The equation of the curve for which the tangent at P (x, y) cuts the y – axis at (0, y 3…
Mathematics · JEE Main · NTA Exams — Differential Equations
The equation of the curve for which the tangent at P (x, y) cuts the y – axis at (0, y3) is
- xy2 = x + y
- x2 (y2 –1) = cy2
- y (x2 –1) = cx2
- yx + x2 = c
Answer
(B) x 2 (y 2 –1) = cy 2
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- The general solution of the differential equation (1 + y 2 ) dx + (1 + x 2 ) dy = 0 is
- A curve passes through the point (0, 1) and the gradient at (x, y) on it is y (xy – 1). The equation of the…
- Solve ( d y d x ) y= (x+y)+ (x-y) .
- STATEMENT - 1 (Assertion): The differential equation of all circles in a plane must be of order 3. STATEMENT…
- The solution of the differential equation y_ 1 +3 x y=x which passes through (0,4) is,
- The integrating factor of the differentiable equation (xy – 1) dy dx + y ^ 2 =0 is
- The order and degree of the differential equation (1+3 d y d x )^ 2 3 =4 d^ 3 y d x^ 3 is,
- The solution of the differential equation cos y log (sec x + tan x) dx = cos x log (sec y + tan y ) dy is
More Differential Equations questions · Browse all practice questions