If ${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$; y(1) = 1; then a value of x satisfying y(x) = e is :

${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$ <br><br>Let y = vx <br><br>$\therefore$ ${{dy} \over {dx}}$ = v + x.${{dv} \over {dx}}$ <br><br>$\Rightarrow$ v + x.${{dv} \over {dx}}$ = ${{x\left( {vx} \right)} \over {{x^2} + {v^2}{x^2}}}$ = ${v \over {1 + {v^2}}}$ <br><br>$\Rightarrow$ x.${{dv} \over {dx}}$ = ${v \over {1 + {v^2}}}$ - v = ${{v - v - {v^3}} \over {1 + {v^2}}}$ = $- {{{v^3}} \over {1 + {v^2}}}$ <br><br>$\Rightarrow$ $\int {{{1 + {v^2}} \over {{v^3}}}} dv = - \int {{{dx} \over x}}$ <br><br>$\Rightarrow$ $- {1 \over {2{v^2}}} + \log v$ = $- \log x + C$ <br><br>$\Rightarrow$ $- {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$ = $- \log x + C$ ......(1) <br><br>putting x = 1, y = 1 we get <br><br>$\Rightarrow$ C = $- {1 \over 2}$ <br><br>From eq. (1) <br><br>$- {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$ = $- \log x - {1 \over 2}$ <br><br>Put y = e <br><br>$- {1 \over 2}{{{x^2}} \over {{e^2}}} + \log \left( {{e \over x}} \right)$ = $- \log x - {1 \over 2}$ <br><br>$\Rightarrow$ x² = 3e² <br><br>$\Rightarrow$ x = $\pm$3$\sqrt e$