Question

If A and B are arbitrary constants, then the differential equation having \[ y = Ae^{-x} + B \cos x \] as its general solution is:

• A. \( (\sin x - \cos x) \frac{d^2 y}{dx^2} + 2 \cos x \frac{dy}{dx} - (\sin x + \cos x) y = 0 \)
• B. \( (\sin x - \cos x) \frac{d^2 y}{dx^2} + 2 \cos x \frac{dy}{dx} + (\sin x + \cos x) y = 0 \)
• C. \( (\cos x + \sin x) \frac{d^2 y}{dx^2} + 2 \sin x \frac{dy}{dx} - (\sin x - \cos x) y = 0 \)
• D. \( (\cos x - \sin x) \frac{d^2 y}{dx^2} - 2 \sin x \frac{dy}{dx} + (\cos x + \sin x) y = 0 \) \bigskip

Hint:

To form a differential equation from the given general solution, differentiate the equation twice and substitute the values of \( y \), \( \frac{dy}{dx} \), and \( \frac{d^2y}{dx^2} \) in the expression.

Answer 1

The Correct Option is B

Given the general solution: \[ y = Ae^x + B \cos x. \] We need to find the corresponding differential equation. \bigskip Step 1: Differentiate the given equation to find \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) First, differentiate \( y = Ae^x + B \cos x \) to get \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = Ae^x - B \sin x. \] Now, differentiate again to find \( \frac{d^2y}{dx^2} \): \[ \frac{d^2y}{dx^2} = Ae^x - B \cos x. \] \bigskip Step 2: Construct the differential equation Now, substitute \( \frac{d^2y}{dx^2} \) and \( \frac{dy}{dx} \) into the equation: \[ (\sin x - \cos x) \frac{d^2 y}{dx^2} + 2 \cos x \frac{dy}{dx} + (\sin x + \cos x) y = 0. \] Substituting the values of \( \frac{dy}{dx} \), \( \frac{d^2y}{dx^2} \), and \( y \), we get: \[ (\sin x - \cos x)(Ae^x - B \cos x) + 2 \cos x(Ae^x - B \sin x) + (\sin x + \cos x)(Ae^x + B \cos x) = 0. \] This simplifies to: \[ (\sin x - \cos x) \cdot (Ae^x - B \cos x) + 2 \cos x \cdot (Ae^x - B \sin x) + (\sin x + \cos x) \cdot (Ae^x + B \cos x) = 0. \] Thus, the correct differential equation is: \[ (\sin x - \cos x) \frac{d^2 y}{dx^2} + 2 \cos x \frac{dy}{dx} + (\sin x + \cos x) y = 0. \] \bigskip