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EEJ MAIN Mathematics QUESTION #6013
Question 1

The solution of the differential equation $\dfrac{dy}{dx} = (x-y)^2$, subject to $y(1)=1$, is:

  • $\ln\left|\dfrac{2-x}{2-y}\right| = x-y$
  • $-\ln\left|\dfrac{1-x+y}{1+x-y}\right| = 2(x-1)$✔️
  • $\ln\left|\dfrac{1+x-y}{1-x+y}\right| = x+y-2$
  • $-\ln\left|\dfrac{2-y}{2-x}\right| = 2(y-1)$
Correct Answer Explanation

Let $v = x - y$, so $\frac{dv}{dx} = 1 - \frac{dy}{dx} = 1 - v^2$.

Separate: $\frac{dv}{1-v^2} = dx \Rightarrow \frac{1}{2}\ln\left|\frac{1+v}{1-v}\right| = x + C$

i.e. $\ln\left|\frac{1+x-y}{1-x+y}\right| = 2x + C'$

At $(1,1)$: $v=0$, $\ln(1)=2+C' \Rightarrow C'=-2$

Final: $-\ln\left|\dfrac{1-x+y}{1+x-y}\right| = 2(x-1)$ which matches option B.