jee-main 2023 Q84

jee-main · India · session1_25jan_shift1 First order differential equations (integrating factor)

Let $\mathrm { y } = \mathrm { y } ( \mathrm { x } )$ be the solution curve of the differential equation $\frac { d y } { d x } = \frac { y } { x } \left( 1 + x ^ { 2 } \left( 1 + \log _ { e } x \right) \right) , \mathrm { x } > 0 , \mathrm { y } ( 1 ) = 3$. Then $\frac { \mathrm { y } ^ { 2 } ( \mathrm { x } ) } { 9 }$ is equal to:
(1) $\frac { x ^ { 2 } } { 5 - 2 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 3 } \right) }$
(2) $\frac { x ^ { 2 } } { 2 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 3 } \right) - 3 }$
(3) $\frac { x ^ { 2 } } { 3 x ^ { 3 } \left( 1 + \log _ { e } x ^ { 2 } \right) - 2 }$
(4) $\frac { x ^ { 2 } } { 7 - 3 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 2 } \right) }$

Let $\mathrm { y } = \mathrm { y } ( \mathrm { x } )$ be the solution curve of the differential equation $\frac { d y } { d x } = \frac { y } { x } \left( 1 + x ^ { 2 } \left( 1 + \log _ { e } x \right) \right) , \mathrm { x } > 0 , \mathrm { y } ( 1 ) = 3$. Then $\frac { \mathrm { y } ^ { 2 } ( \mathrm { x } ) } { 9 }$ is equal to:\\
(1) $\frac { x ^ { 2 } } { 5 - 2 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 3 } \right) }$\\
(2) $\frac { x ^ { 2 } } { 2 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 3 } \right) - 3 }$\\
(3) $\frac { x ^ { 2 } } { 3 x ^ { 3 } \left( 1 + \log _ { e } x ^ { 2 } \right) - 2 }$\\
(4) $\frac { x ^ { 2 } } { 7 - 3 x ^ { 3 } \left( 2 + \log _ { e } x ^ { 2 } \right) }$

Paper Questions

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