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JEE Main 2023 · 29 Jan · Shift 1

JEE Main 2023 Mathematics question on Differential Equations

Let y=f(x)y = f(x) be the solution of the differential equation y(x+1)dx−x2dy=0y(x+1)dx - x^2 dy = 0, y(1)=ey(1) = e. Then lim⁡x→0+f(x)\lim_{x \to 0^+} f(x) is equal to
  1. A.
    e2e^2
  2. B.
    1e\dfrac{1}{e}
  3. C.
    1e2\dfrac{1}{e^2}
  4. D.
    00
Show answer and solution

Correct option: D

Solution

Rewrite the ODE as dy/y = (x+1)/x² dx = (1/x + 1/x²)dx.

Integrating: ln y = ln x − 1/x + C.

Apply y(1) = e: ln e = ln 1 − 1 + C → 1 = 0 − 1 + C → C = 2.

So f(x) = x · e^(2 − 1/x).

As x → 0⁺, the exponent 2 − 1/x → −∞, so e^(2−1/x) → 0 far faster than x → 0. Thus the limit is 0.

[JEE Main 2023 · 29 Jan · Shift 1]

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