JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 8)
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 :
4 $$\phi$$ (2)
4$$\phi$$ (1)
2 $$\phi$$ (1)
$$\phi$$ (1)
Explanation
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 $$\varphi ({t^2}) = p$$
$$\therefore$$ $$\varphi '({t^2})2tdt = dp$$
$$ \Rightarrow \int {{{dy} \over {2p}}} = \int {{{dx} \over x}} $$
$${1 \over 2}\ln \varphi ({t^2}) = \ln x + \ln c$$
$$\varphi ({t^2}) = {x^2}k$$
$$\varphi \left( {{{{y^2}} \over {{x^2}}}} \right) = k{x^2},\varphi (1) = k$$
$$\varphi \left( {{{{y^2}} \over 4}} \right) = 4\varphi (1)$$
$${{dy} \over {dx}} = t + x{{dt} \over {dx}}$$
$$\therefore$$ $$tx\left( {t + x{{dt} \over {dx}}} \right) = x\left( {{t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}} \right)$$
$${t^2} + xt{{dt} \over {dx}} = {t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}$$
$$\int {{{t\varphi '({t^2})} \over {\varphi ({t^2})}}} dt = \int {{{dx} \over x}} $$
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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