Let y=y(x) be the solution of the differential equation (1+y²) eᵗᵃⁿˣ d x+cos ² x(1+e²ᵗᵃⁿˣ) d y=0, y(0)=1. Then y(π/4) is equal to

Solution

Maths sir · 5 steps

Divide to separate x and y, then let t = eᵗᵃⁿˣ: both sides become inverse tangents.

1. Separate variables. Divide the equation by (1+y²)cos² x (1+e²ᵗᵃⁿˣ). Since 1/(cos² x) = sec² x, every y is now on one side and every x on the other.

   dy/(1+y²) = -(eᵗᵃⁿˣ sec² x)/(1+(eᵗᵃⁿˣ)²) dx

2. Substitute. Put t = eᵗᵃⁿˣ, so dt = eᵗᵃⁿˣ sec² x dx. The right side becomes -dt/(1+t²). Integrating both sides gives tan⁻¹ y = -tan⁻¹(eᵗᵃⁿˣ) + C.
3. Use y(0) = 1. At x = 0, eᵗᵃⁿ⁰ = 1, so π/4 = -π/4 + C. This gives C = π/2.
4. Put x = π/4. Here e^(tan(π/4)) = e, so tan⁻¹ y = π/2 - tan⁻¹ e. Because e > 0, π/2 - tan⁻¹ e = cot⁻¹ e = tan⁻¹ 1/e.
5. So. y(π/4) = 1/e, which is option (d).

The idea to keep. When you see eᶠ⁽ˣ⁾f'(x) over 1 + e²ᶠ⁽ˣ⁾, put t = eᶠ⁽ˣ⁾. It becomes dt/(1+t²), and that integrates to tan⁻¹ t.

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