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