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 whose solution represents the family $x^2 y = 4e^x + c$,where $c$ is an arbitrary constant,is

The differential equation,having general solution as $A x^2+B y^2=1$,where $A$ and $B$ are arbitrary constants,is

The differential equation of all parabolas whose axes are parallel to the $Y$-axis is:

The order of the differential equation of the family of all concentric circles centered at $(h, k)$ is

If $a$ and $b$ are arbitrary constants,then the differential equation corresponding to the family of curves given by $y=x[a \cos (\log x)+b \sin (\log x)]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module