$\begin{array}{1 1} (A)\;(x^2 - 9)(y')^2 + x^2 = 0 \\ (B)\;(x^2 - 9)(y')^2 - x^2 = 0 \\ (C)\;(x^2 - 9)(y')^2 + x^3 = 0 \\ (D)\;(x^2 - 9)(y')^2 + x^4 = 0 \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- Equation of a circle having centre on the y axis and radius a units is $x^2 + (y-h)^2 = a^2$

Step 1:

From the information in the tool box we obtain the equation of the circle as

$x^2 + (y-h)^2 = 9$-----(1)

On differentiating on both sides with respect to $x$ we get,

$2x + 2(y-h).y' = 0$

$y-h =\large\frac{ x}{y'}$

Step 2:

Substituting for $y-h$ in equ(1) we get

$x^2 + (\large\frac{x}{y'})^2$$ = 9$

on simplifying and rearranging we get

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

This is the required differential equation.

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...