The differential equation of the family of li | vedclass

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

Question

(D) Given the family of lines: $y = mx + \frac{4}{m}$  $(1)$ Differentiating with respect to $x$: $\frac{dy}{dx} = m$ Substituting the value of $m$ in

Vedclass · Accepted Answer

$x\left(\frac{dy}{dx}\right)^2 - y\left(\frac{dy}{dx}\right) + 4 = 0$

Vedclass · Answer

(D) Given the family of lines: $y = mx + \frac{4}{m}$  $(1)$ Differentiating with respect to $x$: $\frac{dy}{dx} = m$ Substituting the value of $m$ into equation $(1)$: $y = \left(\frac{dy}{dx}\right)x + \frac{4}{\left(\frac{dy}{dx}\right)}$ Multiplying both sides by $\frac{dy}{dx}$: $y\left(\frac{dy}{dx}\right) = x\left(\frac{dy}{dx}\right)^2 + 4$ Rearranging the terms: $x\left(\frac{dy}{dx}\right)^2 - y\left(\frac{dy}{dx}\right) + 4 = 0$

The differential equation of the family of lines $y = mx + \frac{4}{m}$ obtained by eliminating the arbitrary constant $m$ is

Similar Questions

The degree of the differential equation obtained from the equation $(y - a)^2 = 4(x - b)$ [where $a$ and $b$ are arbitrary constants] is

The differential equation of the family of circles touching the $y$-axis at the origin is

The differential equation obtained by eliminating the arbitrary constant from the equation $y^2 = (x + c)^3$ is

$y = mx + \frac{2}{m}$ is the general solution of

The differential equation of the family of curves $y=e^{x}(A \cos x+B \sin x)$,where $A$ and $B$ are arbitrary constants,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module