The differential equation representing the fa | vedclass

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

Question

(C) The given equation is ${y^2} = \sqrt{c}(x + 2c)$.Since there is only one arbitrary constant $c$,the order of the differential equation is $1$.To f

Vedclass · Accepted Answer

order $= 1$,degree $= 3$

Vedclass · Answer

(C) The given equation is ${y^2} = \sqrt{c}(x + 2c)$.Since there is only one arbitrary constant $c$,the order of the differential equation is $1$.To find the degree,we differentiate with respect to $x$:$2y \frac{dy}{dx} = \sqrt{c}$.Squaring both sides,we get $4y^2 \left(\frac{dy}{dx}ight)^2 = c$.Substitute $\sqrt{c} = 2y \frac{dy}{dx}$ into the original equation:${y^2} = 2y \frac{dy}{dx} (x + 2(4y^2 \left(\frac{dy}{dx}ight)^2))$.${y^2} = 2xy \frac{dy}{dx} + 16y^3 \left(\frac{dy}{dx}ight)^3$.Dividing by $y$ (assuming $y 
eq 0$):$y = 2x \frac{dy}{dx} + 16y^2 \left(\frac{dy}{dx}ight)^3$.The highest power of the derivative $\frac{dy}{dx}$ is $3$,so the degree is $3$.

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

Similar Questions

The differential equation obtained by eliminating the arbitrary constants from the equation $y^{2}=(2 x+c)^{5}$ is

The differential equation obtained by eliminating the arbitrary constants $a$ and $b$ from $xy = ae^x + be^{-x}$ is

The differential equation corresponding to the family of circles having centres on the $X$-axis and passing through the origin is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module