The differential equation representing the family of curves \(y=A\cos(x+B)\), where A and B are arbitrary constant 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: they are eliminated with the help of two derivatives.
\(y = A\cos(x+B)\)
\( \displaystyle \frac{dy}{dx} = -A\sin(x+B) \)
\( \displaystyle \frac{d^2y}{dx^2} = -A\cos(x+B) \)
The right side is \(-y\), so
\( \displaystyle \frac{d^2y}{dx^2} = -y \)
\( \displaystyle \frac{d^2y}{dx^2} + y = 0 \)