Question

Sol. [Key Concept: Group as $d(y/x)$ (homogeneous); integrate, apply $x(e) = e$, then find $x(e^{2})$.]

$${2y(ydx - xdy) + x^{2}dy = 0 }{\Rightarrow - 2{yx}^{2}\text{ }d\left( \frac{y}{x} \right) + x^{2}dy = 0 }{\Rightarrow - 2yd\left( \frac{y}{x} \right) + dy = 0 }{\Rightarrow - 2\text{ }d\left( \frac{y}{x} \right) + \frac{1}{y}dy = 0 }{\Rightarrow \frac{- 2y}{x} + \log_{e}y = C }$$Given $x(e) = e$

$${\Rightarrow C = - 1 }{\therefore\frac{- 2y}{x} + \log_{e}y = - 1 }{\Rightarrow \frac{2y}{x} - \log_{e}y = 1 }$$Put $y = e^{2}$, we get

$${\Rightarrow \frac{2e^{2}}{x} - 2 = 1 \Rightarrow \frac{2e^{2}}{x} = 3 \Rightarrow x = \frac{2e^{2}}{3} }{\Rightarrow x\left( e^{2} \right) = \frac{2e^{2}}{3}}$$