Let $y = y ( x )$ be the solution of the differential equation $x \log _ { e } x \frac { d y } { d x } + y = x ^ { 2 } \log _ { e } x , ( x > 1 )$. If $y ( 2 ) = 2$, then $y ( e )$ is equal to (1) $\frac { 4 + e ^ { 2 } } { 4 }$ (2) $\frac { 1 + e ^ { 2 } } { 4 }$ (3) $\frac { 2 + e ^ { 2 } } { 2 }$ (4) $\frac { 1 + e ^ { 2 } } { 2 }$

The solution of the differential equation $\frac{dy}{dx} = -\left(\frac{x^{2} + 3y^{2}}{3x^{2} + y^{2}}\right)$, $y(1) = 0$ is
(1) $\log_{e}|x + y| - \frac{xy}{(x+y)^{2}} = 0$
(2) $\log_{e}|x + y| + \frac{xy}{(x+y)^{2}} = 0$
(3) $\log_{e}|x + y| + \frac{2xy}{(x+y)^{2}} = 0$
(4) $\log_{e}|x + y| - \frac{2xy}{(x+y)^{2}} = 0$

If the solution curve of the differential equation $\left( y - 2 \log _ { e } x \right) d x + \left( x \log _ { e } x ^ { 2 } \right) d y = 0 , x > 1$ passes through the points $\left( e , \frac { 4 } { 3 } \right)$ and $\left( e ^ { 4 } , \alpha \right)$, then $\alpha$ is equal to $\_\_\_\_$

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f'(x) + f(x) = \int_0^2 f(t)\, dt$. If $f(0) = e^{-2}$, then $2f(0) - f(2)$ is equal to $\_\_\_\_$.

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

Let $\alpha$ be a non-zero real number. Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function such that $f(0) = 1$ and $\lim_{x \to -\infty} f(x) = 1$. If $f'(x) = \alpha f(x) + 3$, for all $x \in \mathbb{R}$, then $f(-\log_e 2)$ is equal to:
(1) 1
(2) 5
(3) 9
(4) 7

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

Let $y = y(x)$ be the solution of the differential equation $\frac { d y } { d x } = \frac { \tan x + y } { \sin x \sec x - \sin x \tan x } , \quad x \in \left(0, \frac { \pi } { 2 }\right)$ satisfying the condition $y\left(\frac { \pi } { 4 }\right) = 2$. Then, $y\left(\frac { \pi } { 3 }\right)$ is
(1) $\sqrt { 32 } + \log _ { e } \sqrt { 3 }$
(2) $\frac { \sqrt { 3 } } { 2} \left(2 + \log _ { e } 3\right)$
(3) $\sqrt { 3 } \left(1 + 2 \log _ { e } 3\right)$
(4) $\sqrt { 3 } \left(2 + \log _ { e } 3\right)$

A function $y = f ( x )$ satisfies $f ( x ) \sin 2 x + \sin x - \left( 1 + \cos ^ { 2 } x \right) f ^ { \prime } ( x ) = 0$ with condition $f ( 0 ) = 0$. Then $f \left( \frac { \pi } { 2 } \right)$ is equal to
(1) 1
(2) 0
(3) - 1
(4) 2

If $y = y ( x )$ is the solution curve of the differential equation $x ^ { 2 } - 4 d y - y ^ { 2 } - 3 y d x = 0 , x > 2 , y ( 4 ) = \frac { 3 } { 2 }$ and the slope of the curve is never zero, then the value of $\mathrm { y } ( 10 )$ equals :
(1) $\frac { 3 } { 1 + ( 8 ) ^ { \frac { 1 } { 4 } } }$
(2) $\frac { 3 } { 1 + 2 \sqrt { z } }$
(3) $\frac { 3 } { 1 - 2 \sqrt { 2 } }$
(4) $\frac { 3 } { 1 - ( 8 ) ^ { \frac { 1 } { 4 } } }$

Let $y = y ( x )$ be the solution of the differential equation $\left( 1 + y ^ { 2 } \right) e ^ { \tan x } d x + \cos ^ { 2 } x \left( 1 + e ^ { 2 \tan x } \right) d y = 0 , y ( 0 ) = 1$. Then $y \left( \frac { \pi } { 4 } \right)$ is equal to
(1) $\frac { 2 } { e }$
(2) $\frac { 2 } { e ^ { 2 } }$
(3) $\frac { 1 } { e }$
(4) $\frac { 1 } { e ^ { 2 } }$

Let $y = y ( x )$ be the solution curve of the differential equation $\sec y \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 x \sin y = x ^ { 3 } \cos y , y ( 1 ) = 0$. Then $y ( \sqrt { 3 } )$ is equal to : (1) $\frac { \pi } { 3 }$ (2) $\frac { \pi } { 6 }$ (3) $\frac { \pi } { 12 }$ (4) $\frac { \pi } { 4 }$

Let $y = y ( x )$ be the solution of the differential equation $\left( 1 + x ^ { 2 } \right) \frac { d y } { d x } + y = e ^ { \tan ^ { - 1 } x } , y ( 1 ) = 0$. Then $y ( 0 )$ is
(1) $\frac { 1 } { 2 } \left( e ^ { \pi / 2 } - 1 \right)$
(2) $\frac { 1 } { 2 } \left( 1 - e ^ { \pi / 2 } \right)$
(3) $\frac { 1 } { 4 } \left( 1 - e ^ { \pi / 2 } \right)$
(4) $\frac { 1 } { 4 } \left( e ^ { \pi / 2 } - 1 \right)$

