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 $\_\_\_\_$ .

A particle is moving with velocity $\vec { v } = K ( y \hat { i } + x \hat { j } )$, where $K$ is a constant. The general equation for its path is
(1) $y = x ^ { 2 } +$ constant
(2) $y ^ { 2 } = x +$ constant
(3) $x y =$ constant
(4) $y ^ { 2 } = x ^ { 2 } +$ constant

If a curve $y = f(x)$ passes through the point $(1, -1)$ and satisfies the differential equation, $y(1 + xy) dx = x\, dy$, then $f\left(-\frac{1}{2}\right)$ is equal to:
(1) $-\frac{2}{5}$
(2) $-\frac{4}{5}$
(3) $\frac{2}{5}$
(4) $\frac{4}{5}$

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{2}\sec^2 x = \frac{\tan x \sec^2 x}{2y}$, where $y(0) = 1$, is given by:
(1) $y^2 = 1 + \frac{\tan x}{x}$
(2) $y^2 = 1 + \tan x$
(3) $y = 1 - \tan x$
(4) $y^2 = 1 - \tan x$

If a curve $y = f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation, $y(1+xy)dx = x\,dy$, then $f\left(-\frac{1}{2}\right)$ is equal to: (1) $-\frac{2}{5}$ (2) $-\frac{4}{5}$ (3) $\frac{2}{5}$ (4) $\frac{4}{5}$

If $( 2 + \sin x ) \frac { d y } { d x } + ( y + 1 ) \cos x = 0$ and $y ( 0 ) = 1$, then $y \left( \frac { \pi } { 2 } \right)$ is equal to:
(1) $\frac { 1 } { 3 }$
(2) $- \frac { 2 } { 3 }$
(3) $- \frac { 1 } { 3 }$
(4) $\frac { 4 } { 3 }$

If $(2 + \sin x)\dfrac{dy}{dx} + (y + 1)\cos x = 0$ and $y(0) = 1$, then $y\left(\dfrac{\pi}{2}\right)$ is equal to
(1) $\dfrac{1}{3}$
(2) $-\dfrac{2}{3}$
(3) $-\dfrac{1}{3}$
(4) $\dfrac{4}{3}$

A particle is moving with a velocity $\vec { v } = K y \hat { i } + x \hat { j }$, where $K$ is a constant. The general equation for its path is:
(1) $y ^ { 2 } = x +$ constant
(2) $x y =$ constant
(3) $y = x ^ { 2 } +$ constant
(4) $y ^ { 2 } = x ^ { 2 } +$ constant

If a curve passes through the point $( 1 , - 2 )$ and has slope of the tangent at any point $( x , y )$ on it as $\frac { x ^ { 2 } - 2 y } { x }$, then the curve also passes through the point
(1) $( \sqrt { 3 } , 0 )$
(2) $( - 1,2 )$
(3) $( - \sqrt { 2 } , 1 )$
(4) $( 3,0 )$

Given that the slope of the tangent to a curve $y = y ( x )$ at any point $( x , y )$ is $\frac { 2 y } { x ^ { 2 } }$. If the curve passes through the centre of the circle $x ^ { 2 } + y ^ { 2 } - 2 x - 2 y = 0$, then its equation is
(1) $x ^ { 2 } \log _ { e } | y | = - 2 ( x - 1 )$
(2) $x \log _ { e } | y | = 2 ( x - 1 )$
(3) $x \log _ { e } | y | = - 2 ( x - 1 )$
(4) $x \log _ { e } | y | = x - 1$

If $y = y(x)$ is the solution of the differential equation, $e ^ { y } \left( \frac { d y } { d x } - 1 \right) = e ^ { x }$ such that $y(0) = 0$, then $y(1)$ is equal to
(1) $1 + \log _ { e } 2$
(2) $2 + \log _ { e } 2$
(3) $2e$
(4) $\log _ { e } 2$

If $\frac { d y } { d x } = \frac { x y } { x ^ { 2 } + y ^ { 2 } } ; y ( 1 ) = 1$; then a value of $x$ satisfying $y(x) = e$ is:

Let $y = y(x)$ be the solution of the differential equation, $\frac{2 + \sin x}{y + 1} \cdot \frac{dy}{dx} = -\cos x, y > 0, y(0) = 1$. If $y(\pi) = a$ and $\frac{dy}{dx}$ at $x = \pi$ is $b$, then the ordered pair $(a, b)$ is equal to
(1) $\left(2, \frac{3}{2}\right)$
(2) $(1, -1)$
(3) $(1, 1)$
(4) $(2, 1)$

The solution curve of the differential equation, $\left( 1 + e ^ { - x } \right) \left( 1 + y ^ { 2 } \right) \frac { d y } { d x } = y ^ { 2 }$ which passes through the point $( 0,1 )$, is
(1) $y ^ { 2 } + 1 = y \left( \log _ { e } \left( \frac { 1 + e ^ { - x } } { 2 } \right) + 2 \right)$
(2) $y ^ { 2 } + 1 = y \left( \log _ { e } \left( \frac { 1 + e ^ { x } } { 2 } \right) + 2 \right)$
(3) $y ^ { 2 } = 1 + y \log _ { e } \left( \frac { 1 + e ^ { x } } { 2 } \right)$
(4) $y ^ { 2 } = 1 + y \log _ { e } \left( \frac { 1 + e ^ { - x } } { 2 } \right)$

