JEE Main 2020 · 9 January · morning · Numerical answer · +4 / −0

If for x ≥ 0, y = y(x) is the solution of the differential equation
(x + 1)dy = ((x + 1)² + y – 3)dx, y(2) = 0, then y(3) is equal to _______.

Answer and solution

Answer 3

Solution

Maths sir · 4 steps

Divide by dx and by (x+1), push the y term to the left, and this is a plain linear equation.

  1. Standard linear form. Divide both sides by (x+1) dx to get dy/dx = (x+1) + (y-3)/(x+1). Move the y term to the left: dy/dx - y/(x+1) = (x+1) - 3/(x+1).
  2. Integrating factor. Here P = -1/(x+1), so the integrating factor is e⁻ˡⁿ⁽ˣ⁺¹⁾ = 1/(x+1). Since x ≥ 0, x+1 > 0 and no modulus is needed. Multiplying through turns the left side into the derivative of y/(x+1).

    d/dx(y/(x+1)) = 1 - 3/(x+1)² ⇒ y/(x+1) = x + 3/(x+1) + C

  3. Find C. Put x = 2, y = 0: 0 = 2 + 3/3 + C, so C = -3.
  4. So. At x = 3: y/4 = 3 + 3/4 - 3 = 3/4, so y(3) = 4 × 3/4 = 3.

The common slip: 0.75. Stopping at y/(x+1) = 3/4 and not multiplying by x+1 = 4 gives 0.75, not 3.

The idea to keep. When a y sits alone on the right side, divide through to see if it is linear in y. Then the integrating factor e^(∫ P dx) does the rest.

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 2020.