Let x = x(y) be the solution of the differential equation

$2y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 0$ such that x(1) = 0. Then, x(e) is equal to :

<p>Given differential equation</p> <p>$$2y{e^{{x \over {{y^2}}}}}dx + \left( {{y^2} - 4x{e^{{x \over {{y^2}}}}}} \right)dy = 0,\,x(1) = 0$$</p> <p>$\Rightarrow {e^{{x \over {{y^2}}}}}[2ydx - 4xdy] = - {y^2}dy$</p> <p>$$ \Rightarrow {e^{{x \over {{y^2}}}}}\left[ {{{2{y^2}dx - 4xydy} \over {{y^4}}}} \right] = {{ - 1} \over y}dy$$</p> <p>$$ \Rightarrow 2{e^{{x \over {{y^2}}}}}d\left( {{x \over {{y^2}}}} \right) = - {1 \over y}dy$$</p> <p>$\Rightarrow 2{e^{{x \over {{y^2}}}}} = - \ln y + c$ ...... (i)</p> <p>Now, using x(1) = 0, c = 2</p> <p>So, for x(e), Put y = e in (i)</p> <p>$2{e^{{x \over {{e^2}}}}} = - 1 + 2$</p> <p>$$ \Rightarrow {x \over {{e^2}}} = \ln \left( {{1 \over 2}} \right) \Rightarrow x(e) = - {e^2}\ln 2$$</p>