If $\frac{d y}{d x}=\frac{x y}{x^{2}+y^{2}} ; y(1)=1$; then a value of $x$
satisfying $y(x)=e$ is :
$\sqrt{2} \mathrm{e}$
$\frac{\mathrm{e}}{\sqrt{2}}$
$\frac{1}{2} \sqrt{3} \mathrm{e}$
$\sqrt{3} \mathrm{e}$
Correct Option: , 4
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