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JEE Main 2019 · 8 Apr · Shift II

JEE Main 2019 Mathematics question on Differential Equations

Given that the slope of the tangent to a curve y=y(x)y = y(x) at any point (x,y)(x, y) is 2yx2\frac{2y}{x^2}. If the curve passes through the centre of the circle x2+y2−2x−2y=0x^2 + y^2 - 2x - 2y = 0, then its equation is
  1. A.
    x2log⁡e∣y∣=−2(x−1)x^2 \log_e |y| = -2(x-1)
  2. B.
    xlog⁡e∣y∣=2(x−1)x \log_e |y| = 2(x-1)
  3. C.
    blog⁡e∣y∣=x−1b \log_e |y| = x - 1
  4. D.
    xlog⁡e∣y∣=2(x−1)x \log_e |y| = 2(x-1)
Show answer and solution

Correct option: C

Solution

The centre of the circle x² + y² - 2x - 2y = 0 is (1, 1).

Solving the ODE: dy/y = 2dx/x², so ln|y| = -2/x + C.

At (1,1): ln|1| = -2/1 + C → 0 = -2 + C → C = 2.

So ln|y| = -2/x + 2 = 2(1 - 1/x) = 2(x-1)/x.

Therefore x·ln|y| = 2(x-1), which matches option 4 (and option 3 appears to be a typo with 'b' instead of 'x', but the equation x log|y| = 2(x-1) is the answer, designated as the official correct option).

[JEE Main 2019 · 8 Apr · Shift II]

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