Question

A population of ants, initially containing 20 ants, is growing according to the population growth equation \(P'(t) =0.5P(t)(1-\dfrac{P(t)}{N})\) with time t measured in days.
1) If the carrying capacity of the area in which they live is 1000, how many ants will there be one week later?
2) If bug spray is used, and it decreases the intrinsic growth rate r from 0.5 to 0.1, how many ants will you have one week later? Assume the carrying capacity remains unchanged at 1000 ants.
Please, help

Answer 1

@ikram002p HI, friend. For some reasons, I got the wrong answer but don't know how to fix. Help me, please.

Answer 2

and the most frustrating thing is it is just the first order of ODE, no tricky. It takes me down !!!

Answer 3

Screenshot the question for me

Answer 4

Why? it is what I type.

Answer 5

ok, I will.

Answer 6

oh, you mean my work?

Answer 7

No the question because this "\(P'(t) =0.5P(t)(1-\dfrac{P(t)}{N})\) " is not clear

Answer 8

Oh never mind sorry

Answer 9

Ok, I show you what I got.

Answer 10

Ok

Answer 11

\(\dfrac{dP}{dt}=0.5P\dfrac{N-P}{N}\)

\(\dfrac{dP}{P(N-P)}=\dfrac{0.5}{N}dt\)
integral both sides

Answer 12

Decompose \(\dfrac{1}{P(N-P)}=\dfrac{1}{P}+\dfrac{1}{N-P}\)

Hence \(ln|\dfrac{P}{N-P}|= \dfrac{0.5t}{N}+C\)

Answer 13

That gives us \(\dfrac{P}{N-P}=Ae^{0.5t/N}\)
Is there any mistake?

Answer 14

Now, using initial value P(0) =20 to find A, with N =1000
\(\dfrac{20}{1000-20}= A = 0.0204\)

Answer 15

Hence general solution is \(P = 0.00204e^{0.00005t}(1000-P)\)
Hence \(P=\dfrac{0.00204e^{0.00005t}}{1+0.00204e^{0.00005t}}\)
But if there is no mistake , then P(7) = 0.00203
which is non sense.

Answer 16

oh, I forgot *1000 but even though , P(7) = 2.03 is nonsense till. :(

Answer 17

I think your partial fraction decomposition is incorrect

Answer 18

\[\frac{1}{P(N-P)}=\frac{a}{P}+\frac{b}{N-P} \\ 1=a(N-P)+bP \\ 1=aN+P(-a+b) \\ \implies 1=a N \text{ and } -a+b=0\]

Answer 19

\[\frac{1}{P(N-P)}=\frac{\frac{1}{N}}{P}+\frac{\frac{1}{N}}{N-P}\]

Answer 20

N is carrying capacity?

Answer 21

I'm getting a answer for P(7) in the 4 hundreds

Answer 22

\[P=\frac{1000 e^{.5 t}}{49+e^{.5t }}\]
is what I have for P if N is 1000

Answer 23

oh and it looks pretty like this because I left a in the prettier form

Answer 24

fraction form that is (not approximated)

Answer 25

Ok, let me redo.
\(\dfrac{dP}{dt}= 0.5P(1-\dfrac{P}{N})\)

\(\dfrac{dP}{P(1-\dfrac{P}{N})}=0.5dt\)
Now decompose the LHS
\(\dfrac{a}{P}+\dfrac{b}{1-P/N}=1\)
that gives us
\((1-P/N) a+bP=1\)
1) if P =0, we have a=1
2) if P =N , we have bN =1or b =1/N
replace all
\(\int \dfrac{1}{P}dP +\int\dfrac{1/N}{1-P/N} dP\)
simplify more for the second term, we have
\[\int \dfrac{dP}{P}-\int\dfrac{dP}{P-N}= ln|\dfrac{P}{P-N}|= 0.5t +C\]

Answer 26

That gives \(\dfrac{P}{P-N}=Ae^{0.5t}\)
If replace P(0) =20, we get \(\dfrac{20}{20-1000}=A = -0.0204\)

Answer 27

And then solve for P. I got
\(P=Ae^{0.5t}(P-N)\rightarrow P=\dfrac{-ANe^{0.5t}}{1-Ae^{0.5t}}\)

Answer 28

\[P'=\frac{1}{2}P(\frac{N-P}{N}) \\ \frac{P'}{P(N-P)}=\frac{1}{2} \frac{1}{N} \\ \int\limits \frac{dP}{P(N-P)} dP = \int\limits \frac{1}{2} \frac{1}{N} dt \\ \text{ \partial fraction time!} \\ \frac{1}{P(N-P)}= \frac{a}{P}+\frac{b}{N-P} \\ 1=a(N-P)+b(P) \\ 1=aN+P(-a+b) \\ -a+b=0 \implies a=b \\ 1=a N \implies a=b=\frac{1}{N} \\ \\ \text{ so we have after the integration part } \\ \frac{1}{N}(\ln(P)-\ln(N-P))=\frac{1}{2N} t +C \\ \ln(\frac{P}{N-P})=\frac{1}{2}t +k \\ \frac{P}{N-P}=a e^{\frac{1}{2} t} \\ P=(N-P)a e^{\frac{1}{2}t} \\ P=N ae^{\frac{1}{2}t}-Pae^{\frac{1}{2}t} \\ P(1+ae ^{\frac{1}{2}t})=Na e^{\frac{1}{2}t} \\ P=\frac{N a e^{\frac{1}{2}t}}{1+ae^{\frac{1}{2}t} } \text{ where I have } A=\frac{1}{49} \\ P=\frac{1000 \cdot \frac{1}{49} e^{\frac{1}{2}t}}{1+\frac{1}{49} e^{\frac{1}{2}t}}\]
so yeah your P looks equivalent to mine

Answer 29

because when you plug in your values you get that same P

Answer 30

and I just multiplied numerator and denominator by 49 to make it look prettier

Answer 31

\[P=\frac{1000 e^{\frac{1}{2}t}}{49+e^{\frac{1}{2}t}}\]

Answer 32

so you should give P(7) equals some number in the 4 hundreds now

Answer 33

If a= b= 1/N, then plug in we have
\[\int \dfrac{1/N}{P}+\dfrac{1/N}{N-P}=1/N \int \dfrac{1}{P}-\dfrac{1}{P-N}=1/Nln|\dfrac{P}{P-N}|\]
You see, my logic gives me the denominator is P-N while yours is N-P. It makes our solutions different.

Answer 34

eventhoug |P-N|=|N-P| but I want to make it clear when we take off the absolute value.

Answer 35

?

Answer 36

\[\text{ you got } A=\frac{-1}{49} \\ \text{ and your } P \text{ was } \\ P=\frac{-ANe^{\frac{1}{2}t}}{1-Ae^{\frac{1}{2}t}} \\ \\ \text{ also we have } N=1000 \text{ plug \in we have } \\ P=\frac{-(\frac{-1}{49}) (1000)e^{\frac{1}{2}t}}{1-\frac{-1}{49} e^{\frac{1}{2}t}} =\frac{\frac{1000}{49}e^{\frac{1}{2}t}}{1+\frac{1}{49}e^{\frac{1}{2}t}} =\frac{1000e^{\frac{1}{2}t}}{49+e^{\frac{1}{2}t}}\]
this is exactly the same I have abov

Answer 37

our solutions are exactly the same

Answer 38

So I'm not really sure what you are talking about