The differential equation corresponding to th | vedclass

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

Question

(A) Given the family of parabolas is $y^2 = 4a(x+a) \quad ...(i)$Differentiating with respect to $x$,we get:$2y \frac{dy}{dx} = 4a$$\Rightarrow a = \f

Vedclass · Accepted Answer

$y\left(\frac{dy}{dx}\right)^2+2x\frac{dy}{dx}-y=0$

Vedclass · Answer

(A) Given the family of parabolas is $y^2 = 4a(x+a) \quad ...(i)$Differentiating with respect to $x$,we get:$2y \frac{dy}{dx} = 4a$$\Rightarrow a = \frac{1}{2} y \frac{dy}{dx} \quad ...(ii)$Substituting the value of $a$ from equation $(ii)$ into equation $(i)$:$y^2 = 4 \left( \frac{1}{2} y \frac{dy}{dx} ight) \left( x + \frac{1}{2} y \frac{dy}{dx} ight)$$y^2 = 2y \frac{dy}{dx} \left( x + \frac{1}{2} y \frac{dy}{dx} ight)$Dividing by $y$ (assuming $y 
eq 0$):$y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} ight)^2$Rearranging the terms,we get:$y \left( \frac{dy}{dx} ight)^2 + 2x \frac{dy}{dx} - y = 0$

The differential equation corresponding to the family of parabolas $y^2=4a(x+a)$,where $a$ is the parameter,is

Similar Questions

The differential equation of all circles which pass through the origin and whose centres lie on the $y$-axis is:

The differential equation corresponding to the family of circles in the plane touching the $Y$-axis at the origin is:

The differential equation for which $l x^2+m y^2=x+y$ is the general solution is

The differential equation of the family of parabolas with focus at the origin and the axis as the $X$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module