How to estimate the parameters of a logistic differential equation from the values of its solution at times 0, 1 and 2? How do I solve this system of equations? I received these equations after letting Wolfram Alpha solve the logistic differential equation $$N'(t)=kN(t)(M-N(t)),\qquad N(0)=65,$$ that outputs:
$$N(t)=\frac{65Me^{kMt}}{65e^{kMt}+M-65}$$ 
I also knew that $N(1)=98$ and $N(2)=142$, which gave me this system of equations to solve for $k$ and $M$:
$$
98=\frac{65Me^{kM}}{65e^{kM}+M-65} \qquad
142=\frac{65Me^{2kM}}{65e^{2kM}+M-65}$$
I tried to solve it by hand and I also tried putting the system into Wolfram Alpha which timed out, both the free version and the Pro version.
If anyone has the digital tools or the knowledge of how to solve this by hand, I would be grateful.
 A: $$n'(t)=kn(t)\left(m-n(t)\right)\Longleftrightarrow$$
$$\frac{n'(t)}{n(t)\left(m-n(t)\right)}=k\Longleftrightarrow$$
$$\int\frac{n'(t)}{n(t)\left(m-n(t)\right)}\space\text{d}t=\int k\space\text{d}t\Longleftrightarrow$$
$$\frac{\ln\left|\frac{n(t)}{m\left(n(t)-m\right)}\right|}{m}=kt+\text{C}\Longleftrightarrow$$
$$\left|\frac{n(t)}{m\left(n(t)-m\right)}\right|=\text{C}e^{mkt}\Longleftrightarrow$$
$$\left|\frac{1}{m-\frac{m^2}{n(t)}}\right|=\text{C}e^{mkt}\Longleftrightarrow$$
$$\left|m-\frac{m^2}{n(t)}\right|=\frac{\text{C}}{e^{mkt}}\Longleftrightarrow$$
$$\left|1-\frac{m}{n(t)}\right|=\frac{\text{C}}{|m|e^{mkt}}$$
Now solve $\text{C}$ using $n(0)=65$:
$$\left|1-\frac{m}{65}\right|=\frac{\text{C}}{|m|e^{mk\cdot0}}\Longleftrightarrow\left|1-\frac{m}{65}\right|=\frac{\text{C}}{|m|}\Longleftrightarrow\text{C}=|m|\left|1-\frac{m}{65}\right|$$
So, we get:
$$\left|1-\frac{m}{n(t)}\right|=\frac{\left|1-\frac{m}{65}\right|}{e^{mkt}}$$
Now, when we know that $n(1)=98$ and $n(2)=142$ we get the system:
$$
\begin{cases}
\left|1-\frac{m}{98}\right|=\frac{\left|1-\frac{m}{65}\right|}{e^{mk}}\\
\left|1-\frac{m}{142}\right|=\frac{\left|1-\frac{m}{65}\right|}{e^{2mk}}
\end{cases}\Longleftrightarrow
\begin{cases}
k=\frac{\ln\left(\frac{\left|1-\frac{m}{65}\right|}{\left|1-\frac{m}{98}\right|}\right)}{m}\\
\left|1-\frac{m}{142}\right|=\frac{\left|1-\frac{m}{65}\right|}{e^{2mk}}
\end{cases}
$$
Substitute them into eachother gives:
$$\left|1-\frac{m}{142}\right|=\frac{\left|1-\frac{m}{65}\right|}{\exp\left[2\cdot m\cdot\frac{\ln\left(\frac{\left|1-\frac{m}{65}\right|}{\left|1-\frac{m}{98}\right|}\right)}{m}\right]}\Longleftrightarrow\left|1-\frac{m}{142}\right|=\frac{\left|1-\frac{m}{98}\right|^2}{\left|1-\frac{m}{65}\right|}$$
And the solutions we get are:
$$m=\frac{8134}{17},k=\frac{14\ln\left(\frac{213}{130}\right)}{8134}$$
Now, we get that:
$$\left|1-\frac{8134}{17n(t)}\right|=\frac{7029}{1105\left(\frac{213}{130}\right)^{\frac{14t}{17}}}$$
