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(B) Given the family of curves $y = a \cos(x + b)$,where $a$ and $b$ are arbitrary constants.Differentiating with respect to $x$,we get:$\frac{dy}{dx}

Vedclass · Accepted Answer

$\frac{d^2y}{dx^2} + y = 0$

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(B) Given the family of curves $y = a \cos(x + b)$,where $a$ and $b$ are arbitrary constants.Differentiating with respect to $x$,we get:$\frac{dy}{dx} = -a \sin(x + b)$Differentiating again with respect to $x$,we get:$\frac{d^2y}{dx^2} = -a \cos(x + b)$Since $y = a \cos(x + b)$,we can substitute this into the second derivative:$\frac{d^2y}{dx^2} = -y$Rearranging the terms,we get:$\frac{d^2y}{dx^2} + y = 0$Thus,the correct option is $B$.

The differential equation of the family of curves $y = a \cos(x + b)$ is

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