jee-advanced 2025 Q13

jee-advanced · India · paper1 4 marks Differential equations Solving Separable DEs with Initial Conditions
For all $x > 0$, let $y _ { 1 } ( x ) , y _ { 2 } ( x )$, and $y _ { 3 } ( x )$ be the functions satisfying
$$\begin{aligned} & \frac { d y _ { 1 } } { d x } - ( \sin x ) ^ { 2 } y _ { 1 } = 0 , \quad y _ { 1 } ( 1 ) = 5 \\ & \frac { d y _ { 2 } } { d x } - ( \cos x ) ^ { 2 } y _ { 2 } = 0 , \quad y _ { 2 } ( 1 ) = \frac { 1 } { 3 } \\ & \frac { d y _ { 3 } } { d x } - \left( \frac { 2 - x ^ { 3 } } { x ^ { 3 } } \right) y _ { 3 } = 0 , \quad y _ { 3 } ( 1 ) = \frac { 3 } { 5 e } \end{aligned}$$
respectively. Then
$$\lim _ { x \rightarrow 0 ^ { + } } \frac { y _ { 1 } ( x ) y _ { 2 } ( x ) y _ { 3 } ( x ) + 2 x } { e ^ { 3 x } \sin x }$$
is equal to $\_\_\_\_$ .
For all $x > 0$, let $y _ { 1 } ( x ) , y _ { 2 } ( x )$, and $y _ { 3 } ( x )$ be the functions satisfying

$$\begin{aligned}
& \frac { d y _ { 1 } } { d x } - ( \sin x ) ^ { 2 } y _ { 1 } = 0 , \quad y _ { 1 } ( 1 ) = 5 \\
& \frac { d y _ { 2 } } { d x } - ( \cos x ) ^ { 2 } y _ { 2 } = 0 , \quad y _ { 2 } ( 1 ) = \frac { 1 } { 3 } \\
& \frac { d y _ { 3 } } { d x } - \left( \frac { 2 - x ^ { 3 } } { x ^ { 3 } } \right) y _ { 3 } = 0 , \quad y _ { 3 } ( 1 ) = \frac { 3 } { 5 e }
\end{aligned}$$

respectively. Then

$$\lim _ { x \rightarrow 0 ^ { + } } \frac { y _ { 1 } ( x ) y _ { 2 } ( x ) y _ { 3 } ( x ) + 2 x } { e ^ { 3 x } \sin x }$$

is equal to $\_\_\_\_$ .