If the differential equation representing the | vedclass

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

Question

(C) The equation of the family of all circles touching the $x-$axis at the origin $(0, 0)$ with center $(0, a)$ and radius $|a|$ is given by:$x^2 + (y

Vedclass · Accepted Answer

$2x$

Vedclass · Answer

(C) The equation of the family of all circles touching the $x-$axis at the origin $(0, 0)$ with center $(0, a)$ and radius $|a|$ is given by:$x^2 + (y - a)^2 = a^2$$x^2 + y^2 - 2ay + a^2 = a^2$$x^2 + y^2 - 2ay = 0$ ... $(1)$Differentiating both sides with respect to $x$:$2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0$$x + y\frac{dy}{dx} = a\frac{dy}{dx}$$a = \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}}$Substituting the value of $a$ into equation $(1)$:$x^2 + y^2 - 2y \left( \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}} \right) = 0$$(x^2 + y^2)\frac{dy}{dx} - 2y(x + y\frac{dy}{dx}) = 0$$(x^2 + y^2)\frac{dy}{dx} - 2xy - 2y^2\frac{dy}{dx} = 0$$(x^2 - y^2)\frac{dy}{dx} = 2xy$Comparing this with the given equation $(x^2 - y^2)\frac{dy}{dx} = g(x)y$,we get:$g(x)y = 2xy$$g(x) = 2x$

If the differential equation representing the family of all circles touching the $x-$axis at the origin is $(x^2 - y^2)\frac{dy}{dx} = g(x)y$,then $g(x)$ equals

Similar Questions

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

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

The differential equation which represents the family of curves $y = C_1 e^{C_2 x}$,where $C_1$ and $C_2$ are arbitrary constants,is:

The differential equation of the family of circles touching $y$-axis at the origin is

If $A$ and $B$ are arbitrary constants,then the differential equation having $y=Ae^{x}+B \sin 2 x$ as its general solution is

Vedclass Test Series

Exam Paper Generator

Online Exam Module