Differential Equations
Functional Equations Leading to ODEs
Grade 12
Question:
<p>A function \(f: \mathbb{R} \to \mathbb{R}\) satisfies \(\sin x \cos y (f(2x + 2y) − f(2x − 2y)) = \cos x \sin y (f(2x + 2y) + f(2x − 2y))\). If \(f(0) = \frac{1}{2}\), find the differential equation that \(f\) satisfies.</p>
<p>(a) \(f''(x) - f(x) = 0\)</p>
<p>(b) \(4f''(x) - f(x) = 0\)</p>
<p>(c) \(f''(x) + f(x) = 0\)</p>
<p>(d) \(4f''(x) + f(x) = 0\)</p>
Step-by-Step Solution
Key Concept: Recognize the functional equation form, deduce the solution function using initial conditions, and verify the resulting differential equation.
<p><strong>Step 1:</strong> From the given functional equation, divide both sides by $\cos x \sin y$ (when non-zero):</p><p>$\frac{\sin x \cos y}{\cos x \sin y}(f(2x + 2y) − f(2x − 2y)) = f(2x + 2y) + f(2x − 2y)$</p><p><strong>Step 2:</strong> This simplifies to: $\tan x \cot y (f(2x + 2y) − f(2x − 2y)) = f(2x + 2y) + f(2x − 2y)$</p><p><strong>Step 3:</strong> Setting $y = x$ in the original equation and analyzing the structure suggests $f(x) = K\sin(\frac{x}{2})$ for some constant K.</p><p><strong>Step 4:</strong> With $f(0) = \frac{1}{2}$, we determine $f(x) = \frac{1}{2}\sin(\frac{x}{2})$.</p><p><strong>Step 5:</strong> Computing derivatives: $f'(x) = \frac{1}{4}\cos(\frac{x}{2})$ and $f''(x) = -\frac{1}{8}\sin(\frac{x}{2})$</p><p><strong>Step 6:</strong> Therefore: $4f''(x) = -\frac{1}{2}\sin(\frac{x}{2}) = -f(x)$, giving $4f''(x) + f(x) = 0$</p><p>∴ Answer is (d). [Note: The provided solution indicates (b), but verification suggests (d) is correct based on the analysis.]</p>
Correct Answer: b