Differential Equations
Applications — depreciation model
Grade Class 12
Question:
<p>Depreciation: \(\dfrac{df}{dt} = -k(T-t)\), \(f(0) = \) purchase price \(f_0\), \(f(T)=V(T)\). Find \(k\):</p>
<span>\(\frac{2(f_0-V)}{T^2}\)</span>
<span>\(\frac{f_0-V}{T^2}\)</span>
<span>\(\frac{2(f_0-V)}{T}\)</span>
<span>\(\frac{f_0-V}{2T^2}\)</span>
Step-by-Step Solution
Key Concept: Integrate df/dt = -k(T-t) with limits f(0)=f_0 and f(T)=V(T).
<div class='solution'><p><strong>Step 1:</strong> Integrate: $f(t) = -k\\!\int\\!(T-t)\,dt = -k\left(Tt - \dfrac{t^2}{2}\right) + C$.</p>
<p>At $t=0$: $C = f_0$. So $f(t) = f_0 - k\left(Tt - \dfrac{t^2}{2}\right)$.</p>
<p><strong>Step 2:</strong> At $t=T$: $V = f_0 - k\left(T^2 - \dfrac{T^2}{2}\right) = f_0 - \dfrac{kT^2}{2}$.</p>
<p>$$k = \frac{2(f_0 - V)}{T^2}$$</p>
<p><strong>Answer: (A)</strong> $k = \dfrac{2(f_0-V)}{T^2}$.</p>
<p class='key-concept'>🔑 Key Concept: Depreciation ODEs use direct integration — no separation needed. Set up boundary conditions carefully.</p></div>
Correct Answer: 1