$The solution of primitive integral equation (x^{2} + y^{2}) dy = xydx is y = y (x) . If$ $y (1) = 1 and y (x_{0}) = e then x_{0} is$

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$\sqrt{2 (e^{2} - 1)}$

$\sqrt{2 (e^{2} + 1)}$

$\sqrt{3} e$

$\sqrt{\frac{e^{2} + 1}{2}}$

answer is C.

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Detailed Solution

$\begin{array}{l} Put y = Vx \Rightarrow \frac{d y}{d x} = V + x \frac{dv}{dx} \\ f r o m g i v e n \frac{d y}{d x} = \frac{x y}{x^{2} + y^{2}} \\ V + x \frac{dv}{dx} = \frac{{Vx}^{2}}{x^{2} (1 + V^{2})} = \frac{V}{1 + V^{2}} \\ x \frac{dv}{dx} = \frac{V - V - V^{3}}{1 + V^{2}} = \frac{- V^{3}}{1 + V^{2}} \\ \frac{1 + V^{2}}{V^{3}} d v = - \frac{d x}{x} \\ \to \int (\frac{1}{V^{3}} + \frac{V^{2}}{V^{3}}) d v = - \int \frac{dx}{x} \\ \to \frac{- 1}{2 V^{2}} + \log V = - \log x + C \\ \to \frac{- x^{2}}{2 y^{2}} + \log y - \log x = - \log x + C \end{array}$

$\begin{array}{l} \to \log y - \frac{x^{2}}{2 y^{2}} = C \\ \underline{Pass e s t h r o u g h (1, 1)} \end{array}$

$\begin{array}{l} \log 1 - \frac{1}{2} = C, C = - \frac{1}{2} \\ \Rightarrow \log y - \frac{x^{2}}{2 e^{2}} = - \frac{1}{2} \\ Put y = e \Rightarrow \log e - \frac{x^{2}}{2 e^{2}} = - \frac{1}{2} \\ 1 + \frac{1}{2} = \frac{x^{2}}{2 e^{2}} \\ \Rightarrow \frac{x^{2}}{2 e^{2}} = \frac{3}{2} \Rightarrow x^{2} = 3 e^{2} \\ \Rightarrow x_{0}^{2} = 3 e^{2} \\ \Rightarrow x_{0} = \sqrt{3} e \end{array}$