Question

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

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

Hint:

For \(y=a\sin(x+b)\), differentiating twice gives the relation \(y''=-y\).

Answer 1

The Correct Option is B

Concept: To form a differential equation from a family of curves, differentiate enough times to eliminate arbitrary constants.

Step 1: Given: \[ y=a\sin(x+b) \] Here \(a\) and \(b\) are arbitrary constants.

Step 2: Differentiate once. \[ \frac{dy}{dx}=a\cos(x+b) \]

Step 3: Differentiate again. \[ \frac{d^2y}{dx^2}=-a\sin(x+b) \]

Step 4: But from the original equation: \[ y=a\sin(x+b) \] Therefore: \[ \frac{d^2y}{dx^2}=-y \]

Step 5: Rearrange: \[ \frac{d^2y}{dx^2}+y=0 \] Therefore, \[ \boxed{\frac{d^2y}{dx^2}+y=0} \]