Stuck solving an ODE(IVP) with separation of variables I am trying to solve an IVP which has a particularity of having several variables which should be found with a set of conditions, which is: $y'(t)=k y(t)(P-y(t))$ $y(0)=500$ $y(1)=1000$ $lim_{t->\infty}y(t)=50000$ By using the separation of variables method, I found that, after doing the integrations: $\frac{\ln(y)-\ln(y-P)}{Pk}=t+C$ Which is where I get stuck. By trying to simplify I got: $\frac{1}{kP}(\ln\frac{y}{y-P})=t+C$ $(e^\frac{1}{kp})(\frac{y}{y-P})=Ce^t$ $\frac{y}{y-P}=Ce^{(t-\frac{1}{kP})}$ Where, at the end: $y(t)=\frac{-pCe^{(t-\frac{1}{kP})}}{1-e^{(t-\frac{1}{kP})}}$ Which does not meet the criteria for the IVP, so it should mean that I have messed up along the way, and I guess it's when I started using the logarithmic and exponential simplifications... Any suggestions/corrections are welcome.

$$\frac1k\int \frac1{y(P-y)}\, dy = t+c$$

$$\frac1k\int \frac1{Py}+\frac1{P(P-y)}\, dy = t+c$$

$$\frac1{Pk}\int \frac1{y}+\frac1{(P-y)}\, dy = t+c$$

$$\frac1{kP }\left( \ln y-\ln (P-y)\right)=t+c$$

$$\frac{y}{P-y}= A\exp\left(kPt \right)$$

I think you made a mistake when you take exponential. Note that $e^{AB} \
e e^A e^B$.

Edit:

$$y = \frac{AP\exp(kPT)}{1+A\exp(kPt)}=\frac{P}{1+\frac{\exp(-kPt)}{A}}$$

We are told that as $t \to \infty$, the value converges. In that case, we need $kP>0$ and the exponential term vanishes. Hence $P=50000$.

Use information about $t=0$ to solve for $A$.

Finally use information about $t=1$ to solve for $k$.