Question

The equation of the curve passing through the origin and satisfying the differential equation ${\left(\frac{dy}{dx}\right)}^2=(x-y{)}^2$, is

• A. $e^{2x}(1-x+y)=1+x-y$
• B. $e^{2x}(1+x-y)=1-x+y$
• C. $e^{2x}(1-x+y)=1+x+y$
• D. $e^{2x}(1+x+y)=1-x+y$

Answer 1

The correct option is A $e^{2x}(1-x+y)=1+x-y$

We have, $\frac{dy}{dx}=\pm (x-y)$
Case 1 When $\frac{dy}{dx}=(x-y)$
In this case, we have
$1-\frac{dv}{dx}=v,wherex-y=v\Rightarrow \frac{dv}{dx}=1-v\Rightarrow \frac{1}{1-v}dv=dx\Rightarrow -log(1-v)=x+logc\Rightarrow (1-v{)}^{-}1=c{e}^x\Rightarrow \frac{1}{1-x+y}=c{e}^x$
It passes through the origin.
$\therefore c=1Hence,\frac{1}{1-x+y}=e^x.....\left(i\right)$
Case 2 When $\frac{dy}{dx}=-(x-y)=y-x$
In this case, we have
$\frac{du}{dx}+1=u,$ where $u=y-x$
$\Rightarrow \frac{du}{u-1}=dx\Rightarrow log(u-1)=x+logc\Rightarrow u-1=c{e}^x\Rightarrow y-x-1=c{e}^x$
It passes through the origin.
$\therefore c=-1Hence,y-x-1=-e^x\Rightarrow x-y+1=e^x$
From Eqs. (i) and (ii), we obtain
$\frac{x-y+1}{1-x+y}=e^{2x}\Rightarrow (x-y+1)=(1-x+y)e^{2x}$