diff-equation

Question: $({{x}^{2}}+{{y}^{2}})dy=xydx$. If $y({{x}_{0}})=e$, $y(1)=1$, then value of ${{x}_{0}}=$



1) $\sqrt{3}e$
2) $\sqrt{{{e}^{2}}-\frac{1}{2}}$
3) $\sqrt{\frac{{{e}^{2}}-1}{2}}$
4) $\sqrt{\frac{{{e}^{2}}+1}{2}}$
Solution: Explanation: No Explanation
Exact differential equations

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