Let the population of rabbits surviving at a time $t$ be governed by the differential equation $\frac { d p ( t ) } { d t } = \frac { 1 } { 2 } \{ p ( t ) - 400 \}$. If $p ( 0 ) = 100$, then $p ( t )$ equals
(1) $600 - 500 e ^ { \frac { t } { 2 } }$
(2) $400 - 300 e ^ { \frac { - t } { 2 } }$
(3) $400 - 300 e ^ { t / 2 }$
(4) $300 - 200 e ^ { \frac { - t } { 2 } }$

If the differential equation representing the family of all circles touching $x$-axis at the origin is $\left( x ^ { 2 } - y ^ { 2 } \right) \frac { d y } { d x } = g ( x ) y$, then $g ( x )$ equals
(1) $\frac { 1 } { 2 } x ^ { 2 }$
(2) $2 x$
(3) $\frac { 1 } { 2 } x$
(4) $2 x ^ { 2 }$

For $x \in R , x \neq 0$, if $y ( x )$ is a differentiable function such that $x \int _ { 1 } ^ { x } y ( t ) d t = ( x + 1 ) \int _ { 1 } ^ { x } t y ( t ) d t$, then $y ( x )$ equals (where $C$ is a constant)
(1) $C x ^ { 3 } e ^ { \frac { 1 } { x } }$
(2) $\frac { C } { x ^ { 2 } } e ^ { - \frac { 1 } { x } }$
(3) $\frac { C } { x } e ^ { - \frac { 1 } { x } }$
(4) $\frac { C } { x ^ { 3 } } e ^ { - \frac { 1 } { x } }$

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) $\dfrac{4}{3}$

The curve satisfying the differential equation, $y d x - \left( x + 3 y ^ { 2 } \right) d y = 0$ and passing through the point $( 1,1 )$ also passes through the point
(1) $\left( \frac { 1 } { 4 } , - \frac { 1 } { 2 } \right)$
(2) $\left( - \frac { 1 } { 3 } , \frac { 1 } { 3 } \right)$
(3) $\left( \frac { 1 } { 4 } , \frac { 1 } { 2 } \right)$
(4) $\left( \frac { 1 } { 3 } , - \frac { 1 } { 3 } \right)$

Let $S = \left\{ ( \lambda , \mu ) \in R \times R : f ( t ) = \left( | \lambda | e ^ { | t | } - \mu \right) \sin ( 2 | t | ) , t \in R \right.$ is a differentiable function $\}$. Then, $S$ is a subset of :
(1) $( - \infty , 0 ) \times R$
(2) $R \times [ 0 , \infty )$
(3) $[ 0 , \infty ) \times R$
(4) $R \times ( - \infty , 0 )$

Let $S = \left\{ ( \lambda , \mu ) \in R \times R : f ( t ) = \left( | \lambda | e ^ { t } - \mu \right) \cdot \sin ( 2 | t | ) , t \in R \right.$, is a differentiable function $\}$. Then $S$ is a subest of?
(1) $R \times [ 0 , \infty )$
(2) $( - \infty , 0 ) \times R$
(3) $[ 0 , \infty ) \times R$
(4) $R \times ( - \infty , 0 )$

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

Let $f:[0,1] \rightarrow R$ be such that $f(xy) = f(x) \cdot f(y)$, for all $x,y \in [0,1]$, and $f(0) \neq 0$. If $y = y(x)$ satisfies the differential equation, $\frac{dy}{dx} = f(x)$ with $y(0) = 1$ then $y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right)$ is equal to:
(1) 5
(2) 2
(3) 3
(4) 4

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 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) $( 3,0 )$

The solution of the differential equation $x \frac { d y } { d x } + 2 y = x ^ { 2 } , ( x \neq 0 )$ with $y ( 1 ) = 1$, is
(1) $y = \frac { x ^ { 3 } } { 5 } + \frac { 1 } { 5 x ^ { 2 } }$
(2) $y = \frac { 3 } { 4 } x ^ { 2 } + \frac { 1 } { 4 x ^ { 2 } }$
(3) $y = \frac { x ^ { 2 } } { 4 } + \frac { 3 } { 4 x ^ { 2 } }$
(4) $y = \frac { 4 } { 5 } x ^ { 3 } + \frac { 1 } { 5 x ^ { 2 } }$

If $\cos x \frac { d y } { d x } - y \sin x = 6 x , \left( 0 < x < \frac { \pi } { 2 } \right)$ and $y \left( \frac { \pi } { 3 } \right) = 0$, then $y \left( \frac { \pi } { 6 } \right)$ is equal to
(1) $- \frac { \pi ^ { 2 } } { 4 \sqrt { 3 } }$
(2) $\frac { \pi ^ { 2 } } { 2 \sqrt { 3 } }$
(3) $- \frac { \pi ^ { 2 } } { 2 }$
(4) $- \frac { \pi ^ { 2 } } { 2 \sqrt { 3 } }$

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) $\log _ { e } 2$

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) $y ^ { 2 } = 1 + y \log _ { e } \left( \frac { 1 + e ^ { - x } } { 2 } \right)$

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 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

Let $f ( x ) = \int _ { 0 } ^ { x } e ^ { t } f ( t ) d t + e ^ { x }$ be a differentiable function for all $x \in R$. Then $f ( x )$ equals:
(1) $e ^ { \left( e ^ { x } - 1 \right) }$
(2) $e ^ { e ^ { x } } - 1$
(3) $2 e ^ { e ^ { x } } - 1$

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 solution curve of the differential equation $\left( 2 x - 10 y ^ { 3 } \right) d y + y d x = 0$, passes through the points $( 0,1 )$ and $( 2 , \beta )$, then $\beta$ is a root of the equation? (1) $y ^ { 5 } - 2 y - 2 = 0$ (2) $y ^ { 5 } - y ^ { 2 } - 1 = 0$ (3) $2 y ^ { 5 } - y ^ { 2 } - 2 = 0$ (4) $2 y ^ { 5 } - 2 y - 1 = 0$

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) $y ( 1 ) = e^{\frac{1}{2}} - 1$

A differential equation representing the family of parabolas with axis parallel to $y$-axis and whose length of latus rectum is the distance of the point $( 2 , - 3 )$ from the line $3 x + 4 y = 5$, is given by: (1) $11 \frac { d ^ { 2 } x } { d y ^ { 2 } } = 10$ (2) $11 \frac { d ^ { 2 } y } { d x ^ { 2 } } = 10$ (3) $10 \frac { d ^ { 2 } y } { d x ^ { 2 } } = 11$ (4) $10 \frac { d ^ { 2 } x } { d y ^ { 2 } } = 11$

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) $\log _ { 2 } ( 1 + e )$