If c is a parameter,then the differential equ | vedclass

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

Question

(D) Given the family of curves: $x^2 = c(y+c)^2$. Taking the square root on both sides,we get $x = \sqrt{c}(y+c)$ (assuming $x > 0$ for simplicity,as

Vedclass · Accepted Answer

$x\left(\frac{dy}{dx}\right)^3-y\left(\frac{dy}{dx}\right)^2-1=0$

Vedclass · Answer

(D) Given the family of curves: $x^2 = c(y+c)^2$. Taking the square root on both sides,we get $x = \sqrt{c}(y+c)$ (assuming $x > 0$ for simplicity,as the parameter $c$ absorbs the sign). Differentiating with respect to $x$: $1 = \sqrt{c} \frac{dy}{dx}$,which implies $\sqrt{c} = \frac{dx}{dy}$. Substituting $\sqrt{c} = \frac{dx}{dy}$ into the equation $x = \sqrt{c}(y+c)$: $x = \frac{dx}{dy} \left( y + \left( \frac{dx}{dy} \right)^2 \right)$. Multiplying by $\left( \frac{dy}{dx} \right)^3$: $x \left( \frac{dy}{dx} \right)^3 = \left( \frac{dx}{dy} \cdot \frac{dy}{dx} \right) \left( y \left( \frac{dy}{dx} \right)^2 + 1 \right)$. Since $\frac{dx}{dy} \cdot \frac{dy}{dx} = 1$,we have: $x \left( \frac{dy}{dx} \right)^3 = y \left( \frac{dy}{dx} \right)^2 + 1$. Rearranging the terms,we get $x \left( \frac{dy}{dx} \right)^3 - y \left( \frac{dy}{dx} \right)^2 - 1 = 0$. Thus,option $D$ is correct.

If $c$ is a parameter,then the differential equation of the family of curves $x^2=c(y+c)^2$ is

Similar Questions

The differential equation of all straight lines passing through the point $(1, -1)$ is

The difference between the degree and the order of the differential equation that represents the family of curves given by $y^{2}=a\left(x+\frac{\sqrt{a}}{2}\right)$,where $a>0$,is:

The degree of the differential equation whose solution is $y^2=8a(x+a)$ is

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

The differential equation representing the family of curves ${y^2} = \sqrt{c}(x + 2c)$,where $c$ is a positive parameter,is of

Vedclass Test Series

Exam Paper Generator

Online Exam Module