If y = y(x) is the solution of the differential equation
xdy/dx + 2y = x eˣ, y(1) = 0 then the local maximum value
of the function z(x) = x²y(x) - eˣ, x ∈ R is :

Solution

Maths sir · 4 steps

Use the integrating factor x²: it hands you x²y directly, which is exactly what z needs.

1. Standard linear form. Divide by x to get dy/dx + 2/xy = eˣ. The integrating factor is e^(∫ 2/x dx) = x², so the left side becomes d/dx(x² y).

   x² y = ∫ x² eˣ dx = eˣ(x² - 2x + 2) + C

2. Use y(1) = 0. At x = 1: 0 = e(1 - 2 + 2) + C, so C = -e. Then z = x² y - eˣ = eˣ(x² - 2x + 1) - e = eˣ(x-1)² - e.
3. Find the maximum. z'(x) = eˣ[(x-1)² + 2(x-1)] = eˣ(x-1)(x+1). The sign of z' goes from positive to negative at x = -1, so that is the local maximum. At x = 1 it goes from negative to positive, so that is the local minimum.
4. So. The local maximum value is z(-1) = e⁻¹(-2)² - e = 4/e - e, option (d).

The idea to keep. When the function to study contains x²y, and the integrating factor is x², do not solve for y alone. Use x²y as it comes, then check the sign change of z' to tell maximum from minimum.

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