Question

If \( y = A \sin 2x + B \cos 2x \) and \[ \frac{d^2y}{dx^2} - k y = 0, \text{ find the value of } k. \]

Hint:

When solving for constants in differential equations, substitute the function and its derivatives into the equation, then equate coefficients to determine the unknowns.

Answer 1

Step 1: Differentiate \( y = A \sin 2x + B \cos 2x \) twice. The first derivative of \( y \) is: \[ \frac{dy}{dx} = 2A \cos 2x - 2B \sin 2x. \] The second derivative of \( y \) is: \[ \frac{d^2y}{dx^2} = -4A \sin 2x - 4B \cos 2x. \] Step 2: Substitute \( \frac{d^2y}{dx^2} \) and \( y \) into the given equation. The given equation is: \[ \frac{d^2y}{dx^2} - k y = 0. \] Substitute \( \frac{d^2y}{dx^2} = -4A \sin 2x - 4B \cos 2x \) and \( y = A \sin 2x + B \cos 2x \): \[ -4A \sin 2x - 4B \cos 2x - k (A \sin 2x + B \cos 2x) = 0. \] Step 3: Group and simplify terms. Combine terms involving \( \sin 2x \) and \( \cos 2x \): \[ (-4A - kA) \sin 2x + (-4B - kB) \cos 2x = 0. \] For this equation to hold for all \( x \), the coefficients of \( \sin 2x \) and \( \cos 2x \) must be zero: \[ -4A - kA = 0 \quad \text{and} \quad -4B - kB = 0. \] Step 4: Solve for \( k \). From the first equation: \[ A(-4 - k) = 0 \quad \Rightarrow \quad k = -4 \quad \text{(since \( A \neq 0 \))}. \] From the second equation: \[ B(-4 - k) = 0 \quad \Rightarrow \quad k = -4 \quad \text{(since \( B \neq 0 \))}. \] Final Answer: \[ \boxed{k = -4.} \]