Differential Equations
Linear ODE — finding solution function
Grade Class 12
Question:
<p>A function \(y=f(x)\) satisfies \((x+1)f'(x) - 2(x^2+x)f(x)=0\), \(f(0)=1\). Let \(S = \{x: f(x) > 1\}\). Then \(S\) is:</p>
<span>\((-\infty,0)\)</span>
<span>\((-\infty,0)\cup(0,\infty)\)</span>
<span>\((-1,0)\cup(0,1)\)</span>
<span>\((0,\infty)\)</span>
Step-by-Step Solution
Key Concept: Solve the linear ODE for f(x) explicitly, then find where f(x) > 1.
<div class='solution'><p><strong>Step 1:</strong> Rewrite: \(\dfrac{f'}{f} = \dfrac{2x(x+1)}{x+1} = 2x\) (for \(x \ne -1\)).</p>
<p>\[\ln f = x^2 + C \implies f(x) = Ae^{x^2}\]</p>
<p><strong>Step 2:</strong> \(f(0)=1\): \(A = 1\). So \(f(x) = e^{x^2}\).</p>
<p><strong>Step 3:</strong> \(f(x) > 1 \iff e^{x^2} > 1 \iff x^2 > 0 \iff x \ne 0\).</p>
<p>\(S = (-\infty, 0) \cup (0, \infty) = \mathbb{R} \setminus \{0\}\). <strong>Answer: (B)</strong>.</p>
<p class='key-concept'>🔑 Key Concept: Factor the ODE to get \(f'/f = 2x\), integrate to get \(f(x) = e^{x^2}\). Since \(e^{x^2} \ge 1\) with equality only at \(x=0\), the set is all reals except 0.</p>
<p class='trap-warning'>⚠️ Trap: Students write \(f'/f = 2(x^2+x)/(x+1) = 2x^2/(x+1)\) by failing to factor \(x^2+x = x(x+1)\). Always factorise before simplifying.</p></div>
Correct Answer: 2