JEE Main 2022 · 26 June · evening · Single correct · +4 / −1

If the solution of the differential equation
dy/dx + eˣ( x² - 2 )y = ( x² - 2x )( x² - 2 )e²ˣ satisfies y(0) = 0, then the value of y(2) is _______________.

  1. (a) -1
  2. (b) 1
  3. (c) 0
  4. (d) e
Answer and solution

Answer (c) 0

Solution

Maths sir · 5 steps

The exponent of the integrating factor, eˣ(x²-2x), is the right side in disguise, so call it u.

  1. Integrating factor. Here P = eˣ(x²-2). Since d/dx[eˣ(x²-2x)] = eˣ(2x-2) + eˣ(x²-2x) = eˣ(x²-2), we get ∫ P dx = eˣ(x²-2x). So the integrating factor is eᵘ with u = eˣ(x²-2x).
  2. Spot the pattern. The right side is (x²-2x)(x²-2)e²ˣ = [eˣ(x²-2x)][eˣ(x²-2)] = u du/dx. So the equation becomes d/dx(y eᵘ) = u eᵘdu/dx.
  3. Integrate. Integration by parts gives ∫ u eᵘ du = (u-1)eᵘ + C. So y eᵘ = (u-1)eᵘ + C.

    y = u - 1 + C e⁻ᵘ, u = eˣ(x²-2x)

  4. Use y(0) = 0. At x = 0, u = 0, so 0 = 0 - 1 + C, which gives C = 1. The solution is y = u - 1 + e⁻ᵘ.
  5. So. At x = 2, u = e²(4-4) = 0, so y(2) = 0 - 1 + e⁰ = 0, option (c).

The idea to keep. In a linear equation, first find ∫ P dx. If the right side looks like that integral times its derivative, substitute u for it and the equation becomes easy to integrate.

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

Study Differential Equations

Asked about 200 times in JEE Main since 2020. This question is from 2022.