Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>The given differential equation is \(\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200\). If the initial number of rabbits is 100 and is decreasing, then \(p(t)\) is:</p>
<p>\(p(t) = 400 + 300e^{t/2}\)</p>
<p>\(p(t) = 400 - 300e^{t/2}\)</p>
<p>\(p(t) = 400 + 300e^{-t/2}\)</p>
<p>\(p(t) = 400 - 300e^{-t/2}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a first-order linear ODE; solve using integrating factor method or separation of variables after finding the equilibrium point. The equilibrium p = 400 determines whether p(t) increases or decreases from initial condition p(0) = 100.
<p><strong>Step 1:</strong> Rewrite the ODE in standard form: $\frac{dp}{dt} - \frac{1}{2}p = -200$</p><p><strong>Step 2:</strong> Find the equilibrium point by setting $\frac{dp}{dt} = 0$: $\frac{1}{2}p - 200 = 0 \Rightarrow p = 400$</p><p><strong>Step 3:</strong> Use integrating factor $\mu(t) = e^{-\frac{t}{2}}$. Multiply through and integrate: $e^{-\frac{t}{2}}\frac{dp}{dt} - \frac{1}{2}e^{-\frac{t}{2}}p = -200e^{-\frac{t}{2}}$</p><p><strong>Step 4:</strong> This gives $\frac{d}{dt}(e^{-\frac{t}{2}}p) = -200e^{-\frac{t}{2}}$. Integrating: $e^{-\frac{t}{2}}p = 400e^{-\frac{t}{2}} + C$</p><p><strong>Step 5:</strong> Apply initial condition $p(0) = 100$: $100 = 400 + C \Rightarrow C = -300$</p><p><strong>Step 6:</strong> Therefore: $p(t) = 400 - 300e^{\frac{t}{2}}$</p><p>∴ Answer: B</p>
Correct Answer: B