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: Concept
Differentiate the equation twice to eliminate the two arbitrary constants $a$ and $b$.

Step 2: Meaning
First derivative: $\frac{dy}{dx} = a \cos(x + b)$.

Step 3: Analysis
Second derivative: $\frac{d^2y}{dx^2} = -a \sin(x + b)$. Substituting the original $y = a \sin(x+b)$ gives $\frac{d^2y}{dx^2} = -y$.

Step 4: Conclusion
Rearranging the terms, we get $\frac{d^{2}y}{dx^{2}} + y = 0$.
Final Answer: (B)