Yes, it's nonlinear, but separable. Separate and integrate:
$$\frac{dy}{y(N-y)}=kdt$$
$$\int \frac{dy}{y(N-y)}= k \int dt $$
$$\ln(y)-\ln(y-N) = Nkt + C. $$
Use the two conditions given, namely $y(0)=N/10$ and $y(1)=N/5$ to determine $k$ and $C$.
Hope this helps some,
Cheers,
Paul Safier
Since the population increases at a rate proportional to the number of people present at time t (read: the rate of change of $P$ is proportional to $P$.)
$$\frac{\mathrm{d}P}{\mathrm{d}t} = kP$$
Separating the variables and integrating yields
$$\int \frac{1}{P} \, \mathrm{d}P = \int k \, \mathrm{d}t$$
So that we get
$$\ln P = kt + c$$ or equivalently, letting the initial population be $P_0$ so that when $t=0$, $P= P_0$ to find the arbitrary constant and re-arranging yields
$$P = P_0e^{kt}$$
We are given, that the population doubles ($2P_0$) in five years ($t=5$) so:
$$2P_0 = P_0 e^{5k}$$
Solving for $k$ yields
$$e^{5k} = 2 \implies k = \frac{\ln 2}{5}$$
Our population growth equation is then
$$P = P_0 e^{\frac{\ln 2}{5}t}$$
So for the population to triple, we need to solve $3P_0 = P_0e^{kt}$ for $t$, where $k$ is what we found above. This yields
$$e^{kt} = 3 \implies kt = \ln 3 \implies t = \frac{5\ln 3}{\ln 2}$$
The same can be done to find the time it takes for the population to quadruple, that is solve $4P_0 = P_0e^{kt}$ for t.
For the second question, is we have that the population after three years ($t=3$) is $P = 10000$
then substituting this into our population growth equation gives
$$10000 = P_0e^{3k}$$
so that we solve for $P_0$ to get
$$P_0 = \frac{10000}{e^{3k}} = \frac{10000}{\exp{\left(\frac{3\ln 2}{5}\right)}}$$
Best Answer
This is a typical differential equation problem. It leads to what is known as a "logistics model".
Let $N(t)$ denote the number of infected people at time $t$ after the first infection, with $t$ measured in days. The rate at which the disease spreads is the rate of change of $N(t)$, which is the derivative $N'(t)$. We are told that the rate of change is proportional to $N(t)$ (the number of people who already have the disease) times $10000-N(t)$ (the number of people who don't have the disease). That is: $$N'(t) = kN(t)(10000-N(t))$$ for some constant $k$.
We are also told that $N(0) = 1$ (one person gets sick) and $N(5)=50$. We are asked for the value of $N(10)$.
Note that we cannot have $N(t)=0$ for all $t$, nor can we have $N(t)=10000$ for all $t$.
This differential equation is of a kind called "separable". We can solve it as follows: $$\begin{align*} \frac{dN}{dt} &= kN(10000-N)\\ \frac{dN}{N(10000-N)} &= k\,dt\\ \int\frac{dN}{N(10000-N)} &= \int k\,dt\\ \int\frac{dN}{N(10000-N)} &= kt + C. \end{align*}$$ for some constant $C$. To solve the integral on the left, we can use Partial Fractions: $$\frac{1}{N(10000-N)} = \frac{\quad\frac{1}{10000}\quad}{N} + \frac{\quad\frac{1}{10000}\quad}{10000-N}$$ so $$\begin{align*} \int\frac{dN}{N(10000-N)} &= \frac{1}{10000}\int\left(\frac{1}{N} + \frac{1}{10000-N}\right)\,dN\\ &= \frac{1}{10000}\left(\ln|N| - \ln|10000-N|\right) + D. \end{align*}$$ Putting it all together, we get $$\frac{1}{10000}\left(\ln |N| - \ln|10000-N|\right) = kt + E,$$ where $E$ is some constant. Rewriting, we have: $$\ln\left|\frac{N}{10000-N}\right| = Kt + F,$$ where $K$ and $F$ are constants. Since $N$ is always between $1$ and $10000$, we can drop the absolute values.
Plugging in $t=0$, we know that $N=1$, so we have $\ln\frac{1}{9999} = F$. Plugging in $N=5$, we know that $N(t)=50$, so we have $$\ln\frac{50}{9950} = 50K + \ln\frac{1}{9999}.$$
From here, we can plug in $t=10$ and solve for $N$.