Let y = y(x) be the solution of the differential equation (1 - x²)dy = ( xy + (x³ + 2)√(1 - x²) )dx, - 1 < x < 1, and y(0) = 0. If ∫_((- 1)/2)^(1/2) √(1 - x²) y(x)dx = k, then k⁻¹ is equal to _____________.

Solution

Maths sir · 5 steps

The integrating factor is exactly √(1-x²), so the integral asks for y√(1-x²) directly.

1. Standard linear form. Divide by (1-x²) dx to get dy/dx - x/(1-x²) y = (x³+2)/√(1-x²). Here P = -x/(1-x²).
2. Integrating factor. ∫ P dx = 1/2 ln(1-x²), so the integrating factor is e^(1/2 ln(1-x²)) = √(1-x²). Multiplying by it gives d/dx(y√(1-x²)) = x³ + 2.
3. Integrate and fix C. y√(1-x²) = x⁴/4 + 2x + C. Putting x = 0, y = 0 gives C = 0.

   k = ∫_(-1/2)^(1/2)(x⁴/4 + 2x)dx = 2∫₀^(1/2)x⁴/4 dx

4. Use symmetry. The odd term 2x integrates to zero over [-1/2, 1/2]. The even term gives 2 × 1/20(1/2)⁵ = 2 × 1/640 = 1/320.
5. So. k = 1/320, so k⁻¹ = 320.

The common slip: 640. Integrating only from 0 to 1/2 and forgetting to double the even part gives k = 1/640, so k⁻¹ = 640.

The idea to keep. In a linear DE, always check whether the integrating factor already appears in the asked integral. If it does, you never need y alone, only y × IF.

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