# Solve differential equation explicitly: dP/dt = 0.2P (1 - P/1000) - 48, using P_0 = 350

#### MarkFL

##### Super Moderator
Staff member
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 $$P_0$$. #### Dr.Peterson

##### Elite Member
I think you took too large a step in getting to $$\displaystyle \left|\frac{P-600}{P-400}\right| = \frac{1}{200}e^{-\frac{1}{5000}t + C}$$. Try writing an extra step or two and see what you get.