jee-main 2024 Q75

jee-main · India · session2_08apr_shift1 Differential equations Higher-Order and Special DEs (Proof/Theory)
Let $f ( x )$ be a positive function such that the area bounded by $y = f ( x ) , y = 0$ from $x = 0$ to $x = a > 0$ is $e ^ { - a } + 4 a ^ { 2 } + a - 1$. Then the differential equation, whose general solution is $y = c _ { 1 } f ( x ) + c _ { 2 }$, where $c _ { 1 }$ and $c _ { 2 }$ are arbitrary constants, is
(1) $\left( 8 e ^ { x } - 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } = 0$
(2) $\left( 8 e ^ { x } - 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - \frac { d y } { d x } = 0$
(3) $\left( 8 e ^ { x } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - \frac { d y } { d x } = 0$
(4) $\left( 8 e ^ { x } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } = 0$
Let $f ( x )$ be a positive function such that the area bounded by $y = f ( x ) , y = 0$ from $x = 0$ to $x = a > 0$ is $e ^ { - a } + 4 a ^ { 2 } + a - 1$. Then the differential equation, whose general solution is $y = c _ { 1 } f ( x ) + c _ { 2 }$, where $c _ { 1 }$ and $c _ { 2 }$ are arbitrary constants, is\\
(1) $\left( 8 e ^ { x } - 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } = 0$\\
(2) $\left( 8 e ^ { x } - 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - \frac { d y } { d x } = 0$\\
(3) $\left( 8 e ^ { x } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - \frac { d y } { d x } = 0$\\
(4) $\left( 8 e ^ { x } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } = 0$