The differential equation representing the family of curves \(y=a\sin(x+b)\), where a and b are arbitrary constants is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{d^2y}{dx^2}+y = 0\)
The family has two arbitrary constants, a and b, so its differential equation is of order 2; it is found by differentiating twice and eliminating a and b.
\(y = a\sin(x+b)\)
\( \displaystyle \frac{dy}{dx} = a\cos(x+b) \)
\( \displaystyle \frac{d^2y}{dx^2} = -a\sin(x+b) \)
\( \displaystyle \frac{d^2y}{dx^2} = -y \)
Hence
\( \displaystyle \frac{d^2y}{dx^2} + y = 0 \)