What is the general solution of this differential equation? $$ \frac{dy}{dt} = k \enspace y \enspace \ln(\frac{a}{y})$$ where $a$ and $k$ are positive constants.
So far, my solution is: $$ \frac{dy}{y \enspace \ln(\frac{a}{y})} = k \enspace dt$$ When I let $u=lny$, $$ \int \frac{1}{y(\ln a-\ln y)} \,dy =\int k dt $$ $$ \int \frac{1}{\ln a-u} \,du= kt + C $$
How to continue this?
On
$$y'=k y (\log a-\log y)$$ $$kdt=\frac{dy}{y(\log a-\log y)}$$ Let $\log y=u\to\frac{dy}{y}=du$ $$kt=\int \frac{du}{\log a-u}$$ $$kt=-\log (\log a-u)+C$$ $$C-kt=\log (\log a-\log y)$$ $$C-kt=\log\log\frac{a}{y}$$ $$e^{C-kt}=\log\frac{a}{y}$$
The general solution is $$y=a \exp(-e^{C-k t})$$
On
$$ -\frac{(\ln a-\ln y)'}{(\ln a-\ln y)} = k dt\ \ \Rightarrow \ \ -\ln(\ln a-\ln y) = k t + C_0 $$
so
$$ \ln a-\ln y = C_1 e^{-k t}\ \ \Rightarrow y = a e^{-C_1 e^{-k t}} $$
On
$$\frac{dy}{dt}=ky\ln\left(\frac ay\right)$$ $$\frac{dy}{dt}=ky\left[\ln(a)-\ln(y)\right]$$ now make the substitution: $y=e^z$ so $\frac{dy}{dt}=e^z\frac{dz}{dt}=y\frac{dz}{dt}$ and so: $$y\frac{dz}{dt}=ky\left[\ln(a)-z\right]$$ $$\frac{dz}{dt}=k\ln(a)-kz$$ if we let $\alpha=k\ln(a)$ you will see we have a simple equation: $$z'=-kz+\alpha$$ $$\int\frac{dz}{kz-\alpha}=-\int dt$$ $$\frac1k\ln(kz-\alpha)=-t+C_1$$ $$\ln(kz-\alpha)=-kt+C_2$$ $$kz-\alpha=C_3e^{-kt}$$ $$z=C_4e^{-kt}+\ln(a)$$ $$y=a\exp\left(Ce^{-kt}\right)$$
$$\frac{dy}{dt} = k \, y \, \log\left(\frac{a}{y}\right)$$ Let $y=e^z$ to make $$z' + k z=k \log (a)$$ which looks to be simple.