The differential equation corresponding to al | vedclass

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

Question

(A) Let the radius of the circle be $a$. Then the center of the circle is $(a, a)$. Hence, the equation of the circle is:$(x-a)^2 + (y-a)^2 = a^2$$x^2

Vedclass · Accepted Answer

$(x-y)^2\left[1+\left(\frac{d y}{d x}\right)^2\right]=\left(x+y \frac{d y}{d x}\right)^2$

Vedclass · Answer

(A) Let the radius of the circle be $a$. Then the center of the circle is $(a, a)$. Hence, the equation of the circle is:$(x-a)^2 + (y-a)^2 = a^2$$x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = a^2$$x^2 + y^2 - 2ax - 2ay + a^2 = 0$  --- $(i)$Differentiating equation $(i)$ with respect to $x$, we get:$2x + 2y \frac{dy}{dx} - 2a - 2a \frac{dy}{dx} = 0$$x + y \frac{dy}{dx} - a(1 + \frac{dy}{dx}) = 0$$a = \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}}$Substituting the value of $a$ into equation $(i)$:$(x-a)^2 + (y-a)^2 = a^2$$(x-a)^2 + (y-a)^2 = (\frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}})^2$Note that $(x-a) = x - \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{x + x \frac{dy}{dx} - x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{(x-y) \frac{dy}{dx}}{1 + \frac{dy}{dx}}$And $(y-a) = y - \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{y + y \frac{dy}{dx} - x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{y-x}{1 + \frac{dy}{dx}}$Substituting these into $(x-a)^2 + (y-a)^2 = a^2$:$(\frac{(x-y) \frac{dy}{dx}}{1 + \frac{dy}{dx}})^2 + (\frac{y-x}{1 + \frac{dy}{dx}})^2 = (\frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}})^2$$(x-y)^2 (\frac{dy}{dx})^2 + (x-y)^2 = (x + y \frac{dy}{dx})^2$$(x-y)^2 [1 + (\frac{dy}{dx})^2] = (x + y \frac{dy}{dx})^2$

The differential equation corresponding to all the circles lying in the first quadrant and touching the coordinate axes is

Similar Questions

The differential equation of all parabolas with axis parallel to the $y$-axis is

If $m$ and $n$ are respectively the order and the degree of the differential equation representing the family of curves $y^2 - 5ax - 5a^{3/2} = 0$ ($a > 0$ is a parameter), then the value of $m - n$ is

The differential equation of which $xy = ae^x + be^{-x} + x^2$ is a solution, is

The differential equation of all straight lines passing through the point $(1, -1)$ is

The differential equation of the family of curves for which the length of the normal is equal to a constant $k$,is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module