The differential equation representing the fa | vedclass

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

Question

(C) Given, $y^{2}=2 d(x+\sqrt{d})$  $(i)$Differentiating with respect to $x$, we get:$2y y_{1} = 2d \Rightarrow d = y y_{1}$Substituting $d = y y_{1}$

Vedclass · Accepted Answer

degree $3$

Vedclass · Answer

(C) Given, $y^{2}=2 d(x+\sqrt{d})$  $(i)$Differentiating with respect to $x$, we get:$2y y_{1} = 2d \Rightarrow d = y y_{1}$Substituting $d = y y_{1}$ into equation $(i)$:$y^{2} = 2(y y_{1})(x + \sqrt{y y_{1}})$Rearranging the terms to isolate the radical:$y^{2} - 2y y_{1} x = 2y y_{1} \sqrt{y y_{1}}$Squaring both sides:$(y^{2} - 2y y_{1} x)^{2} = (2y y_{1})^{2} (y y_{1})$$(y^{2} - 2y y_{1} x)^{2} = 4(y y_{1})^{3}$The highest derivative present is $y_{1}$ (order $1$), and the highest power of the highest derivative is $3$. Therefore, the degree of the differential equation is $3$.

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

Similar Questions

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

For the differential equation $\frac{d^3 y}{d x^3}=0$,$y=a x^2+b x+c$ is

The order and degree of the differential equation $x \left( \frac{dy}{dx} \right)^3 + 2 \left( \frac{d^2y}{dx^2} \right)^2 + 3y + x = 0$ are respectively

The degree of the differential equation $1+\left(\frac{dy}{dx}\right)^2+\left(\frac{d^2y}{dx^2}\right)^2=\sqrt[3]{\frac{d^2y}{dx^2}+1}$ is

The differential equation of all parabolas having their axis of symmetry coinciding with the $x$-axis has its order and degree respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module