If a and b are arbitrary constants,then the d | vedclass

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

Question

(C) Given the equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \

Vedclass · Accepted Answer

$x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0$

Vedclass · Answer

(C) Given the equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{y}{b^2} \frac{dy}{dx} = -\frac{x}{a^2} \implies \frac{y}{x} \frac{dy}{dx} = -\frac{b^2}{a^2}$. Differentiating again with respect to $x$: $\frac{d}{dx} \left( \frac{y}{x} \frac{dy}{dx} \right) = \frac{d}{dx} \left( -\frac{b^2}{a^2} \right) = 0$. Using the product rule: $\frac{y}{x} \frac{d^2y}{dx^2} + \frac{dy}{dx} \left( \frac{x \frac{dy}{dx} - y}{x^2} \right) = 0$. Multiplying by $x^2$: $xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0$.

If $a$ and $b$ are arbitrary constants,then the differential equation having $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as its general solution is

Similar Questions

The differential equation of all straight lines passing through the origin is

The differential equation of all circles,passing through the origin and having their centres on the $X$-axis,is

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

The differential equation of a family of hyperbolas whose axes are parallel to coordinate axes,centres lie on the line $y=2x$ and eccentricity is $\sqrt{3}$ is

Form the differential equation of the family of parabolas having vertex at origin and axis along the positive $y$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module