Hard differential equation problem?

...

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.

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?

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.

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

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

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.

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

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