Let $x = x ( t )$ and $y = y ( t )$ be solutions of the differential equations $\frac { \mathrm { dx } } { \mathrm { dt } } + \mathrm { ax } = 0$ and $\frac { \mathrm { dy } } { \mathrm { dt } } + $ by $= 0$ respectively, $\mathrm { a } , \mathrm { b } \in \mathrm { R }$. Given that $x ( 0 ) = 2 ; y ( 0 ) = 1$ and $3 y ( 1 ) = 2 x ( 1 )$, the value of $t$, for which $x ( t ) = y ( t )$, is:
(1) $\log _ { \frac { 2 } { 3 } } 2$
(2) $\log _ { 4 } 3$
(3) $\log _ { 3 } 4$
(4) $\log _ { \frac { 4 } { 3 } } 2$

Let $y = y ( x )$ be the solution of the differential equation $\left( x ^ { 2 } + 4 \right) ^ { 2 } d y + \left( 2 x ^ { 3 } y + 8 x y - 2 \right) d x = 0$. If $y ( 0 ) = 0$, then $y ( 2 )$ is equal to
(1) $\frac { \pi } { 32 }$
(2) $2 \pi$
(3) $\frac { \pi } { 8 }$
(4) $\frac { \pi } { 16 }$

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

If the solution $y ( x )$ of the given differential equation $\left( \mathrm { e } ^ { y } + 1 \right) \cos x \mathrm {~d} x + \mathrm { e } ^ { y } \sin x \mathrm {~d} y = 0$ passes through the point $\left( \frac { \pi } { 2 } , 0 \right)$, then the value of $\mathrm { e } ^ { y \left( \frac { \pi } { 6 } \right) }$ is equal to $\_\_\_\_$

For a differentiable function $f : \mathbb { R } \rightarrow \mathbb { R }$, suppose $f ^ { \prime } ( x ) = 3 f ( x ) + \alpha$, where $\alpha \in \mathbb { R } , f ( 0 ) = 1$ and $\lim _ { x \rightarrow - \infty } f ( x ) = 7$. Then $9 f \left( - \log _ { \mathrm { e } } 3 \right)$ is equal to $\_\_\_\_$

If the solution curve $y = y ( x )$ of the differential equation $\left( 1 + y ^ { 2 } \right) \left( 1 + \log _ { e } x \right) d x + x d y = 0 , x > 0$ passes through the point $( 1,1 )$ and $y ( e ) = \frac { \alpha - \tan \left( \frac { 3 } { 2 } \right) } { \beta + \tan \left( \frac { 3 } { 2 } \right) }$, then $\alpha + 2 \beta$ is

If the solution curve, of the differential equation $\frac { d y } { d x } = \frac { x + y - 2 } { x - y }$ passing through the point $( 2,1 )$ is $\tan ^ { - 1 } \frac { y - 1 } { x - 1 } - \frac { 1 } { \beta } \log _ { e } \alpha + \frac { y - 1 } { x - 1 } ^ { 2 } = \log _ { e } x - 1$, then $5 \beta + \alpha$ is equal to $\_\_\_\_$.

Let $y = y ( x )$ be the solution of the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } y = x \mathrm { e } ^ { \frac { 1 } { \left( 1 + x ^ { 2 } \right) } } ; y ( 0 ) = 0$. Then the area enclosed by the curve $f ( x ) = y ( x ) \mathrm { e } ^ { - \frac { 1 } { \left( 1 + x ^ { 2 } \right) } }$ and the line $y - x = 4$ is $\_\_\_\_$

Let $x = x ( y )$ be the solution of the differential equation $y ^ { 2 } \mathrm {~d} x + \left( x - \frac { 1 } { y } \right) \mathrm { d } y = 0$. If $x ( 1 ) = 1$, then $x \left( \frac { 1 } { 2 } \right)$ is :
(1) $\frac { 1 } { 2 } + \mathrm { e }$
(2) $3 + e$
(3) $3 - e$
(4) $\frac { 3 } { 2 } + e$

If for the solution curve $y = f ( x )$ of the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \tan x ) y = \frac { 2 + \sec x } { ( 1 + 2 \sec x ) ^ { 2 } }$, $x \in \left( \frac { - \pi } { 2 } , \frac { \pi } { 2 } \right) , f \left( \frac { \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 10 }$, then $f \left( \frac { \pi } { 4 } \right)$ is equal to :
(1) $\frac { \sqrt { 3 } + 1 } { 10 ( 4 + \sqrt { 3 } ) }$
(2) $\frac { 5 - \sqrt { 3 } } { 2 \sqrt { 2 } }$
(3) $\frac { 9 \sqrt { 3 } + 3 } { 10 ( 4 + \sqrt { 3 } ) }$
(4) $\frac { 4 - \sqrt { 2 } } { 14 }$

Let $x = x ( y )$ be the solution of the differential equation $y = \left( x - y \frac { \mathrm {~d} x } { \mathrm {~d} y } \right) \sin \left( \frac { x } { y } \right) , y > 0$ and $x ( 1 ) = \frac { \pi } { 2 }$. Then $\cos ( x ( 2 ) )$ is equal to :
(1) $1 - 2 \left( \log _ { e } 2 \right) ^ { 2 }$
(2) $1 - 2 \left( \log _ { \mathrm { e } } 2 \right)$
(3) $2 \left( \log _ { e } 2 \right) - 1$
(4) $2 \left( \log _ { e } 2 \right) ^ { 2 } - 1$

If $x = f ( y )$ is the solution of the differential equation $\left( 1 + y ^ { 2 } \right) + \left( x - 2 \mathrm { e } ^ { \tan ^ { - 1 } y } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = 0 , y \in \left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$ with $f ( 0 ) = 1$, then $f \left( \frac { 1 } { \sqrt { 3 } } \right)$ is equal to :
(1) $e ^ { \pi / 12 }$
(2) $e ^ { \pi / 4 }$
(3) $e ^ { \pi / 3 }$
(4) $e ^ { \pi / 6 }$