Differential Equations
Rate of Change of Quantities
Grade 12
Question:
<p>The population \(p(t)\) at time <em>t</em> of a certain mouse species satisfies the differential equation \(\dfrac{dp(t)}{dt} = 0.5\,p(t) - 450\). If \(p(0) = 850\), then the time at which the population becomes zero is</p>
<p>\(2\ln 18\)</p>
<p>\(\ln 9\)</p>
<p>\(\dfrac{1}{2}\ln 18\)</p>
<p>\(\ln 18\)</p>
Step-by-Step Solution
Key Concept: Find the equilibrium point where dp/dt = 0, then solve the linear first-order ODE by recognizing that the population approaches this equilibrium. The population hits zero when it crosses the equilibrium from above.
<p><strong>Step 1: Find the equilibrium point</strong></p><p>Set dp/dt = 0: 0.5p - 450 = 0 ⟹ p = 900</p><p><strong>Step 2: Rewrite the differential equation</strong></p><p>dp/dt = 0.5(p - 900)</p><p><strong>Step 3: Separate variables</strong></p><p>dp/(p - 900) = 0.5 dt</p><p><strong>Step 4: Integrate both sides</strong></p><p>ln|p - 900| = 0.5t + C</p><p><strong>Step 5: Apply initial condition p(0) = 850</strong></p><p>ln|850 - 900| = 0 + C ⟹ ln(50) = C</p><p><strong>Step 6: Solve for p(t)</strong></p><p>ln|p - 900| = 0.5t + ln(50)</p><p>|p - 900| = 50e^(0.5t)</p><p>Since p(0) = 850 < 900: p - 900 = -50e^(0.5t)</p><p>p(t) = 900 - 50e^(0.5t)</p><p><strong>Step 7: Find when p(t) = 0</strong></p><p>900 - 50e^(0.5t) = 0</p><p>e^(0.5t) = 18</p><p>0.5t = ln(18)</p><p>t = 2ln(18) = ln(324)</p><p>∴ Answer: A</p>
Correct Answer: A