Question

If the slope of normal at any point on the curve $y=y\left(x\right)$ is propotional to the difference in its abscissa and ordinate and passing through origin, then the curve is ?
(Assume propotionality constant $=1$)

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

Answer 1

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

Given : $-\frac{dx}{dy}=k(x-y),k=1$
$\Rightarrow \frac{dx}{dy}+x=y$ (linear)
$I.F.=e^{\int 1dy}=e^y$
solution is given by :
$x\left(e^y\right)=\int \left(y{e}^y\right)dy=e^y(y-1)+C$
$\Rightarrow x=y-1+C{e}^{-y}$
As $y\left(0\right)=0;C=1$
$\Rightarrow x=y-1+e^{-y}\Rightarrow x+1=y+e^{-y}$