The particular solution of the differential equation $e^x\sqrt{1-y^2}dx +\frac{y}{x} dy = 0$, given that $y = 1$, when $x = 0$, is:

$\sqrt{1-y^2}=(x-1)e^x +1$

The correct answer is Option (1) → $\sqrt{1-y^2}=(x-1)e^x +1$

Given differential equation

$e^x\sqrt{1-y^2}\,dx+\frac{y}{x}\,dy=0$

Rearrange

$\frac{y}{x}\,dy=-e^x\sqrt{1-y^2}\,dx$

Separate variables

$\frac{y}{\sqrt{1-y^2}}\,dy=-xe^x\,dx$

Integrate

$\int\frac{y}{\sqrt{1-y^2}}\,dy=-\int xe^x\,dx$

$-\sqrt{1-y^2}=-(x-1)e^x+C$

$\sqrt{1-y^2}=(x-1)e^x+C$

Given $y=1$ at $x=0$

$\sqrt{1-1}=(-1)e^0+C$

$0=-1+C$

$C=1$

Hence particular solution

$\sqrt{1-y^2}=(x-1)e^x+1$

The correct answer is $\sqrt{1-y^2}=(x-1)e^x+1$.