Differential Equations
Differential inequalities
Grade None
Question:
<p>Given inequality can be written as: \(f''(x) - 2f'(x) \geq 3(f'(x) - 2f(x))\).<br>Let \(f'(x) - 2f(x) = g(x)\). It is given that \(g(x)e^{-3x}\) is non-decreasing, \(g(0) = f'(0) - 2f(0) = -2\). If \(f(x) \geq 3e^{2x} - 2e^{3x},\ \forall x \geq 0\), and comparing \(ah(bx) - bh(ax)\) with \(3e^{2x} - 2e^{3x}\), find \((a+b)h(0)\).</p>
Step-by-Step Solution
Key Concept: Recognize that the inequality f''(x) - 2f'(x) ≥ 3(f'(x) - 2f(x)) can be rewritten as d/dx[g(x)e^(-3x)] ≥ 0 where g(x) = f'(x) - 2f(x), meaning g(x)e^(-3x) is non-decreasing. This constrains g(x) and allows us to find the minimal f(x).
<p><strong>Step 1:</strong> Rewrite the given inequality with substitution g(x) = f'(x) - 2f(x).</p><p>The inequality becomes: g'(x) ≥ 3g(x), which means d/dx[g(x)e^(-3x)] ≥ 0.</p><p><strong>Step 2:</strong> Since g(x)e^(-3x) is non-decreasing and g(0) = -2, we have g(x)e^(-3x) ≥ -2e^0 = -2.</p><p>Thus: g(x) ≥ -2e^(3x), with equality when g(x)e^(-3x) is constant.</p><p><strong>Step 3:</strong> Solve f'(x) - 2f(x) = -2e^(3x) (the equality case for minimum f).</p><p>This is a linear ODE: f'(x) - 2f(x) = -2e^(3x).</p><p>Integrating factor: e^(-2x). Solution: f(x) = Ae^(2x) + e^(3x).</p><p><strong>Step 4:</strong> The lower bound is f(x) = 3e^(2x) - 2e^(3x), giving A = 3.</p><p>So the minimal function is f(x) = 3e^(2x) - 2e^(3x).</p><p><strong>Step 5:</strong> Compare ah(bx) - bh(ax) = 3e^(2x) - 2e^(3x).</p><p>Setting h(x) = e^x: ah(bx) - bh(ax) = ae^(bx) - be^(ax).</p><p>Match coefficients: a = 3, b = 2 (from 3e^(2x) - 2e^(3x)).</p><p><strong>Step 6:</strong> h(0) = e^0 = 1.</p><p>∴ (a+b)h(0) = (3+2)(1) = <strong>5</strong></p>
Correct Answer: 5