Question

$\frac{f(2 x+2 y)-f(2 x-2 y)}{f(2 x+2 y)+f(2 x-2 y)}=\frac{\cos x \sin y}{\sin x \cos y}$ Applying C \& D, $\frac{\mathrm{f}(2 \mathrm{x}+2 \mathrm{y})}{\mathrm{f}(2 \mathrm{x}-2 \mathrm{y})}=\frac{\sin (\mathrm{x}+\mathrm{y})}{\sin (\mathrm{x}-\mathrm{y})}$ \(\frac{\mathrm{f}(2 \mathrm{x}+2 \mathrm{y})}{\sin (\mathrm{x}+\mathrm{y})}\) is constant \(\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{C} \sin \frac{\mathrm{x}}{2}\) \(\mathrm{C}=1\) using \(\mathrm{f}^{\prime}(0)=\frac{1}{2}\) \(\Rightarrow \mathrm{f}(\mathrm{x})=\sin \frac{\mathrm{x}}{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \cos \frac{\mathrm{x}}{2}\) $\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-1}{4} \sin \frac{\mathrm{x}}{2}$ $\Rightarrow 4 \mathrm{f}^{\prime \prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})=0$

Short answer

Question: Determine a function f(x) that satisfies the given functional equation \(\frac{f(2 x+2 y)-f(2 x-2 y)}{f(2 x+2 y)+f(2 x-2 y)}=\frac{\cos x \sin y}{\sin x \cos y}\), with the following properties: 1. The function f(x) is continuous and differentiable for all x ∈ R. 2. There exists a function C(x) and D(x) such that applying function C(x) and D(x) to the given functional equation, we obtain the equation \(\frac{f(2 x+2 y)}{f(2 x-2 y)} = \frac{\sin (x+y)}{\sin (x-y)}\). 3. The first derivative of f(x) at 0 is 1/2, i.e., f'(0) = 1/2. 4. The second derivative of f(x) satisfies the equation 4f''(x) + f(x) = 0. Answer: The function f(x) that satisfies the given functional equation with the desired properties is given by \(f(x) = \sin \frac{x}{2}\).

Step-by-step solution

Write the given exercise in terms of f(x) and trigonometric terms

We are given the equation: \(\frac{f(2 x+2 y)-f(2 x-2 y)}{f(2 x+2 y)+f(2 x-2 y)}=\frac{\cos x \sin y}{\sin x \cos y}\)

Apply the C & D relationship

We are given that after applying C & D, the relationship becomes: \(\frac{f(2 x+2 y)}{f(2 x-2 y)} = \frac{\sin (x+y)}{\sin (x-y)}\)

Observe that f(x) is proportional to the sine function

We see that the relationship implies that: \(\frac{f(2 x+2 y)}{\sin (x+y)}\) is constant Thus, we can write it as: \(f(x) = C \sin \frac{x}{2}\), where C is a constant.

Determine the value of C using the given f'(0)

We are given that f'(0) = 1/2, so we differentiate f(x) with respect to x: \(f'(x) = \frac{C}{2} \cos \frac{x}{2}\) Now, substituting x = 0: \(f'(0) = \frac{C}{2}\) Since f'(0) = 1/2, we have: \(C = 1\)

Write f(x) and find its second derivative

Now that we have determined the constant C, we can write f(x): \(f(x) = \sin \frac{x}{2}\) Now, let us find the second derivative, f''(x): \(f'(x) = \frac{1}{2} \cos \frac{x}{2}\) \(f''(x) = -\frac{1}{4} \sin \frac{x}{2}\)

Show that 4f''(x) + f(x) = 0

We want to verify that 4f''(x) + f(x) = 0 for the determined function f(x). Let us compute the sum: \(4f''(x) + f(x) = 4\left(-\frac{1}{4} \sin \frac{x}{2}\right) + \sin \frac{x}{2} = -\sin \frac{x}{2} + \sin \frac{x}{2} = 0\) We have thus shown that the given functional equation is satisfied by the function f(x) = \(\sin \frac{x}{2}\) with the desired properties.