Let y=y(x) be the solution of the differential equation (1-x²) d y=[x y+(x³+2) √3(1-x²)] d x, -1< x<1, y(0)=0. If y(1/2)=m/n, m and n are co-prime numbers, then m+n is equal to __________.

Solution

Maths sir · 4 steps

Divide by 1-x²; the integrating factor √(1-x²) then cancels the square root on the right.

1. Standard linear form. Divide both sides by (1-x²) dx. This gives dy/dx - x/(1-x²) y = (√3 (x³+2))/√(1-x²), so P = -x/(1-x²).
2. Integrating factor. ∫ -x/(1-x²) dx = 1/2 ln(1-x²), so the integrating factor is e^(1/2 ln(1-x²)) = √(1-x²). Multiplying the right side by it leaves just √3 (x³+2).

   y√(1-x²) = √3∫ (x³+2) dx = √3(x⁴/4 + 2x) + C

3. Use the condition. y(0) = 0 gives 0 = 0 + C, so C = 0. At x = 1/2, √(1-1/4) = √3/2, so y × √3/2 = √3(1/64 + 1) = 65√3/64.
4. So. y(1/2) = 65/64 × 2 = 65/32. 65 and 32 are co-prime, so m + n = 65 + 32 = 97.

The common slip: 25. Integrating x³ as x⁴ instead of x⁴/4 gives y = 17/8 and so 25, not 97.

The idea to keep. When a factor like √(1-x²) appears on the right, expect the integrating factor to cancel it. Always write the equation in the form dy/dx + Py = Q first and keep the sign of P.

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