If \(\left(x^2+y^2\right) \mathrm{d} y=x y \mathrm{~d} x\), with \(y\left(x_0\right)=e, y(1)=1\), then \(x_0\) has the value

\(\begin{aligned}
& \left.\left(x^2+y^2\right) \mathrm{d} y=x y \mathrm{~d} x \quad \text { [Let } y=v x\right] \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x y}{x^2+y^2} \\
& \Rightarrow v+x \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{v}{1+v} \\
& \Rightarrow-\int \frac{1+v^2}{v^3} \mathrm{~d} v=\int \frac{\mathrm{d} x}{x} \\
& \Rightarrow \frac{1}{2 v^2}-\log v=\log x+\log c \\
& \Rightarrow c y=e^{\frac{x^2}{2 y^2}}
\end{aligned}\)
[Let \(y=v x\) ]
Putting \(x=1\) and \(y=1\) we get \(c=e^{-\frac{1}{2}}\)
\(\Rightarrow y=e^{\frac{x^2-y^2}{2 y^2}}\), now putting \(x=x_0\) and \(y=e\) we get \(x_0=\sqrt{3} e\)