Form the differential equation of the family | vedclass

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

Question

(B) Let the centre of the circle on the $y$-axis be $(0, b)$.The equation of the family of circles with centre $(0, b)$ and radius $3$ is:$x^2 + (y -

Vedclass · Accepted Answer

$(x^2 - 9)(y')^2 + x^2 = 0$

Vedclass · Answer

(B) Let the centre of the circle on the $y$-axis be $(0, b)$.The equation of the family of circles with centre $(0, b)$ and radius $3$ is:$x^2 + (y - b)^2 = 3^2$$x^2 + (y - b)^2 = 9$  --- $(1)$Differentiating equation $(1)$ with respect to $x$,we get:$2x + 2(y - b) \cdot y' = 0$$(y - b) \cdot y' = -x$$(y - b) = -\frac{x}{y'}$Substituting the value of $(y - b)$ in equation $(1)$:$x^2 + \left(-\frac{x}{y'}\right)^2 = 9$$x^2 + \frac{x^2}{(y')^2} = 9$Multiplying by $(y')^2$:$x^2(y')^2 + x^2 = 9(y')^2$$x^2(y')^2 - 9(y')^2 + x^2 = 0$$(x^2 - 9)(y')^2 + x^2 = 0$

Form the differential equation of the family of circles having centre on the $y$-axis and radius $3$ units.

Similar Questions

The differential equation obtained by eliminating $A$ and $B$ from $y = A \cos \omega t + B \sin \omega t$ is:

If the order and degree of the differential equation corresponding to the family of curves $y^2=4a(x+a)$ (where $a$ is a parameter) are $m$ and $n$ respectively,then $m+n^2=$

The differential equation of $y=e^x(a+bx+x^2)$ 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

If $k$ and $l$ respectively are the order and degree of the differential equation whose general solution represents the family of circles of constant radius,then $k^2+l^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module