Question

If $a$ and $b$ are arbitrary constants, then the differential equation representing the family of curves $y = a \sin(x + b)$ is

• A. $\frac{d^{2}y}{dx^{2}} - y = 0$
• B. $\frac{d^{2}y}{dx^{2}} + y = 0$
• C. $\frac{d^{2}y}{dx^{2}} - y^{2} = 0$
• D. $\frac{dy}{dx} - y = 0$

Hint:

For $y = A\sin(kx + \phi)$, the second derivative is always $-k^2y$. Here $k=1$, so $y'' = -y$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The order of a differential equation is equal to the number of arbitrary constants in the given equation. Since there are two constants (\(a\) and \(b\)), we differentiate twice to eliminate them.
Step 2: Key Formula or Approach:
1. Differentiate \(y\) with respect to \(x\).
2. Differentiate again to find \(\frac{d^2y}{dx^2}\).
3. Substitute the original expression of \(y\) back into the result.
Step 3: Detailed Explanation:
Given \( y = a \sin(x + b) \). First derivative: \[ \frac{dy}{dx} = a \cos(x + b) \] Second derivative: \[ \frac{d^2y}{dx^2} = -a \sin(x + b) \] Since \( y = a \sin(x + b) \), we can substitute: \[ \frac{d^2y}{dx^2} = -y \] \[ \frac{d^2y}{dx^2} + y = 0 \]
Step 4: Final Answer:
The differential equation is \( \frac{d^2 y}{dx^2} + y = 0 \).