The differential equation of the family of cu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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

Similar Questions

The differential equation of all parabolas whose axis is the $y$-axis is:

The differential equation of all parabolas with axis parallel to the $y$-axis is

The differential equation of the family of curves given by $y = a e^{2x} + b e^{5x}$,where $a$ and $b$ are parameters,is:

The differential equation of all parabolas each of which has a latus rectum $4a$ and whose axes are parallel to the $x$-axis is:

The differential equation of $y=e^x(a+bx+x^2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module