differential equations: Suppose dN/dt = 1.5N (1 - N/50)....

[refid]

Hello
theres this question:

Suppose N(t) denote the size of the population at time and that

$\displaystyle \frac{dN}{dt}=\ 1.5N ( 1 -\frac{N}{50})$

solve this differential equation when
$\displaystyle N(0) = 10$

Do just take the integral

$\displaystyle \frac{dN}{dt}=\ 1.5N -\frac{1.5N^2}{50}$

$\displaystyle \frac{dN}{dt}=\ 1.5(N -\frac{N^2}{50})$

\(\displaystyle \L\int {\frac{{dN}}{N -\frac{N^2}{50}}} = \L\int {1.5} dt\)

do approach this such shown above?

 

[Unco]

Keep it factorised so you can use partial fractions (with the cover-up rule, or however you're used to).

 

[refid]

I still dont understand what to do

\(\displaystyle \L\int {\frac{{dN}}{N -\frac{N^2}{50}}} = \L\int {1.5} dt\)

\(\displaystyle \L\int \frac{1}{N} -\frac{50}{N^2} = {1.5t} + C1\)

\(\displaystyle \L\ln(N) + \frac{50}{N} = {1.5t} + C1\)

\(\displaystyle \L N + \frac{50}{N} = e^{{1.5t} + C1}\)

\(\displaystyle \L N + \frac{50}{N} = Ce^{{1.5t}}\)

not sure if if my process is right

 

[pka]

Here is a hint using Unco post.
\(\displaystyle \L
\frac{{50}}{{50N - N^2 }} = \frac{1}{N} - \frac{1}{{N - 50}}\)

 

[soroban]

Re: differential equations: Suppose dN/dt = 1.5N (1 - N/50).

Hello, refid!

Suppose $\displaystyle N(t)$ denote the size of the population at time $\displaystyle t$and that:

\(\displaystyle \L\;\;\frac{dN}{dt}\:= \:1.5N\left(1\,-\,\frac{N}{50}\right)\)

Solve this differential equation when: $\displaystyle \,N(0) \: =\:10$

Get rid of the fractions and decimals as soon as possible . . .


We have: \(\displaystyle \L\,\frac{dN}{dt}\:=\:\frac{3}{2}N\left(1\,-\,\frac{N}{50}\right) \;= \;\frac{3}{2}\cdot N\left(\frac{50\,-\,N}{50}\right) \;=\;\frac{3}{100}N(50\,-\,N)\)


Multiply through by \(\displaystyle 100:\L\;\;100\frac{d\!N}{dt}\:=\:3N(50\,-\,N)\)

Separate variables: \(\displaystyle \L\,\frac{100}{N(50\,-\,N)}\,d\!N \;= \; 3\,dt\)

Partial fractions: \(\displaystyle \L\,\left(\frac{2}{N}\,+\,\frac{2}{50\,-\,N\right)dN\;=\;3\,dt\)

Can you finish it now?

 

[refid]

yes I can solve it im wondering a would need to factor -1 from the left side?

\(\displaystyle \L\,-2( -lnN +ln({50-N})) =\;3t +C1\)

\(\displaystyle \L\, -lnN +ln({50-N}) =\;-1.5t +C1\)


why is it
\(\displaystyle \L\, \frac{50-N}{N} =\;Ce^{-1.5t}\)

and not
\(\displaystyle \L\, \frac{N}{50-N} =\;Ce^{1.5t }\)