Let the solution curve y=y(x) of the differential equation (1+e²ˣ)((d y)/(d x)+y)=1 pass through the point (0, π/2). Then, lim(x → ∞) eˣ y(x) is equal to :

Solution

Maths sir · 4 steps

Divide by 1+e²ˣ: the integrating factor eˣ is exactly the factor in the limit.

1. Standard linear form. Divide both sides by 1+e²ˣ to get dy/dx + y = 1/(1+e²ˣ). Here P = 1, so the integrating factor is e^(∫ 1 dx) = eˣ.

   d/dx(y eˣ) = eˣ/(1+e²ˣ)

2. Integrate. Put t = eˣ, so dt = eˣ dx and the right side becomes ∫ dt/(1+t²) = tan⁻¹ t. This gives y eˣ = tan⁻¹(eˣ) + C.
3. Find C. At x = 0, y = π/2, so π/2 = tan⁻¹(1) + C = π/4 + C. That gives C = π/4.
4. So. As x → ∞, eˣ → ∞ and tan⁻¹(eˣ) → π/2. So eˣy(x) → π/2 + π/4 = 3π/4, which is option (b).

The idea to keep. When a limit asks for (integrating factor) × y, do not solve for y fully. Stop at y · IF = ∫ Q · IF dx + C and take the limit right there.

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