Question

The solution of the differential equation , $\frac{dy}{dx} = \left(x-y\right)^{2} , $ when $y(1) = 1$, is :

• A. $\log_{e} \left|\frac{2-y}{2-x}\right| = 2 \left(y-1\right)$
• B. $\log_{e} \left|\frac{2-x}{2-y}\right| = x -y $
• C. $- \log_{e} \left|\frac{1+x-y}{1-x+y}\right| =x + y - 2 $
• D. $- \log_{e} \left| \frac{1 - x + y}{1 + x - y}\right| = 2 (x - 1)$

Answer 1

Answer: (D)

$x-y =t \Rightarrow \frac{dy}{dx} =1 - \frac{dt}{dx} $
$ \Rightarrow 1- \frac{dt}{dx} =t^{2} \Rightarrow \int \frac{dt}{1-t^{2}} =\int 1dx $
$ \Rightarrow \frac{1}{2} \ell n \left(\frac{1+t}{1-t}\right) =x +\lambda $
$ \Rightarrow \frac{1}{2} \ell n \left(\frac{1+x-y}{1-x+y}\right) =x +\lambda $ given y(1) = 1
$ \Rightarrow \frac{1}{2} \ell n\left(1\right) = 1+\lambda \Rightarrow \lambda = - 1 $
$ \Rightarrow \ell n \left(\frac{1+x-y}{1-x+y}\right) = 2\left(x-1\right) $
$\Rightarrow -\ell n \left(\frac{1-x+y}{1+x-y}\right) =2\left(x-1\right) $