The solution of the differential equation $\frac { d y } { d x } - \frac { y + 3 x } { \log _ { e } ( y + 3 x ) } + 3 = 0$ is (where $C$ is a constant of integration)
(1) $x - \frac { 1 } { 2 } \left( \log _ { e } ( y + 3 x ) \right) ^ { 2 } = C$
(2) $x - \log _ { e } ( y + 3 x ) = C$
(3) $y + 3 x - \frac { 1 } { 2 } \left( \log _ { e } x \right) ^ { 2 } = C$
(4) $x - 2 \log _ { e } ( y + 3 x ) = C$

If $y = y ( x )$ is the solution of the differential equation $\frac { 5 + e ^ { x } } { 2 + y } \cdot \frac { d y } { d x } + e ^ { x } = 0$ satisfying $y ( 0 ) = 1$ then value of $y \left( \log _ { e } 13 \right)$ is
(1) 1
(2) - 1
(3) 0
(4) 2

Which of the following is true for $y ( x )$ that satisfies the differential equation $\frac { d y } { d x } = x y - 1 + x - y ; y ( 0 ) = 0$
(1) $y ( 1 ) = \mathrm { e } ^ { - \frac { 1 } { 2 } } - 1$
(2) $y ( 1 ) = e ^ { \frac { 1 } { 2 } } - e ^ { - \frac { 1 } { 2 } }$
(3) $y ( 1 ) = 1$
(4) $y ( 1 ) = e^{\frac{1}{2}} - 1$

If a curve passes through the origin and the slope of the tangent to it at any point $( x , y )$ is $\frac { x ^ { 2 } - 4 x + y + 8 } { x - 2 }$, then this curve also passes through the point:
(1) $( 5,4 )$
(2) $( 4,4 )$
(3) $( 4,5 )$
(4) $( 5,5 )$

If the curve $y = y ( x )$ represented by the solution of the differential equation $\left( 2 x y ^ { 2 } - y \right) d x + x \, d y = 0$, passes through the intersection of the lines $2 x - 3 y = 1$ and $3 x + 2 y = 8$, then $| y ( 1 ) |$ is equal to $\underline{\hspace{1cm}}$.

Let a curve $y = y ( x )$ be given by the solution of the differential equation $\cos \left( \frac { 1 } { 2 } \cos ^ { - 1 } \left( e ^ { - x } \right) \right) dx = \left( \sqrt { e ^ { 2 x } - 1 } \right) dy$. If it intersects $y$-axis at $y = - 1$, and the intersection point of the curve with $x$-axis is $( \alpha , 0 )$, then $e ^ { \alpha }$ is equal to $\underline{\hspace{1cm}}$.

If $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ^ { x + y } - 2 ^ { x } } { 2 ^ { y } } , y ( 0 ) = 1$, then $y ( 1 )$ is equal to :
(1) $\log _ { 2 } \left( 1 + \mathrm { e } ^ { 2 } \right)$
(2) $\log _ { 2 } ( 2 \mathrm { e } )$
(3) $\log _ { 2 } ( 2 + e )$
(4) $\log _ { 2 } ( 1 + e )$

The slope of normal at any point $( x , y ) , x > 0 , y > 0$ on the curve $y = y ( x )$ is given by $\frac { x ^ { 2 } } { x y - x ^ { 2 } y ^ { 2 } - 1 }$. If the curve passes through the point $( 1,1 )$, then $e \cdot y ( e )$ is equal to
(1) $\frac { 1 - \tan ( 1 ) } { 1 + \tan ( 1 ) }$
(2) $\tan ( 1 )$
(3) 1
(4) $\frac { 1 + \tan ( 1 ) } { 1 - \tan ( 1 ) }$

Let a smooth curve $y = f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1, 2)$ and $(8, 1)$, then $\left|y\left(\frac{1}{8}\right)\right|$ is equal to
(1) $2\log_e 2$
(2) 4
(3) 1
(4) $4\log_e 2$

Let $S = ( 0 , 2 \pi ) - \left\{ \frac { \pi } { 2 } , \frac { 3 \pi } { 4 } , \frac { 3 \pi } { 2 } , \frac { 7 \pi } { 4 } \right\}$. Let $y = y ( x ) , x \in S$, be the solution curve of the differential equation $\frac { dy } { dx } = \frac { 1 } { 1 + \sin 2 x } , y \left( \frac { \pi } { 4 } \right) = \frac { 1 } { 2 }$. If the sum of abscissas of all the points of intersection of the curve $y = y ( x )$ with the curve $y = \sqrt { 2 } \sin x$ is $\frac { k \pi } { 12 }$, then $k$ is equal to $\_\_\_\_$.

If $\sin\left(\frac{y}{x}\right) = \log_e|x| + \frac{\alpha}{2}$ is the solution of the differential equation $x\cos\left(\frac{y}{x}\right)\frac{dy}{dx} = y\cos\left(\frac{y}{x}\right) + x$ and $y(1) = \frac{\pi}{3}$, then $\alpha^2$ is equal to
(1) 12
(2) 3
(3) 4
(4) 9