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JEE Main 2019 · 11 Jan · Shift II

JEE Main 2019 Mathematics question on Differential Equations

The solution of the differential equation, dydx=(x−y)2\frac{dy}{dx} = (x - y)^2, when y(1)=1y(1) = 1, is
  1. A.
    log⁡e∣2−y2−x∣=2(y−1)\log_e \left|\frac{2-y}{2-x}\right| = 2(y-1)
  2. B.
    −log⁡e∣1+x−y1−x+y∣=x+y−2-\log_e \left|\frac{1+x-y}{1-x+y}\right| = x + y - 2
  3. C.
    log⁡e∣2−x2−y∣=x−y\log_e \left|\frac{2-x}{2-y}\right| = x - y
  4. D.
    −log⁡e∣1−x+y1+x−y∣=2(x−1)-\log_e \left|\frac{1-x+y}{1+x-y}\right| = 2(x-1)
Show answer and solution

Correct option: D

Solution

Let t = x - y, so dt/dx = 1 - dy/dx = 1 - t². This separable ODE gives dt/(1-t²) = dx.

Integrating: (1/2)log|(1+t)/(1-t)| = x + C, i.e., log|(1+x-y)/(1-x+y)| = 2x + C.

Applying y(1)=1: t=0 at x=1, so log(1) = 2(1) + C → C = -2.

Thus log|(1+x-y)/(1-x+y)| = 2x - 2 = 2(x-1), which is exactly -log|(1-x+y)/(1+x-y)| = 2(x-1), confirming option 4.

[JEE Main 2019 · 11 Jan · Shift II]

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