# Solve this differential equation explicitly, either by using partial fractions, or with a computer algebra system. Use the initial populations 350 and 450. Can anyone please help me with this problem and explain how you got the answers?

##### 1 Answer
Jun 24, 2018

For the IV: ${P}_{o} = 350$

• $P = 100 \left(\frac{{e}^{\frac{t}{50}} + 5}{{e}^{\frac{t}{50}} - 5} + 5\right)$

For the IV: ${P}_{o} = 450$

• $P = 100 \left(\frac{{e}^{\frac{t}{50}} - 3}{{e}^{\frac{t}{50}} + 3} + 5\right)$

#### Explanation:

$\dot{P} = \frac{P}{10} - {P}^{2} / 10000 - 24$

Complete square:

$= 1 - {\left(P - 500\right)}^{2} / 10000$

Let $Q = P - 500 q \quad \dot{Q} = \dot{P}$

So the DE is:

$\dot{Q} = 1 - {\left(\frac{Q}{100}\right)}^{2}$

Let $R = \frac{Q}{100} q \quad \dot{R} = \frac{\dot{Q}}{100}$

$100 \dot{R} = 1 - {R}^{2}$

This separates:

$\frac{\mathrm{dR}}{1 - {R}^{2}} = \frac{1}{100} \mathrm{dt}$

Factor and use partial fractions:

$= \int \mathrm{dR} \setminus \frac{1}{\left(1 - R\right) \left(1 + R\right)} = \frac{1}{100} \int \setminus \mathrm{dt}$

$= \frac{1}{2} \int \mathrm{dR} \setminus \frac{1}{R + 1} - \frac{1}{R - 1} = \frac{t}{100} + C$

$= \frac{1}{2} \left(\ln \left(R + 1\right) - \ln \left(R - 1\right)\right) = \frac{t}{100} + C$

$= \ln \left(\frac{R + 1}{R - 1}\right) = \frac{t}{50} + C$

$R = \frac{C {e}^{\frac{t}{50}} + 1}{C {e}^{\frac{t}{50}} - 1}$

Reverse the subs:

$Q = 100 \frac{C {e}^{\frac{t}{50}} + 1}{C {e}^{\frac{t}{50}} - 1}$

bb( P= 100 ((Ce^( t/50) + 1)/(Ce^( t/50) - 1) + 5 )

For the IV: ${P}_{o} = 350$

$350 = 100 \left(\frac{C + 1}{C - 1} + 5\right) \implies C = \frac{1}{5}$

$P = 100 \left(\frac{{e}^{\frac{t}{50}} + 5}{{e}^{\frac{t}{50}} - 5} + 5\right)$

For the IV: ${P}_{o} = 450$

$450 = 100 \left(\frac{C + 1}{C - 1} + 5\right) \implies C = - \frac{1}{3}$

$P = 100 \left(\frac{{e}^{\frac{t}{50}} - 3}{{e}^{\frac{t}{50}} + 3} + 5\right)$