The solution of the differential equation, $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^{2}$, when $y(1)=1,$ is:

The solution of the differential equation, $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^{2}$, when $y(1)=1,$ is:
  1. $\quad \log _{e}\left|\frac{2-x}{2-y}\right|=x-y$
  2. $-\log _{e}\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$
  3. $-\log _{e}\left|\frac{1+x-y}{1-x+y}\right|=x+y-2$
  4. $\log _{\mathrm{e}}\left|\frac{2-y}{2-x}\right|=2(y-1)$

Solution

The given differential equation $\frac{d y}{d x}=(x-y)^{2}$ Let $x-y=t \Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}$ $\Rightarrow \frac{d y}{d x}=1-\frac{d t}{d x}$ Now, from equation (1) $\left(1-\frac{d t}{d x}\right)=(t)^{2}$ $\Rightarrow \quad 1-t^{2}=\frac{d t}{d x} \Rightarrow \int d x=\int \frac{d t}{1-t^{2}}$ $\Rightarrow-x=\frac{1}{2 \times 1} \ln \left|\frac{t-1}{t+1}\right|+c$ $\Rightarrow-x=\frac{1}{2} \ln \left|\frac{x-y-1}{x-y+1}\right|+c$ $\because$ The given condition $y(1)=1$ $-1=\frac{1}{2} \ln \left|\frac{1-1-1}{1-1+1}\right|+c \Rightarrow c=-1$ Hence, $2(x-1)=-\ln \left|\frac{1-x+y}{1-y+x}\right|$

Asked in: JEE Main 2019 (11 Jan Shift 2)

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