The differential equation representing the fa | vedclass

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

Question

(D) Given equation: $y^2=2 c(x+\sqrt{c}) \dots (i)$Differentiating with respect to $x$:$2 y \frac{dy}{dx} = 2c \implies c = y \frac{dy}{dx} \dots (ii)

Vedclass · Accepted Answer

order $1$,degree $3$

Vedclass · Answer

(D) Given equation: $y^2=2 c(x+\sqrt{c}) \dots (i)$Differentiating with respect to $x$:$2 y \frac{dy}{dx} = 2c \implies c = y \frac{dy}{dx} \dots (ii)$Substituting $(ii)$ into $(i)$:$y^2 = 2 \left(y \frac{dy}{dx}ight) \left(x + \sqrt{y \frac{dy}{dx}}ight)$Dividing by $y$ (assuming $y 
eq 0$):$y = 2 \frac{dy}{dx} \left(x + \sqrt{y \frac{dy}{dx}}ight)$$y - 2x \frac{dy}{dx} = 2 \frac{dy}{dx} \sqrt{y \frac{dy}{dx}}$Squaring both sides:$(y - 2x \frac{dy}{dx})^2 = 4 \left(\frac{dy}{dx}ight)^2 \left(y \frac{dy}{dx}ight)$$(y - 2x \frac{dy}{dx})^2 = 4y \left(\frac{dy}{dx}ight)^3$The highest order derivative is $\frac{dy}{dx}$,so the order is $1$. The power of the highest order derivative is $3$,so the degree is $3$.

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

Similar Questions

The differential equation representing the family of parabolas having vertex at the origin and axis along the positive $Y$-axis is

The differential equation of all circles having center on the $x$-axis and touching the $y$-axis at the origin is:

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

If $l$ and $m$ are the order and degree of the differential equation of all straight lines at a constant distance of $P$ units from the origin,then $l m^2+l^2 m=$

$y=A e^x+B e^{-2 x}$ satisfies which of the following differential equations?

Vedclass Test Series

Exam Paper Generator

Online Exam Module