to progress past this with any correct answer.

Secondly, pluggin the values in gives: \frac{1}{2}\ln \vert 2\cdot 500 - 100 \vert = c = \frac{1}{2}\ln(900)

Right so now if I do

ln|kn-p|= kt + kc

e^(kt+kc)= kn-p

therefore n= (e^(kt+kc) +p)/k

And if I plug in the values we are given... I still dont get the right answer
Yes that would be the expression for n but of not much use if you bury the constant of integration - the thing you want to find - that deep into the expression.

Anyway, plugging in the values nevertheless would give you 500=\frac{e^{2c}+100}{2} = \frac{A+100}{2}.


You can just skip this entirely if you just consider kn-p=Ae^{kt} in which case you only need to find A which is a whole lot quicker... I did this for you in the line above with A+100
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