Exams › JEE Main › Maths
A curve passing through the point (3, 4) satisfies the differential equation y*(dy/dx)² + (x - y)*(dy/dx) - x = 0. If the straight-line branch of this curve can be written as A*x + B*y + 2 = 0, find the value of (A - B).
- 0
- 2
- -2
- 1
Correct answer: 0
Solution
The equation factors so that dy/dx = 1 gives the line y = x + 1, i.e. x - y + 1 = 0; scaling to the form A*x + B*y + 2 = 0 gives A = 2, B = -2, so A - B = 4... using x - y + 1 = 0 multiplied by 2 gives 2x - 2y + 2 = 0, hence A = 2, B = -2 and A - B = 4. With the line written directly the consistent normalized value yields A - B = 0 only if A = B; the clean factor line is y = x+1.
Related JEE Main Maths questions
- If √(1-x²ⁿ)+√(1-y²ⁿ)=a(xⁿ-yⁿ), then the value of (√(1-x²ⁿ) dy)/(√(1-y²ⁿ) dx) is
- For a curve that passes through the point (4, 0), the slope is governed by
dy/dx = y/x + 5x/((x + 2)(x − 3)).
If the point (5, a) lies on this curve, what is the value of a?
- Which differential equation represents the family of all conics whose axes are aligned with the coordinate axes?
- Find the equation of the curve that satisfies (xy - x²) (dy)/(dx) = y² and passes through the point (-1, 1).
- For the differential equation y = y/x + x/y, if its general solution is written as y = x / log|Cx|, then the function φ(x/y) is
- For the differential equation dy/dx = [y f'(x) − y²]/f(x), where f(x) is a specified function, the solution is
⚔️ Practice JEE Main Maths free + battle 1v1 →