Solving for a differential equation Gompertz growth equation 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?
 A: $$\frac{dy}{dt} = k \,  y \, \log\left(\frac{a}{y}\right)$$ Let $y=e^z$ to make
$$z' + k z=k \log (a)$$ which looks to be simple.
A: $$y'=k y (\log a-\log y)$$
$$kdt=\frac{dy}{y(\log a-\log y)}$$
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 $$y=a \exp(-e^{C-k t})$$
A: $$
-\frac{(\ln a-\ln y)'}{(\ln a-\ln y)} = k dt\ \ \Rightarrow \ \ -\ln(\ln a-\ln y) = k t + C_0
$$
so
$$
\ln a-\ln y = C_1 e^{-k t}\ \ \Rightarrow y = a e^{-C_1 e^{-k t}}
$$
A: $$\frac{dy}{dt}=ky\ln\left(\frac ay\right)$$
$$\frac{dy}{dt}=ky\left[\ln(a)-\ln(y)\right]$$
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:
$$z'=-kz+\alpha$$
$$\int\frac{dz}{kz-\alpha}=-\int dt$$
$$\frac1k\ln(kz-\alpha)=-t+C_1$$
$$\ln(kz-\alpha)=-kt+C_2$$
$$kz-\alpha=C_3e^{-kt}$$
$$z=C_4e^{-kt}+\ln(a)$$
$$y=a\exp\left(Ce^{-kt}\right)$$
