Let the solution curve y=y(x) of the differential equation
(d y)/(d x)-(3 x⁵ tan ⁻¹(x³))/(1+x⁶)^(3 / 2) y=2 x exp {(x³-tan ⁻¹ x³)/√(1+x⁶)} pass through the origin. Then y(1) is equal to :

Solution

Maths sir · 5 steps

The ugly exponent on the right is exactly what the integrating factor will cancel, so find it first.

1. Linear form. The equation is dy/dx + P y = Q with P = -(3x⁵ tan⁻¹(x³))/(1+x⁶)^(3/2). The integrating factor is e^(∫ P dx), so the work is in ∫ P dx.
2. Two substitutions. Put t = x³, so 3x⁵ dx = t dt. Then put t = tan θ. Since (1+t²)^(3/2) = sec³ θ and dt = sec² θ dθ, the integrand becomes θ sin θ dθ.

   ∫ θ sin θ dθ = sin θ - θ cos θ = (x³ - tan⁻¹ x³)/√(1+x⁶) = F(x)

3. Integrating factor. Here sin θ = t/√(1+t²) and cos θ = 1/√(1+t²). Because P carries a minus sign, ∫ P dx = -F(x) and the integrating factor is e^(-F(x)). The right side is 2x e^(F(x)), so the exponentials cancel.
4. Solve. d/dx(y e^(-F)) = 2x, so y e^(-F) = x² + C. At the origin, F(0) = 0 and y = 0, so C = 0 and y = x² e^(F(x)).
5. So. F(1) = (1 - π/4)/√2 = (4-π)/4√2, so y(1) = exp((4-π)/4√2), which is option (b).

The idea to keep. When the right side of a linear DE has a strange exponential, expect the integrating factor to be its reciprocal. For t/(1+t²)^(3/2) forms, put t = tan θ.

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