Let f(x) = x - 1 and g(x) = eˣ for x ∈ ℝ. If dy/dx = ( e^(-2√x) g(f(f(x))) - y/√x ), y(0) = 0, then y(1) is

Solution

Maths sir · 4 steps

Simplify g(f(f(x))) first, then the integrating factor e^(2√x) cancels the ugly exponential.

1. Simplify the composition. f(f(x)) = (x-1) - 1 = x - 2, so g(f(f(x))) = eˣ⁻². The equation becomes dy/dx + y/√x = e^(-2√x) eˣ⁻², which is linear in y.
2. Integrating factor. Here P = 1/√x, and ∫ x^(-1/2) dx = 2√x. So the integrating factor is e^(2√x). Multiplying by it cancels e^(-2√x) on the right.

   d/dx(y e^(2√x)) = eˣ⁻² ⇒ y e^(2√x) = eˣ⁻² + C

3. Use the condition. At x = 0, y = 0: 0 = e⁻² + C, so C = -e⁻². Then y e^(2√x) = eˣ⁻² - e⁻².
4. So. At x = 1: y(1) e² = e⁻¹ - e⁻² = (e-1)/e², so y(1) = (e-1)/e⁴, which is option (b).

The idea to keep. In a linear equation, look for an integrating factor that cancels a factor already on the right side. When it does, the integration becomes a one-liner.

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