My old eyes have a hard time with your image.

We are given to solve the IVP:

\(\displaystyle \d{P}{t}=0.2P\left(1-\frac{P}{1000}\right)-48\) where \(P(0)=P_0\)

Let's write the ODE as:

\(\displaystyle \d{P}{t}=-\frac{(P-400)(P-600)}{5000}\)

Thus, separating variables and switching dummy variables, and integrating on the given boundaries:

\(\displaystyle 5000\int_{P_0}^{P}\frac{1}{(P-400)(P-600)}\,du=-\int_0^t\,dv\)

\(\displaystyle \ln\left(\frac{(P-600)(P_0-400)}{(P-400)(P_0-600)}\right)=-\frac{2}{50}t\)

\(\displaystyle \frac{(P-600)(P_0-400)}{(P-400)(P_0-600)}=e^{-\frac{2}{50}t}\)

\(\displaystyle \frac{P-600}{P-400}=\frac{P_0-600}{P_0-400}e^{-\frac{2}{50}t}\)

\(\displaystyle P-600=\frac{P_0-600}{P_0-400}e^{-\frac{2}{50}t}(P-400)\)

\(\displaystyle P\left(1-\frac{P_0-600}{P_0-400}e^{-\frac{2}{50}t}\right)=600-400\frac{P_0-600}{P_0-400}e^{-\frac{2}{50}t}\)

\(\displaystyle P\left((P_0-400)-(P_0-600)e^{-\frac{2}{50}t}\right)=600(P_0-400)-400(P_0-600)e^{-\frac{2}{50}t}\)

\(\displaystyle P(t)=\frac{600(P_0-400)-400(P_0-600)e^{-\frac{2}{50}t}}{(P_0-400)-(P_0-600)e^{-\frac{2}{50}t}}\)

\(\displaystyle P(t)=\frac{600(P_0-400)e^{\frac{2}{50}t}-400(P_0-600)}{(P_0-400)e^{\frac{2}{50}t}-(P_0-600)}\)

Now, you can just plug in for \(P_0\).