Let y=f(x) be the solution of the differential equation (d y)/(d x)+(x y)/(x²-1)=(x⁶+4 x)/√(1-x²),-1< x<1 such that f(0)=0. If 6 ∫_(-1 / 2)^(1 / 2) f(x) d x=2 π-α then α² is equal to _________ .

Solution

Maths sir · 5 steps

Find the integrating factor, then let the odd part of f vanish over the symmetric limits.

1. Integrating factor. The equation is linear with P(x) = x/(x²-1), so ∫ P dx = 1/2 ln|x²-1|. On -1<x<1 this gives IF = √(1-x²).
2. Solve for f. Multiplying by the IF cancels the square root on the right: d/dx(y√(1-x²)) = x⁶+4x. Integrating, y√(1-x²) = x⁷/7 + 2x² + C, and f(0)=0 gives C=0.

   f(x) = x⁷/7√(1-x²) + 2x²/√(1-x²)

3. Use symmetry. The first term is odd, so its integral over [-1/2,1/2] is 0. The second is even, so ∫_(-1/2)^(1/2) f(x) dx = 4∫₀^(1/2)x²/√(1-x²) dx.
4. Trig substitution. Put x = sin θ, so the limits become 0 to π/6 and the integral becomes ∫₀^(π/6)sin² θ dθ = [θ/2 - (sin 2θ)/4]₀^(π/6) = π/12 - √3/8. Multiplying by 4 gives π/3 - √3/2.
5. So. 6∫_(-1/2)^(1/2) f(x) dx = 2π - 3√3, so α = 3√3 and α² = 27.

The idea to keep. When the limits are symmetric, split the integrand into odd and even parts first. The odd part drops out, and often the hardest-looking piece goes with it.

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