For the differential equation $(x^2 + y^2)\,dy = xy\,dx$, it is given that $y(1) = 1$ and $y(x_0) = e$, then the value of $x_0$ is _____

The given differential equation is $(x^2 + y^2)dy = xy dx$.

This can be rewritten as $\dfrac{dx}{dy} = \dfrac{x^2 + y^2}{xy} = \dfrac{x}{y} + \dfrac{y}{x}$.

Let $x = vy$, then $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$.

Substituting this into the differential equation, we get:

$v + y\dfrac{dv}{dy} = v + \dfrac{1}{v}$

$y\dfrac{dv}{dy} = \dfrac{1}{v}$

$v dv = \dfrac{dy}{y}$

Integrating both sides, we get:

$\dfrac{v^2}{2} = \ln|y| + C$

Substituting $v = \dfrac{x}{y}$:

$\dfrac{x^2}{2y^2} = \ln|y| + C$

Given $y(1) = 1$, substituting $x = 1$ and $y = 1$:

$\dfrac{1}{2} = \ln(1) + C \Rightarrow C = \dfrac{1}{2}$

The particular solution is:

$\dfrac{x^2}{2y^2} = \ln|y| + \dfrac{1}{2}$

We are given $y(x_0) = e$, so substituting $x = x_0$ and $y = e$:

$\dfrac{x_0^2}{2e^2} = \ln(e) + \dfrac{1}{2}$

$\dfrac{x_0^2}{2e^2} = 1 + \dfrac{1}{2} = \dfrac{3}{2}$

$x_0^2 = 3e^2$

$x_0 = \pm\sqrt{3}e$

Answer: $\pm\sqrt{3}\,e$