Explanation

(b)
Given, $\frac{dy}{dx} = \frac{xy}{x^{2} + y^{2}}$
Put $y = vx$
$\Rightarrow$ $\frac{dy}{dx} = v + x\ \frac{dv}{dx}$
$\therefore$ $v + x\frac{dv}{dx} = \frac{x^{2}v}{x^{2}(1 + v^{2})}$
$\Rightarrow$ $\int_{}^{}\frac{1 + v^{2}}{v^{3}}\ dv = - \int_{}^{}\frac{dx}{x}$
$\Rightarrow$ $- \frac{1}{2v^{2}} + \log v = - \log x + \log c$
$\Rightarrow$ $- \frac{1}{2}.\ \frac{x^{2}}{y^{2}} + \log{|y|} = \log c$
$\therefore$ $y(1) = 1,\ \ - \frac{1}{2} = \log c$
$\therefore$ $- \frac{1}{2}.\frac{x^{2}}{y^{2}} + \log{|y|} = - \frac{1}{2}$
$\Rightarrow$ $\log_{e}{|y|} + \frac{1}{2} = \frac{x^{2}}{2y^{2}}$
Again, when $x = x_{0},\ y = e$
$1 + \frac{1}{2} = \frac{x_{0}^{2}}{2e^{2}} \Rightarrow x_{0} = \sqrt{3}e$