If $$y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$$, x > 0, $\phi$ > 0, and y(1) = $-$1, then $\phi \left( {{{{y^2}} \over 4}} \right)$ is equal to :

Let, $y = tx$<br><br>${{dy} \over {dx}} = t + x{{dt} \over {dx}}$<br><br>$\therefore$ $$tx\left( {t + x{{dt} \over {dx}}} \right) = x\left( {{t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}} \right)$$<br><br>$${t^2} + xt{{dt} \over {dx}} = {t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}$$<br><br>$\int {{{t\varphi '({t^2})} \over {\varphi ({t^2})}}} dt = \int {{{dx} \over x}}$<br><br>Let $\varphi ({t^2}) = p$<br><br>$\therefore$ $\varphi '({t^2})2tdt = dp$<br><br>$\Rightarrow \int {{{dy} \over {2p}}} = \int {{{dx} \over x}}$<br><br>${1 \over 2}\ln \varphi ({t^2}) = \ln x + \ln c$<br><br>$\varphi ({t^2}) = {x^2}k$<br><br>$\varphi \left( {{{{y^2}} \over {{x^2}}}} \right) = k{x^2},\varphi (1) = k$<br><br>$\varphi \left( {{{{y^2}} \over 4}} \right) = 4\varphi (1)$