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

The area enclosed by the closed curve $C$ given by the differential equation $\frac{dy}{dx} + \frac{x + a}{y - 2} = 0$, $y(1) = 0$ is $4\pi$. Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is
(1) $2\sqrt{3}$
(2) $\frac{2\sqrt{3}}{3}$
(3) 2
(4) $\frac{4\sqrt{3}}{3}$

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

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 $\_\_\_\_$

The slope of tangent at any point $(x, y)$ on a curve $y = y(x)$ is $\frac{x^2 + y^2}{2xy}$, $x > 0$. If $y(2) = 0$, then a value of $y(8)$ is
(1) $-4\sqrt{2}$
(2) $2\sqrt{3}$
(3) $-2\sqrt{3}$
(4) $4\sqrt{3}$

Let $y = y ( x ) , y > 0$, be a solution curve of the differential equation $\left( 1 + x ^ { 2 } \right) d y = y ( x - y ) d x$. If $y ( 0 ) = 1$ and $y ( 2 \sqrt { 2 } ) = \beta$, then
(1) $e ^ { 3 \beta - 1 } = e ( 3 + 2 \sqrt { 2 } )$
(2) $e ^ { 3 \beta - 1 } = e ( 5 + \sqrt { 2 } )$
(3) $e ^ { \beta - 1 } = e ^ { - 2 } ( 3 + 2 \sqrt { 2 } )$
(4) $e ^ { \beta - 1 } = e ^ { - 2 } ( 5 + \sqrt { 2 } )$

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

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) $2 \log _ { e } \frac { x } { y } = y + 1$

If the solution of the differential equation $( 2 x + 3 y - 2 ) d x + ( 4 x + 6 y - 7 ) d y = 0 , y ( 0 ) = 3$, is $\alpha x + \beta y + 3 \log _ { e } | 2 x + 3 y - \gamma | = 6$, then $\alpha + 2 \beta + 3 \gamma$ is equal to $\_\_\_\_$.