change in a coyote population I am having a problem with this calculus problem:

The rate of change of the number of coyotes $N(t)$ in a population is
  directly proportional to $650−N(t)$, where $t$ is the time in years. That
  is,
$dN/dt=k(650−N)$.
When $t=0$, the population is $270$, and when $t=2$, the population has
  increased to $550$. Find the population when $t=3.$ (Round your answer to
  the nearest whole number.)

can you please help walk me through this problem
 A: This can be integrated by multiplying both sides with a suitable integrating factor:
$$
650 k = k N + dN/dt \Rightarrow \\
650k e^{kt} = k e^{kt} N + e^{kt} dN/dt = (d/dt)(e^{kt} N) \Rightarrow \\
650 \int\limits_0^t k e^{k\tau} d\tau = 
650 \left[ e^{k\tau} \right]_{\tau=0}^{\tau=t} = 
650 \left( e^{kt} - 1 \right) =
e^{kt} N(t) - N(0) \Rightarrow \\
N(t) 
= e^{-kt} \left( N(0) + 650 \left( e^{kt}-1 \right) \right)
= e^{-kt} \left( N(0) - 650 \right)+ 650 
$$
Inserting $N(0) = 270$ and $N(2) = 550$ we get
$$
550 = e^{-2k}(270 - 650) + 650 \iff \\
e^{2k} = \frac{650-270}{650-550} = \frac{380}{100} \Rightarrow \\
k = \ln(3.8)/ 2 = 0.66750053336617\dotsb
$$ 
This gives $N(3) =598.7\dotsb \approx 599$.
A: The differential equation is 
$\frac{dN}{dt}=k\cdot (650−N)$
Separating the variables
$\frac{dN}{ (650−N)}=k\cdot dt$
Integrating both sides
$\int \frac{dN}{ (650−N(t))}=k\cdot\int dt$
$-ln(650-N(t))=k\cdot t+C$
$ln(650-N(t))=-k\cdot t+C$
$650-N(t)=C\cdot e^{-k\cdot t }$
$N(t)=650-C\cdot e^{-k\cdot t }$
$N(0)=270=650-C\Rightarrow C=380$
$N(2)=550=650-380\cdot e^{-k\cdot 2 }$
$100=380\cdot e^{-k\cdot 2 }$
$\ln(\frac{100}{380})=-2\cdot k$
$\ln(\frac{380}{100})=2\cdot k$
$k=\frac{ln(3.8)}{2}$
Thus $N(t)=650-380\cdot e^{-\frac{ln(3.8)}{2}\cdot t }\approx 650-380\cdot e^{-0.6675\cdot t }$
A: You have to solve the ODE (ordinary differential equation) for the unknown function $N(t)$
$$
\frac{dN}{dt}=k(w−N)
\\
N(0) = N_0
$$
The additional information $N(0) = N_0$ is called "initial condition": you are requiring that at the initial instant $t=0$ your function $N(t)$ assumes a certain given value $N_0$.
To solve this particular kind of ODE  you can use a trick (it is called "separable ODE"): write it as
$$
\frac{1}{ N(t)-w }\frac{dN }{dt}  =-k
$$
and integrate both sides in time from $t=0$ to a certain final time $t_f$. You should obtain
$$
\int_{0}^{t_f} \frac{1}{ N(t)-w }\frac{dN}{dt} dt \, = \, 
\int_{0}^{t_f}(-k) dt = -k\, t
$$
Now that we solved the LHS of the equation, we can work the RHS. We can change variable: instead of integrating in $dt$ we can integrate in $dN$ by using the usual formal trick
$$
\frac{dN}{dt} \, dt \, = \, dt 
$$
We also have to adjust the integration limits. The time integral extends from $t=0$ to $t=t_f$. Those conditions are equivalent:
$$
  t=0 \quad   \sim \quad   N=N(t=0)=N_0
\\
  t=t_f \quad   \sim \quad   N=N(t=t_f)
$$
Therefore, our integral reads
$$
\int_{0}^{t_f} \frac{1}{ N(t)-w }\frac{dN}{dt} dt \, = \,
\int_{N_0}^{N(t_f)} \frac{1}{ N -w } dN 
\, = \, \log(N(t_f) -w)- \log(N_0 -w)
$$
Finally,
$$
 \log( N(t_f) -w)- \log(N_0 -w) =  \log  \frac{N(t_f) -w}{N_0 -w}  =-k \, t_f
$$
meaning that
$$
 N(t_f) =(N_0 -w)e^{-k \, t_f} + w
$$
The final time $t_f$ can be any generic time, so the subscript $f$ is typically omitted. What is important is that we found the unknown function $N(t)$:
$$
 N(t) =(N_0 -w)e^{-k \, t} + w
$$
You can check that, for $t=0$ we have $e^{-k \, 0}=1$, so that  $N(t=0) = N_0 -w  + w=N_0$, as it should be.
Physical interpretation of the solution
Since we are modelling a population, we must require that  $N_0>0$. In fact, $N$ is the number of animals, and it can not be negative for any $t$. As we will see, $w$ is the asymptotic value of the population, so we have to require also $w>0$.
For positive $k>0$ the population goes to a constant value: $N(t) =(N_0 -w)e^{-|k| \, t} + w$, so that for $t\gg 1/|k|$ you have that $N(t)\approx w$ and its plot as a function of time is flat for large times. This gives an interpretation of $w$ as $N(t\rightarrow \infty) =w$.

*

*if $0<w<N_0$: the population $N(t)$  decreases exponentially in time from the initial value $N_0$ to the asymptotic value $w$ (i.e. the value reached in the limit $t \rightarrow \infty$).


*if $0<N_0<w$:  the population increases in time from the initial value $N_0$ to the asymptotic value $w$.
For negative $k$ the population explodes: $N(t) =(N_0 -w)e^{|k| \, t} + w$

*

*if $0<w <N_0$ the population grows always and exponentially diverges, $N\rightarrow  \infty$ for $t\rightarrow \infty$. In the real world, this is impossible: your model lacks some extra term in the ODE that damps this indefinite growth (even though it can be a good model to describe the early exponential growth of a population before the damping effect kicks in).


*if $0<N_0 <w$ the population goes negative and exponentially diverges, $N\rightarrow -\infty$. This is non-physical, meaning that this set of parameters is forbidden.
Numerical results
If you want to answer your question, you have to use the formula for $N(t)$ above with $N_0 = 270$, $w=650$. You do not have $k$. To find $k$ you have to solve this (according to what you wrote in the question)
$$
N(t=2)  =550 \quad \Rightarrow \quad (270 -650)e^{-k \, 2} + 650 =550 \, .
$$
