# The solution of the differential equation,

Question:

The solution of the differential equation,

$\frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{x}-\mathrm{y})^{2}$, when $\mathrm{y}(1)=1$, is :-

1. $\log _{\mathrm{e}}\left|\frac{2-\mathrm{y}}{2-\mathrm{x}}\right|=2(\mathrm{y}-1)$

2. $\log _{e}\left|\frac{2-x}{2-y}\right|=x-y$

3. $-\log _{\mathrm{e}}\left|\frac{1+\mathrm{x}-\mathrm{y}}{1-\mathrm{x}+\mathrm{y}}\right|=\mathrm{x}+\mathrm{y}-2$

4. $-\log _{e}\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$

Correct Option: , 4

Solution:

$x-y=t \Rightarrow \frac{d y}{d x}=1-\frac{d t}{d x}$

$\Rightarrow 1-\frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{t}^{2} \Rightarrow \int \frac{\mathrm{dt}}{1-\mathrm{t}^{2}}=\int 1 \mathrm{dx}$

$\Rightarrow \frac{1}{2} \ln \left(\frac{1+\mathrm{t}}{1-\mathrm{t}}\right)=\mathrm{x}+\lambda$

$\Rightarrow \frac{1}{2} \ln \left(\frac{1+x-y}{1-x+y}\right)=x+\lambda \quad$ given $\quad y(1)=1$

$\Rightarrow \frac{1}{2} \ell n(1)=1+\lambda \Rightarrow \lambda=-1$

$\Rightarrow \ln \left(\frac{1+x-y}{1-x+y}\right)=2(x-1)$

$\Rightarrow-\ell n\left(\frac{1-x+y}{1+x-y}\right)=2(x-1)$