Question 414 of 490MediumMCQ
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 :
Solutions · 1
Official solution
Let, y=tx
\[{{dy} \over {dx}} = t + x{{dt} \over {dx}}\]
\[\therefore\] \[tx\left( {t + x{{dt} \over {dx}}} \right) = x\left( {{t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}} \right)\]
\[{t^2} + xt{{dt} \over {dx}} = {t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}\]
\[\int {{{t\varphi '({t^2})} \over {\varphi ({t^2})}}} dt = \int {{{dx} \over x}} \]
Let \[\varphi ({t^2}) = p\]
\[\therefore\] \[\varphi '({t^2})2tdt = dp\]
\[ \Rightarrow \int {{{dy} \over {2p}}} = \int {{{dx} \over x}} \]
\[{1 \over 2}\ln \varphi ({t^2}) = \ln x + \ln c\]
\[\varphi ({t^2}) = {x^2}k\]
\[\varphi \left( {{{{y^2}} \over {{x^2}}}} \right) = k{x^2},\varphi (1) = k\]
\[\varphi \left( {{{{y^2}} \over 4}} \right) = 4\varphi (1)\] Answer the question to unlock