I just rediscovered a program I made a few years ago on Khan Academy to model the Malthusian population curve, based on a differential equation, where $K$ is the population limit, $N$ is the population, and $t$ is time. $$\frac{dN}{dt}=\frac1{8\log{t}}+\frac{(K-N)}{1.69}\frac{\sin(t)}t$$ It's sort of like a logistic curve, but with some tweaks to better fit real-life.
I was wondering how I should go about finding a function for $y$, the population, with respect to $x$, time. I already know the general logistic growth equation $$\frac{dN}{dt}=rt(1-\frac{N}{K})$$ and how to get to it, but I'm not sure how to tackle my differential equation.
Thanks in advance.
The numerical constants in the first equation are mostly just arbitrary - feel free to replace them with simpler integers if you like.
Let $N_K = N-K$, then we have a slightly simpler equation
$$ \frac{dN_K}{dt} + \frac{a\sin t}{t}N_K = \frac{b}{\ln t} $$
Using the standard first-order method, an integrating factor can be found as
$$ \mu(a,t) = \exp\left(a\int_0^t \frac{\sin \tau}{\tau}\ d\tau\right) = \exp\big(a \operatorname{Si}(t) \big) $$
where $\operatorname{Si}(t)$ is the sine integral. Then
$$ \mu \frac{dN_K}{dt} + \frac{d\mu}{dt}N_K = \frac{d}{dt}\big(\mu N_K\big) = \frac{b\mu}{\ln t} $$
$$ \implies N_K = \frac{1}{\mu}\int \frac{b\mu}{\ln t} dt = be^{-a\operatorname{Si}(t)}\left[\int_0^t \frac{e^{a\operatorname{Si}(\tau)}}{\ln \tau} d\tau + c\right] $$
If $N(0) = N_0$ then we have the solution
$$ N(t) = K + (N_0-K)e^{-a\operatorname{Si}(t)} + be^{-a\operatorname{Si}(t)}\int_0^t \frac{e^{a\operatorname{Si}(\tau)}}{\ln \tau} d\tau $$
The singularity at $t=1$ can be resolved by the Cauchy principal value. You may notice that this population model grows without bounds, similar to the logarithmic integral.