Saved Bookmarks
| 1. |
The differential equation of all circles which passes through the origin and whose centers lie on Y-axis isA. `(x^2-y^2)(dy)/(dx)-2xy=0`B. `(x^2-y^2)(dy)/(dx)+2xy=0`C. `(x^2-y^2)(dy)/(dx)-xy=0`D. `(x^2-y^2)(dy)/(dx)+xy=0` |
|
Answer» Correct Answer - A Let `x^2+y^2-2ky=0` .......(i) `rArr 2x+2y(dy)/(dx)-2k(dy)/(dx)=0` `rArr k=(x)/(((dy)/(dx)))+y` From Eq.(i) , we get `x^2+y^2-2((x)/((dy//dx))+y)y=0` `rArr (x^2-y^2)(dy)/(dx)-2xy=0` |
|