Let y=y(x) be the solution curve of the differential equation
x(x²+eˣ) d y+(eˣ(x-2) y-x³) d x=0, x>0, passing through the point (1,0). Then y(2) is equal to :

Solution

Maths sir · 4 steps

Put it in linear form, and the integrating factor 1 + eˣ/x² cancels the right side neatly.

1. Linear form. Divide by x(x²+eˣ) dx to get dy/dx + Py = Q with P = eˣ(x-2)/x(x²+eˣ) and Q = x²/(x²+eˣ).
2. Integrating factor. Try ln(x²+eˣ) - 2 ln x. Its derivative is (2x+eˣ)/(x²+eˣ) - 2/x = (xeˣ - 2eˣ)/x(x²+eˣ), which is exactly P. So the integrating factor is (x²+eˣ)/x².

   y·(x²+eˣ)/x² = ∫ x²/(x²+eˣ)·(x²+eˣ)/x² dx = x + C

3. Use the point. At (1, 0) the left side is 0, so 0 = 1 + C and C = -1. This gives y = x²(x-1)/(x²+eˣ).
4. So. y(2) = (4 × 1)/(4+e²) = 4/(4+e²), which is option (c).

The idea to keep. When P is a messy fraction, guess ln(denominator pieces) and differentiate to check. The right integrating factor usually makes Q times it collapse to something trivial.

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