MHT CET202527 Apr 2025Evening ShiftMathematicsDifferential EquationsActual
The slope of the tangent at (x, y) to the curve passing through (2,1) is x^2+y^2 2 x y , then the equation of the curve is
Options
- Aa ^2 -a -a +c ( . where . = ⁻¹ ( x a ) ) and c is the constant of integration
- Ba ^2 - a + a + c ( . where . = ⁻¹ ( x a ) ) and c is the constant of integration
- Ca ^2 +a -a +c ( where = ⁻¹ ( x a ) ) and c is the constant of integration
- Da ^2 +a +a +c ( . where . = ⁻¹ ( x a ) ) and c is the constant of integration
Correct answer
A. a ^2 -a -a +c ( . where . = ⁻¹ ( x a ) ) and c is the constant of integration
Step-by-step solution
Given the homogeneous differential equation: dy dx = x^2 + y^2 2xy Substitute y = vx so that dy dx = v + x dv dx . Substituting into the equation yields: v + x dv dx = x^2 + v^2x^2 2vx^2 Simplifying the right-hand side gives: v + x dv dx = 1 + v^2 2v Separating variables: x dv dx = 1 - v^2 2v 2v 1 - v^2 dv = 1 x dx Integrate both sides: 2v 1 - v^2 dv = 1 x dx Let u = 1 - v^2 , so du = -2v dv : - 1 u du = 1 x dx - |1 - v^2| = |x| + C₁ Exponentiating both sides and setting C₁ = |C| : 1 |1 - v^2| = C|x| Substitute v =