Among the options given below,from which opti | vedclass

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

Question

(A) The order of a differential equation is equal to the number of arbitrary constants present in the general equation of the family of curves. For op

Vedclass · Accepted Answer

All circles passing through the origin

Vedclass · Answer

(A) The order of a differential equation is equal to the number of arbitrary constants present in the general equation of the family of curves. For option $A$,the equation of a circle passing through the origin is $x^2 + y^2 + 2gx + 2fy = 0$. Here,$g$ and $f$ are two independent arbitrary constants. Since there are $2$ arbitrary constants,the differential equation formed will be of order $2$. For option $B$,the equation is $y^2 = 4a(x-h)$,which involves two constants,but the condition of passing through the origin and focus on the $x$-axis restricts it. For option $C$,the equation is $y = mx$,which has only $1$ arbitrary constant. For option $D$,the equation $x^2 - y^2 = k^2$ has only $1$ arbitrary constant $k$. Therefore,the correct option is $A$.

Among the options given below,from which option can a differential equation of order two be formed?

Similar Questions

The order of the differential equation whose solution is $y=a \cos x+b \sin x+c e^{-x}$ is

Verify that the given function $xy = \log y + C$ is a solution of the differential equation $y' = \frac{y^2}{1 - xy}$ $(xy \neq 1)$.

The order of the differential equation corresponding to the family of parabolas whose axes are along the $X$-axis and whose foci are at the origin,is

The differential equation corresponding to the family of circles in the plane touching the $Y$-axis at the origin is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module