If x=f(y) is the solution of the differential equation (1+y²)+(x-2 e^(tan ⁻¹ y)) (d y)/(d x)=0, y ∈(-π/2, π/2) with f(0)=1, then f(1/√3) is equal to :

Solution

Maths sir · 4 steps

Flip it to dx/dy and it becomes linear in x, with e^(tan⁻¹ y) as the integrating factor.

1. Make x the unknown. Multiply the equation by dx/dy to get (1+y²)dx/dy + x = 2e^(tan⁻¹ y). Dividing by 1+y² gives a linear equation in x, with y as the variable.

   dx/dy + x/(1+y²) = 2e^(tan⁻¹ y)/(1+y²)

2. Integrating factor. The integrating factor is e^(∫ dy/(1+y²)) = e^(tan⁻¹ y). So x e^(tan⁻¹ y) = ∫ 2e^(2 tan⁻¹ y)/(1+y²) dy. Put t = tan⁻¹ y and the integral becomes ∫ 2e²ᵗ dt = e^(2 tan⁻¹ y) + C.
3. Use f(0) = 1. At y = 0 we have tan⁻¹ 0 = 0, so 1 × 1 = 1 + C and C = 0. Dividing through by e^(tan⁻¹ y) gives x = f(y) = e^(tan⁻¹ y).
4. So. tan⁻¹ 1/√3 = π/6, so f(1/√3) = e^(π/6), which is option (c).

The idea to keep. When the equation is linear in x but not in y, treat x as the dependent variable and work with dx/dy + P(y) x = Q(y).

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