Differential Equations Question - None

Question

Population $P(t)$ satisfies $\dfrac{dP}{dt} = 0.5P - 450$, $P(0)=850$. When does $P=0$?

Accepted Answer

Step 1: $\dfrac{dP}{dt} = 0.5(P - 900)$. Separate:

$$\frac{dP}{P-900} = 0.5\,dt \implies \ln|P-900| = 0.5t + C$$

Step 2: $P = 900 + Ae^{0.5t}$. At $t=0$: $850 = 900 + A \implies A = -50$.

$$P = 900 - 50e^{0.5t}$$

Step 3: $P = 0$: $900 = 50e^{0.5t} \implies e^{0.5t} = 18 \implies 0.5t = \ln 18 \implies t = 2\ln 18$.

Answer: (B) $t = 2\log 18$ (natural log).

🔑 Key Concept: The equilibrium point is P* = 900. Since P(0)=850 < 900 and dP/dt < 0, population decreases to zero in finite time.

⚠️ Trap: Forgetting the factor of 2 and writing t = log 18 (option A). The 0.5 in the exponent gives t = 2 ln 18.

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