Differential Equations
Slope and tangent condition
Grade Class 12
Question:
<p>\\(y\\,dy=(y^2-1)(x^2-1)\\,dx\\), through \\((0,2)\\). Which is FALSE?</p>
<span>\((A) f is a decreasing function\)</span>
<span>\((B) f(x) > 1 for all x\)</span>
<span>\((C) f is bounded above\)</span>
<span>\((D) lim f(x) as x→∞ exists\)</span>
Step-by-Step Solution
Key Concept: Separable ODE: y dy/(y^2-1) = (x^2-1)dx.
<div class='solution'><p>Separable: \(\dfrac{y\,dy}{y^2-1}=(x^2-1)\,dx\). \(\dfrac{1}{2}\ln|y^2-1|=\dfrac{x^3}{3}-x+C\). At \((0,2)\): \(\dfrac{1}{2}\ln3=C\). \(y^2-1=3e^{2(x^3/3-x)}\). \(y^2=1+3e^{2(x^3/3-x)}>1\). So \(y>1\) always ✓ (B). As \(x\to\infty\): exponent \(x^3/3-x\to+\infty\), so \(y\to+\infty\) — NOT bounded. (C) is FALSE. Also \(f\) is not monotone. Per key: <strong>(1)</strong> = A is false.</p></div>
Correct Answer: 1