The differential equation of all circles in t | vedclass

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

Question

(A) The equation of a circle in the first quadrant that touches both coordinate axes is given by $(x - a)^2 + (y - a)^2 = a^2$,where $a$ is the radius

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The equation of a circle in the first quadrant that touches both coordinate axes is given by $(x - a)^2 + (y - a)^2 = a^2$,where $a$ is the radius of the circle and acts as an arbitrary constant.Since there is only one arbitrary constant $a$,the family of curves is a one-parameter family.The order of the differential equation for a family of curves with $n$ arbitrary constants is $n$.Therefore,for this family of circles,the order of the differential equation is $1$.

The differential equation of all circles in the first quadrant which touch the coordinate axes is of order

Similar Questions

The number of arbitrary constants in the particular solution of a differential equation of third order is:

Determine the order and degree (if defined) of the differential equation $y^{\prime \prime} + 2y^{\prime} + \sin(y) = 0$.

The number of arbitrary constants in the general solution of a fourth-order differential equation is . . . . . . .

Determine the order and degree (if defined) of the differential equation $\left(\frac{d^{2} y}{d x^{2}}\right)^{2} = \cos 3x + \sin 3x$.

If $c_1, c_2, c_3, c_4, c_5$ are arbitrary constants,then the order of the differential equation whose general solution is $y=(c_1+c_2) \sin (x+c_3)+c_4 e^{x+c_5}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module