Let y=y(x) be the solution of the differential equation
(3 y²-5 x²) y d x+2 x(x²-y²) d y=0
such that y(1)=1. Then |(y(2))³-12 y(2)| is equal to :

Solution

Maths sir · 4 steps

Every term has total degree 3, so put y = vx and the equation separates neatly.

1. Substitute y = vx. Rearranged, dy/dx = (5x²-3y²)y/2x(x²-y²), which is homogeneous. With y = vx the right side becomes v(5-3v²)/2(1-v²). Subtracting v gives xdv/dx = v(3-v²)/2(1-v²).

   2(1-v²)/v(3-v²) dv = ((2/3)/v - 4/3·v/(3-v²))dv = dx/x

2. Integrate. This gives 2/3 ln|v| + 2/3 ln|3-v²| = ln x + C, so [v(3-v²)]² = c x³. Putting back v = y/x gives v(3-v²) = y(3x²-y²)/x³, so the solution is y²(3x²-y²)² = c x⁹.
3. Use y(1) = 1. At x = 1, y = 1: 1 × (3-1)² = 4 = c. So the curve is [y(3x²-y²)]² = 4x⁹.
4. So. At x = 2: [y(12-y²)]² = 4 × 512 = 2048. Since |y³-12y| = |y(12-y²)|, it equals √2048 = 32√2, option (d).

The idea to keep. When every term has the same total degree, y = vx always separates the equation. Keep the constants in front of the logs; they turn into powers in the final curve.

Answer from the NTA final answer key. Solution by PrepWiser.