QJEEM21Mac47d: JEE Main 2021 Mathematics - Class 12 JEEM21Mac47d Solution: If , x > 0, > 0, and y(1) = 1, then is equal to :

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

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)

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