The differential equation d y d x = 1-y^ 2 y determines a family of circles with
Mathematics · JEE Main · NTA Exams — Differential Equations
The differential equation \(\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{y}\) determines a family of circles with
- variable radii and a fixed centre at (0, 1)
- variable radii and a fixed centre at (0, -1)
- fixed radius 1 and variable centres along the x-axis
- fixed radius 1 and variable centres along the y-axis
Answer
(C) fixed radius 1 and variable centres along the x-axis
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- The degree of the differential equation [1+ ( d y d x )^ 2 ]^ 5 3 = d^ 2 y d x^ 2 is :
- Consider the differential equation y^ 2 d x+ (x- 1 y ) d y=0 . If y (1) = 1, then x is given by
- The solution of the differential equation y dx + (x + x 2 y) dy = 0 is
- A tangent to the curve y = f (x) cuts the line y = x at a point which is at a distance of 1 unit from y–axis…
- The solution of the differential equation [y (1+x^ -1 )+ y ] d x+(x+ x+x y) d y=0 is,
- The differential equation representing the family of curves y ^ 2 =2 c ( x + c ) . Where c>0 , is a parameter…
- The solution of the differential equation d y d x + 1+x^ 2 x =0 is
- The solution of the differential equation d y d x = x+y x satisfying the condition y (1) = 1 is
More Differential Equations questions · Browse all practice questions