Question

The solution of: $e^x\sqrt{1-y^2}dx+\frac{y}{x}dy=0$

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

Answer 1

The correct option is D $(x-1)e^x=\sqrt{1-y^2}+c$

$e^x\sqrt{1-y^2}dx+\frac{y}{x}dy=0$
$\Rightarrow x{e}^{x}dx=-\frac{y}{\sqrt{1-y^2}}dy$
$\Rightarrow \int x{e}^{x}dx=\int \frac{-y}{\sqrt{1-y^2}}dy+c$
$\Rightarrow (x{e}^x-e^x)=\sqrt{1-y^2}+c$
$\Rightarrow e^x(x-1)=\sqrt{1-y^2}+c$