The differential equation of the family of st | vedclass

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

Question

(A) The equation of a straight line with slope $m$ and $y$-intercept $c$ is given by $y = mx + c$. Given that the slope is equal to the $y$-intercept,

Vedclass · Accepted Answer

$(x+1) \frac{dy}{dx} - y = 0$

Vedclass · Answer

(A) The equation of a straight line with slope $m$ and $y$-intercept $c$ is given by $y = mx + c$. Given that the slope is equal to the $y$-intercept,we have $m = c$. Substituting this into the line equation,we get $y = cx + c = c(x+1)$. Differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = c$. Substituting the value of $c$ from the first equation into the derivative,we have $\frac{dy}{dx} = \frac{y}{x+1}$. Rearranging the terms,we get $(x+1) \frac{dy}{dx} - y = 0$.

The differential equation of the family of straight lines whose slope is equal to $y$-intercept is

Similar Questions

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

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

The differential equation corresponding to the family of ellipses $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$, where '$a$' is an arbitrary constant, is:

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

The differential equation for which $\sin^{-1} x + \sin^{-1} y = c$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module