Differential Equations
Initial value problem — graphical interpretation
Grade Class 12
Question:
<p>Which curve represents the solution of the IVP \(\dfrac{dy}{dx} = 100 - y\), where \(y(0) = 50\)?</p>
<span>\(\text{Decreasing from 50 to 0}\)</span>
<span>\(\text{Increasing past 100}\)</span>
<span>\(\text{Increasing from 50, asymptote at }y=100\)</span>
<span>\(\text{Oscillating around }y=100\)</span>
Step-by-Step Solution
Key Concept: Solve the linear ODE; y \to 100 as x \to \infty since y(0) = 50 < 100.
<div class='solution'><p><strong>Step 1:</strong> Linear ODE: $\dfrac{dy}{dx} + y = 100$.</p>
<p>IF $= e^x$. Solution: $ye^x = 100e^x + C \implies y = 100 + Ce^{-x}$.</p>
<p><strong>Step 2:</strong> At $x=0, y=50$: $50 = 100 + C \implies C = -50$.</p>
<p>$$y = 100 - 50e^{-x}$$</p>
<p><strong>Step 3:</strong> At $x=0$: $y=50$. As $x\to\infty$: $y\to 100$. The curve <em>increases</em> from 50 and approaches 100 asymptotically. <strong>Answer: (C)</strong>.</p>
<p class='key-concept'>🔑 Key Concept: The equilibrium (steady state) is $y^* = 100$. Since $y(0) < y^*$, the solution increases toward $y^*$.</p>
<p class='trap-warning'>⚠️ Trap: Thinking the curve overshoots 100 — for linear first-order ODEs there is no overshoot; the approach is monotonic.</p></div>
Correct Answer: 3