Let y=y(x) be the solution of the differential equation (1+x²) (d y)/(d x)+y=e^(tan ⁻¹ x), y(1)=0. Then y(0) is

Solution

Maths sir · 4 steps

Divide by 1+x² first, and the integrating factor e^(tan⁻¹ x) almost hands you the answer.

1. Standard form. Divide through by 1+x² to get dy/dx + 1/(1+x²) y = e^(tan⁻¹ x)/(1+x²). This is linear with P = 1/(1+x²), so the integrating factor is e^(∫ P dx) = e^(tan⁻¹ x).

   y e^(tan⁻¹ x) = ∫ e^(2 tan⁻¹ x)/(1+x²) dx = 1/2e^(2 tan⁻¹ x) + C

2. Integrate. Put t = tan⁻¹ x, so dt = dx/(1+x²). The integral becomes ∫ e²ᵗ dt = 1/2e²ᵗ, which gives the line above.
3. Use y(1) = 0. At x = 1, tan⁻¹ 1 = π/4, so 0 = 1/2e^(π/2) + C and C = -1/2e^(π/2).
4. So. At x = 0, tan⁻¹ 0 = 0, so y(0) × 1 = 1/2 - 1/2e^(π/2) = 1/2(1 - e^(π/2)). The answer is (b).

The idea to keep. Always bring a linear DE to the form y' + Py = Q before finding the integrating factor. When P = 1/(1+x²), the factor is e^(tan⁻¹ x) and the substitution t = tan⁻¹ x does the rest.

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