MHT CET20239 May 2023Evening ShiftMathematicsDifferential EquationsActual
The differential equation of all circles which pass through the origin and whose centres lie on Y -axis is
Options
- A(x^2-y^2 ) d y d x -2 x y=0
- B(x^2-y^2 ) d y d x +2 x y=0
- C(x^2-y^2 ) d y d x +x y=0
- D(x^2-y^2 ) d y d x -x y=0
Correct answer
A. (x^2-y^2 ) d y d x -2 x y=0
Step-by-step solution
Circle passes through origin and centre lie on Y -axis. Let (0, k) be centre and ' k ' be radius Equation of circle is aligned & (x-0)^2+(y- k )^2= k ^2 & x^2+y^2-2 y k + k ^2= k ^2 & x^2+y^2-2 k y=0 & x^2+y^2=2 k y ...(i) & x^2+y^2 2 y = k ...(ii) aligned Differentiating equation (i) with respect to x , we get aligned & 2 x+2 y d y ~d x =2 k d y ~d x & 2 x+2 y d y ~d x -2 k d y ~d x =0 & 2 x+2(y- k ) d y ~d x =0 & 2 x+2 [y- ( x^2+y^2 2 y ) ] d y ~d x =0 ...[From(ii)] & 2 x+2 [ 2 y^2-x^2-y^2 2 y ] d y ~d x =0 & 2 x