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

(A) Given the equation of the family of curves: ${x^2}y = a$. Differentiating both sides with respect to $x$ using the product rule: $\frac{d}{dx}({x^

Vedclass · Accepted Answer

$\frac{dy}{dx} + \frac{2y}{x} = 0$

Vedclass · Answer

(A) Given the equation of the family of curves: ${x^2}y = a$. Differentiating both sides with respect to $x$ using the product rule: $\frac{d}{dx}({x^2}y) = \frac{d}{dx}(a)$ ${x^2} \frac{dy}{dx} + y \frac{d}{dx}({x^2}) = 0$ ${x^2} \frac{dy}{dx} + y(2x) = 0$ ${x^2} \frac{dy}{dx} + 2xy = 0$ Dividing the entire equation by ${x^2}$ (assuming $x 
eq 0$): $\frac{dy}{dx} + \frac{2xy}{x^2} = 0$ $\frac{dy}{dx} + \frac{2y}{x} = 0$. Thus,the correct option is $A$.

The differential equation of the family of curves represented by the equation ${x^2}y = a$ is:

Similar Questions

The differential equation of the family of circles whose center lies on the $x$-axis is:

If $Ax^3+Bxy=4$ (where $A$ and $B$ are arbitrary constants) is the general solution of the differential equation $F(x) \frac{d^2 y}{d x^2}+G(x) \frac{d y}{d x}-2 y=0$,then $F(1)+G(1)=$

The differential equation of which $xy = ae^x + be^{-x} + x^2$ is a solution,is

The differential equation having the general solution $y=c(x-c)^2$ ($c$ is an arbitrary constant) is

The differential equation of the family of curves $v = \frac{A}{r} + B$,where $A$ and $B$ are arbitrary constants,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module